Proof. We will prove the if part; the only if part is trivial. Suppose first that \(\require{upgreek} N(\uptheta )= -1\) and that the series (?? ) is not greedy. Then the greedy expansion of \(x\) must differ from (?? ) at some power \(\require{upgreek} \uptheta^{i}\). Without loss of generality, we may assume that this power is \(i=m\): thus the greedy coefficient at \(\require{upgreek} \uptheta^{m}\) is larger than \(b_{m}\). Since the hypothesis on (?? ) is that all the partial sums are greedy, the only way to increase the \(m\)th coefficient from \(b_{m}\) to \(b_{m}+1\) is if \(x\) contains the infinite sum \(\require{upgreek} a(\uptheta^{m-1}+\uptheta^{m-3}+\cdots )\). In fact, it must be equal to this sum, for any additional terms would produce either carries or Fibonacci relations.

Similarly, when \(\require{upgreek} N(\uptheta )= 1\), the only way to increase the \(m\)th coefficient from \(b_{m}\) to \(b_{m}+1\) and conserve the hypothesis that all partial sums are greedy, is the existence of an infinite forbidden block in (?? ): and, again, \(x\) must be equal to such a block. ◻

Proof. For \(i=1\), since \(\require{upgreek} a+1>\uptheta\), (8 ) implies \[\require{upgreek} \uptheta^{3} < (a+1)\uptheta^{2} < 2\uptheta^{3} .\] Thus the greedy expansion of \(\require{upgreek} (a+1)\uptheta^{2}\) begins with \(\require{upgreek} \uptheta^{3}\): \[\require{upgreek} (a+1)\uptheta^{2} =_{\rm gr} \uptheta^{3}+x = a\uptheta^{2}+ \uptheta + x.\] Solving for \(x\), we find \[\require{upgreek} x= \uptheta^{2}-\uptheta =_{\rm gr} (a-1)\uptheta +1,\] where the right-hand side is greedy by Example 1. Thus \[\require{upgreek} (a+1)\uptheta^{2} =_{\rm gr} \uptheta^{3} + (a-1)\uptheta +1,\] from which we deduce \[\require{upgreek} (a+1)\uptheta^{n} =_{\rm gr} \uptheta^{n+1} + (a-1)\uptheta^{n-1} + \uptheta^{n-2}.\] Adding \(\require{upgreek} (i-1) \uptheta^{n}\) to both sides gives the desired formula. ◻

Proof. Adding \(\require{upgreek} (i-1)\uptheta^{n}\) to both sides of (14 ) gives (?? ), greedy by Example 1. ◻

Proof. If \(k=1\), then one sees using the Fibonacci relation \(\require{upgreek} a\uptheta = \uptheta^{2} +1\) that \[\require{upgreek} (a-1) \uptheta^{n} + (a-1) \uptheta^{n+1} = \uptheta^{n-1} -\uptheta^{n} + a\uptheta^{n+1} = \uptheta^{n-1}+ \uptheta^{n+2} .\] The general result follows by induction. ◻

Proof. The carrying formulas show that \(\require{upgreek} \mathbb{Z}_{\uptheta}\) is not closed with respect to the sum. For \(\require{upgreek} N(\uptheta )=-1\) and \(a\geq 2\), \[\require{upgreek} a(2+\uptheta) = (2a-1) + \uptheta^{2} =_{\rm gr} \uptheta^{-2} + (a-1)\uptheta^{-1} + (a-2) +\uptheta +\uptheta^{2} \not\in\mathbb{Z}_{\uptheta}\] and for \(a=1\) and \(\require{upgreek} \uptheta =\upvarphi\), \[\require{upgreek} (1+\upvarphi ^{2})^{2} = 1+ 2\upvarphi ^{2} +\upvarphi^{4} = 2+\upvarphi^{5} =_{\rm gr} \upvarphi^{-2} +\upvarphi +\upvarphi^{5} \not\in\mathbb{Z}_{\uptheta} .\] When \(\require{upgreek} N(\uptheta )=-1\), \[\require{upgreek} (a-1)(1+\uptheta) = \uptheta^{-1} +\uptheta^{2}\not\in\mathbb{Z}_{\uptheta}.\] ◻

Proof. By Theorem 3, the set of \(x>0\) for which \(x\) has a finite greedy \(\require{upgreek} \uptheta\)-expansion (15 ) with \(m\geq 0\) may be identified with the positive part of \({\sf Mod}(\mathcal{O}_{K}; V_{1},V_{2}; {\sf W})\) where \(\require{upgreek} {\sf W}=\left[ \frac{a\uptheta'}{1-\uptheta'^{2}}, \frac{a}{1-\uptheta'^{2}}\right)=[-1,\uptheta)\). Now given \(0<x \in \mathcal{O}_{K}\), there exists a multiple \(\require{upgreek} \uptheta^{M}x>0\) with \(\require{upgreek} \uptheta'^{M}x'\in {\sf W}\), hence \(\require{upgreek} \uptheta^{M}x =_{\rm gr} \sum_{i=m}^{N} b_{i}\uptheta^{i}\) for \(m\geq 0\), and since the greedy property is preserved under multiplication by a power of \(\require{upgreek} \uptheta\), \[\require{upgreek} x=_{\rm gr} \sum_{i=m-M}^{N-M} b_{i+M}\uptheta^{i} .\] Thus every element of \(\mathcal{O}_{K}\) has a finite greedy \(\require{upgreek} \uptheta\)-expansion. Since \(\require{upgreek} \uptheta\in\mathcal{O}^{\times}_{K}\), every greedy Laurent \(\require{upgreek} \uptheta\)-polynomial is in \(\mathcal{O}_{K}\). ◻

Proof. Follows trivially from \[\require{upgreek} x =_{\rm gr} \sum_{i=m}^{M} b_{i}\uptheta^{i} \;\; \Longleftrightarrow \;\;\uptheta^{n} x =_{\rm gr} \sum_{i=m}^{M} b_{i}\uptheta^{i+n}.\] ◻

Proof. Assume first \(f,g>0\). By Proposition 5, we may suppose (after scaling by an appropriate power \(\require{upgreek} \uptheta^{n}\)) that \(\require{upgreek} f,g\in\mathbb{Z}_{\uptheta}\), and that furthermore, \(\require{upgreek} |f|_{\uptheta}=1\). Thus, \(\require{upgreek} f=_{\rm gr}\sum_{i=0}^{M} b_{i}\uptheta^{i}>0\), \(\require{upgreek} g=\sum_{i=n}^{N} c_{i}\uptheta^{i}>0\), so that \(\require{upgreek} \max \{ |f|_{\uptheta}, |g|_{\uptheta} \}=1\) and \[f+g\in {\sf Mod}(\mathcal{O}_{K};V_{1}, V_{2}; 2{\sf W})^{+},\] where \(\require{upgreek} {\sf W}=[-1,\uptheta )\). Since \(\require{upgreek} 0<\uptheta'^{2}=\uptheta^{-2}<1/2\), if we scale further by \(\require{upgreek} \uptheta^{2}\), we obtain \(\require{upgreek} \uptheta^{2} (f+g) \in {\sf Mod}(\mathcal{O}_{K};V_{1}, V_{2}; {\sf W})\). Hence there exist \(d_{i}\in \{ 0,\dots ,a\}\) and \(r,R\in\mathbb{N}\) such that \[\require{upgreek} f+g =_{\rm gr} \uptheta^{-2}\sum^{R}_{i= r} d_{i}\uptheta^{i},\] from which the infratriangle inequality follows for \(f+g\). Suppose now that one of the two is negative, i.e., we consider \(f>0\) and \(-g<0\). We may assume without loss of generality that \(f-g>0\) and that both \(\require{upgreek} f,g\in\mathbb{Z}_{\uptheta}\). Then \[\require{upgreek} (f-g)' \in {\sf W}-{\sf W} = (-1 - \uptheta, 1+\uptheta) .\] In the above, we note that although \({\sf W}\) is half-open, half-closed, \({\sf W}-{\sf W}\) is in fact open. Hence \(\require{upgreek} \uptheta^{2}(f-g)\) has conjugate \[\require{upgreek} \uptheta^{-2}(f'-g')\in \left( -\left(\uptheta^{-2} + \uptheta^{-1}\right), \uptheta^{-2} + \uptheta^{-1} \right)\subset {\sf W},\] since \(\require{upgreek} \uptheta^{2}\geq \uptheta+1\). Again, the infratriangle inequality for \(f-g\) follows. (Note that for \(\require{upgreek} \uptheta\not=\upvarphi\), although \(\require{upgreek} |\uptheta'|<1/2\), there is a negative sign present in \(\require{upgreek} \uptheta'\), which means we cannot simply scale by \(\require{upgreek} \uptheta\) in the above argument.) ◻

Proof. It suffices by Proposition 5 to take \(\require{upgreek} |f_{i}|_{\uptheta}=1\) for \(i=1,\dots ,k\). We have \[\require{upgreek} f_{1}'\cdots f_{k}' \in {\sf W}^{k} = \uptheta^{k-1}(-1, \uptheta),\] and since \(\require{upgreek} \uptheta'^{k}= (-1)^{k}\uptheta^{-k}\), \[\require{upgreek} (\uptheta^{k} f_{1}\cdots f_{k})' \in \left\{ \begin{array}{ll} \uptheta^{-k}{\sf W}^{k} = (-\uptheta^{-1},1)\subset {\sf W} & \text{ if k even} \\ \\ - \uptheta^{-k}{\sf W}^{k} = (-1, \uptheta^{-1})\subset {\sf W} & \text{ if k odd} \end{array} \right.\] Thus \(\require{upgreek} \uptheta^{k} f_{1}\cdots f_{k} =_{\rm gr} \sum_{i=s}^{S} e_{i}\uptheta^{i}\) for \(s,S\in\mathbb{N}\) and appropriate \(e_{i}\in \{ 0,\dots ,a\}\), and the upper bound follows.

Now consider the other inequality. We claim that, for any \(h>0\) with \(h\in\mathcal{O}_{K}\) and \(\require{upgreek} |h|_{\uptheta}=1\), \[\require{upgreek} \begin{align} h' > \uptheta^{-1}. \end{align}\] Indeed, since 1) \(h\) has a nontrivial constant term and 2) only the conjugates of odd powers of \(\require{upgreek} \uptheta\) carry a sign, the greedy form of \(h\) implies the following strict inequality \[\require{upgreek} \begin{align} \label{hyponh39} h' & > 1- (a-1)\uptheta^{-1} -a\uptheta^{-3} - a\uptheta^{-5} - \cdots = \uptheta^{-1}. \end{align}\tag{16}\] In the above, we have made use of

1. the explicit characterization of greedy expansions given in Example 1 to conclude that the coefficient of \(\require{upgreek} \uptheta\) is at worst \(a-1\), and that the remaining coefficients of the odd powers of \(\require{upgreek} \uptheta\) are at worst \(a\). This gives the strict inequality in the first line.

2. the infinite Fibonacci relation (10 ) to conclude the equality in the last line.

Thus for each \(i\), \(\require{upgreek} f_{i}'>\uptheta^{-1}\) and hence \(\require{upgreek} (f_{1}\cdots f_{k})'> \uptheta^{-k}\). On the other hand, if we were to have \(\require{upgreek} |f_{1}\cdots f_{k}|_{\uptheta}= \uptheta^{-r}\) where \(r\geq k+1\), then the greedy expansion of \(f_{1}\cdots f_{k}\) would give \[\require{upgreek} |(f_{1}\cdots f_{k})'| < a\uptheta^{-(k+1)} + a\uptheta^{-(k+3)} + \cdots =\frac{a\uptheta^{-(k+1)}}{1-\uptheta^{-2}} = \uptheta^{-k},\] contradiction. This proves the lower bound. ◻

Proof. The sum is only partially defined: indeed, as observed above, \(\require{upgreek} T=\uptheta-1>0\) does not have a finite greedy expansion since its conjugate is negative. On the other hand, the product is globally defined by the identification (17 ). ◻

Proof. Without loss of generality, assume \(\require{upgreek} \upalpha>0\) so that \(\require{upgreek} \upalpha'<0\). Let \(m\) be the largest integer for which \(\require{upgreek} (\upalpha +\uptheta^{m})'>0\). Then \[\require{upgreek} \upalpha + \uptheta^{m}=_{\rm gr} \sum_{i=n}^{N}b_{i}\uptheta^{i} \quad\text{i.e.}\quad \upalpha= \sum_{i=n}^{N}b_{i}\uptheta^{i} -\uptheta^{m},\] where as usual it is understood that \(b_{n}\not=0\). We claim that \(m=n-1\). Indeed, as we are in the case \(\require{upgreek} N(\uptheta )=1\), the conjugate sum \[\require{upgreek} \sum_{i=n}^{N}b_{i}\uptheta'^{i} =\sum_{i=n}^{N}b_{i}\uptheta^{-i}\] is also greedy, hence is strictly less than \(\require{upgreek} \uptheta^{-(n-1)}\), from which the claim follows. Therefore, \[\require{upgreek} \upalpha=_{\rm gr} \sum_{i=n+1}^{N}b_{i}\uptheta^{i} + (b_{n}-1) \uptheta^{n} +\uptheta^{n-1}T .\] ◻

Proof. Since \[\require{upgreek} \mathcal{O}^{0}_{K} + \mathcal{O}^{0}_{K} \subset \mathcal{O}^{0}_{K} + \bigcup_{N\in\mathbb{Z}} \{ \uptheta^{N}T\},\] it will suffice to show that for all \(N\in\mathbb{Z}\), \(\require{upgreek} \uptheta^{N}T\in \mathcal{O}^{0}_{K} +T\). First assume \(N\geq 0\); then if we write \(\require{upgreek} \upbeta = \uptheta^{N}T-T\), we have \(\require{upgreek} \upbeta >0\) and moreover, \[\require{upgreek} \upbeta' = T' (\uptheta'^{N} -1) = (\uptheta^{-1}-1)( \uptheta^{-N}-1)>0 .\] Thus \(\require{upgreek} \upbeta \in \mathcal{O}^{0}_{K}\) and \(\require{upgreek} \uptheta^{N}T= \upbeta + T\in\mathcal{O}^{0}_{K} +T\). In the case where \(N<0\), the same argument shows \(\require{upgreek} {\rm sgn}(\upbeta) = {\rm sgn}(\upbeta')=-\), which again gives \(\require{upgreek} \upbeta \in \mathcal{O}^{0}_{K}\). ◻

Proof. For \[\require{upgreek} f=_{\rm gr} \sum_{i=m}^{M} b_{i}\uptheta^{i} ,\] \(0<f'\in \mathcal{O}^{0}_{K}\) has greedy expansion \[\require{upgreek} \begin{align} f' =_{\rm gr} \sum_{i=-M}^{-m} b_{-i}\uptheta^{i} \end{align}\] which gives (?? ). ◻

Proof. 1. It will be enough to prove the result in the case \(\require{upgreek} |f|_{\uptheta} =1=|g|_{\uptheta}\) and \(f,g>0\) (thus, in particular, \(\require{upgreek} f,g\in\mathbb{Z}_{\uptheta}^{+}\)). This assumption means that we have \[\require{upgreek} f=\sum_{i=0}^{M} b_{i}\uptheta^{i}, \quad g =\sum_{i=0}^{N} c_{i}\uptheta^{i},\] with \(b_{0}\not = 0 \not= c_{0}\). Since \(\require{upgreek} \uptheta'=\uptheta^{-1}>0\), it follows that for such sums we have \(f', g'\geq 1\). In general, for \(\require{upgreek} h=_{\rm gr}\sum_{i\geq r}^{R} d_{i}\uptheta^{i}\), \(d_{r}\not=0\), we claim that \[\begin{align} \label{f39obversation} h' \geq 1 \Longleftrightarrow r\leq 0. \end{align}\tag{20}\] The direction \(\Longleftarrow\) is immediate. As for \(\Longrightarrow\), if it were the case that \(r\geq 1\), then \(h'\) is dominated by the Rényi expansion of 1 (c.f.equation (11 )): \[\require{upgreek} \begin{align} h' & =_{\rm gr} d_{r}\uptheta^{-r} + \cdots + d_{R}\uptheta^{-R} \\ & =\uptheta^{-r+1} \left( d_{r}\uptheta^{-1} + \cdots + d_{R}\uptheta^{-R+r-1}\right) \\ & \leq d_{r}\uptheta^{-1} + \cdots + d_{R}\uptheta^{-R+r-1} \\ &< (a-1)\uptheta^{-1} + (a-2)(\uptheta^{-2} + \uptheta^{-3}+\cdots ) =1, \end{align}\] contradiction. The inequality \(\require{upgreek} 1\leq |fg|_{\uptheta}\) now follows, since \((fg)'\geq 1\), and the latter implies, by (20 ), that the greedy expansion of \(fg\) is of the form \(\require{upgreek} \sum_{i\geq r}^{R}d_{i}\uptheta^{i}\) with \(r\leq 0\). As for the other inequality, let \(\require{upgreek} f,g\in\mathbb{Z}_{\uptheta}^{+}\). By Theorem 3, item 2., we have \(\require{upgreek} f',g'\in \left[ 0, \uptheta \right)\) implying \(\require{upgreek} (fg)' <\uptheta^{2}\) . Thus, \(\require{upgreek} (\uptheta f g)'<\uptheta\), which implies \(\require{upgreek} \uptheta f g\in \mathbb{Z}_{\uptheta}^{+}\), i.e., \[\require{upgreek} \uptheta^{-1}| f g|_{\uptheta} = |\uptheta f g|_{\uptheta} \leq 1.\]

. Note that \(fg\in\mathcal{O}^{0}_{K}\) and \[\require{upgreek} T^{2} = (\uptheta -1)^{2} = \uptheta^{2}-2\uptheta +1 = (a-2)\uptheta.\] Then, by part 1. above, \[\require{upgreek} \uptheta^{-1} |fg|_{\uptheta} = |(a-2)\uptheta|_{\uptheta} |fg|_{\uptheta} \leq |(a-2)\uptheta fg|_{\uptheta} = |TfTg|_{\uptheta} \leq \uptheta |f|_{\uptheta}|g|_{\uptheta},\] which gives the upper bound for \(\require{upgreek} |fg|_{\uptheta}\). Note that in the last inequality the definition (19 ) is implicit. Similarly \[\require{upgreek} |f|_{\uptheta}|g|_{\uptheta} :=|Tf|_{\uptheta}|Tg|_{\uptheta} \leq |TfTg|_{\uptheta} \leq \uptheta |(a-2)\uptheta|_{\uptheta} |fg|_{\uptheta} = |fg|_{\uptheta}\] giving the lower bound.

. Here we have, again by 1., \[\require{upgreek} |f|_{\uptheta}|g|_{\uptheta} :=|f|_{\uptheta}|Tg|_{\uptheta} \leq |fTg|_{\uptheta} =: |fg|_{\uptheta} \leq \uptheta |f|_{\uptheta} |Tg|_{\uptheta} =: \uptheta |f|_{\uptheta} |g|_{\uptheta}.\] ◻

Proof. 1. First assume that \(f+g\in \mathcal{O}^{0}_{K}\); then the proof is the same as that of Theorem 4, except one works now with the window \(\require{upgreek} W=[0,\uptheta )\) and takes advantage of the fact that \(\require{upgreek} \uptheta' = \uptheta^{-1}<1/2\). In particular, \(\require{upgreek} \uptheta^{-1} 2W\subset W\), which gives the adjustment factor of \(\require{upgreek} \uptheta\) rather than that of Theorem 4, which was \(\require{upgreek} \uptheta^{2}\). Suppose now that \(f,g \in \mathcal{O}^{0}_{K}\) but \(f+g\not\in \mathcal{O}^{0}_{K}\). Then we must have \({\rm sgn}(f)\not={\rm sgn}(g)\) so we are essentially dealing with a difference. Therefore, we will write the norm in question as \(\require{upgreek} |f-g|_{\uptheta}\) where we assume \(f-g>0\) but \(-(f'-g')=g'-f'>0\) and \(f,g>0\) with \(f',g'>0\). Note that the condition \(g'-f'>0\) and the fact that the conjugate of a greedy polynomial is greedy implies \(\require{upgreek} \max \{ |f|_{\uptheta}, |g|_{\uptheta}\} = |g|_{\uptheta}\). With these elements in hand, as well as Lemma 1, we now deduce \[\require{upgreek} \begin{align} |f-g|_{\uptheta} := |T(f-g)|_{\uptheta} & \leq T'(f'-g') = (1-\uptheta^{-1})(g'-f')\\ & < g'-f' < g' \\ &\leq \uptheta |g|_{\uptheta} = \uptheta \max \{ |f|_{\uptheta}, |g|_{\uptheta}\} . \end{align}\]

. Assume first that \(f+g\in \mathcal{O}^{1}_{K}\); then \(\require{upgreek} |f+g|_{\uptheta} := |Tf+ Tg|_{\uptheta}\) with \(Tf, Tg\in \mathcal{O}^{0}_{K}\), so the result in 1. implies \[\require{upgreek} |f+g|_{\uptheta} \leq \uptheta \max\{ |f|_{\uptheta}, |g|_{\uptheta} \} .\] Otherwise, if \(f+g\in \mathcal{O}^{0}_{K}\), by Inframultiplicativity (Theorem 7, Part 3.), \[\require{upgreek} |f+g|_{\uptheta} =\uptheta |T|_{\uptheta} |f+g|_{\uptheta}\leq \uptheta |Tf +Tg|_{\uptheta} \leq \uptheta^{2} \max \{ | f|_{\uptheta}, | g|_{\uptheta} \};\] the last equality follows from 1.of the present Theorem, where we have used that \(Tf, Tg\in\mathcal{O}^{0}_{K}\).

. Assume first that \(f+g\in \mathcal{O}^{0}_{K}\). If \(f,g\) have the same sign, we may assume both are positive, in which case: \[\require{upgreek} \begin{align} |f+g|_{\uptheta} & = |f+ Tg-(\uptheta -2)g|_{\uptheta} . \end{align}\] Since \(\require{upgreek} N(\uptheta )=1\), its minimal polynomial has linear term \(-a\), \(a\geq 3\). In particular, \(\require{upgreek} \uptheta-a+\uptheta^{-1}=0\) implies \(\require{upgreek} \uptheta-2>0\). Thus \(\require{upgreek} \uptheta-2\in \mathcal{O}_{K}^{1}\) and \(\require{upgreek} (\uptheta -2)g\in \mathcal{O}_{K}^{0}\). Since \(f, Tg>0\) and \(Tg\in \mathcal{O}_{K}^{0}\), we have likewise, by Note 1, that \(f+ Tg\in \mathcal{O}_{K}^{0}\). Thus we may apply the result of part 1.of this Theorem, as well as Inframultiplicativity, to get \[\require{upgreek} \begin{align} |f+g|_{\uptheta} & \leq \uptheta\max \left\{ |f+ Tg|_{\uptheta} , |(\uptheta -2)g|_{\uptheta} \right\} \\ &\leq \max \left\{ \uptheta^{2} \max\left\{ |f|_{\uptheta} , |g|_{\uptheta}\right\} , \uptheta^{3} |(\uptheta -2)|_{\uptheta}|g|_{\uptheta} \right\} . \end{align}\] Since \(\require{upgreek} |\uptheta-2|_{\uptheta} := | (\uptheta-2)T|_{\uptheta} = |(a-3)\uptheta +1|_{\uptheta}=1\) (where we recall that \(a\geq 3\) when \(\require{upgreek} N(\uptheta )=1\)), we obtain finally \[\require{upgreek} |f+g|_{\uptheta} \leq \uptheta^{3} \max\left\{ |f|_{\uptheta} , |g|_{\uptheta}\right\}.\]

If \(f,g\) have opposite signs, we are essentially dealing with a difference, so we may assume that \(f,g>0\) and the norm in question which we wish to calculate is \(\require{upgreek} |f-g|_{\uptheta}\). Then we write \[\require{upgreek} \begin{align} |f-g|_{\uptheta} & = |f+(\uptheta -2)g- Tg|_{\uptheta} \end{align}\] and note that \(\require{upgreek} (\uptheta -2)g, f+(\uptheta -2)g, Tg\in \mathcal{O}_{K}^{0}\), wherein Part 1.of the present Theorem is available: with Inframultiplicativity, we deduce \[\require{upgreek} \begin{align} \label{DiffSignPart3TriIneq} |f-g|_{\uptheta} & \leq \uptheta\max \left\{ |f+ (\uptheta -2)g|_{\uptheta} , |g|_{\uptheta} \right\} \nonumber \\ & \leq \uptheta\max \left\{ \uptheta \max\left\{ |f|_{\uptheta}, |(\uptheta -2)g|_{\uptheta}\right\}, |g|_{\uptheta} \right\} \nonumber \\ & \leq \uptheta\max \left\{ \uptheta \max\left\{ |f|_{\uptheta}, \uptheta^{2}|g|_{\uptheta}\right\}, |g|_{\uptheta} \right\}\nonumber \\ & \leq \uptheta^{4}\max \left\{ |f|_{\uptheta}, |g|_{\uptheta}\right\} . \end{align}\tag{21}\] Finally, we consider the case where \(f+g\in \mathcal{O}^{1}_{K}\): then \(\require{upgreek} |f+g|_{\uptheta} := |Tf +Tg|_{\uptheta}\), where now \(Tf\in \mathcal{O}^{1}_{K}\) and \(Tg\in \mathcal{O}^{0}_{K}\). If \(Tf, Tg\) have the same sign, we may utilize what we have just proved to obtain \[\require{upgreek} \begin{align} |f+g|_{\uptheta} & \leq \uptheta^{3} \max\left\{ |Tf|_{\uptheta} , |Tg|_{\uptheta}\right\} \\ & = \uptheta^{3} \max\left\{ |Tf|_{\uptheta} , |g|_{\uptheta}\right\} \\ & \leq \uptheta^{3} \max\left\{ |f|_{\uptheta} , |g|_{\uptheta}\right\} \end{align}\] where we have used the inequality \(\require{upgreek} |Tf|_{\uptheta}\leq |f|_{\uptheta}\), in its turn a consequence of Inframultiplicativity. Similarly, if \(Tf, Tg\) have the opposite sign, we have, using (21 ), \[\require{upgreek} |f+g|_{\uptheta} := |Tf+Tg|_{\uptheta} \leq \uptheta^{4} \max\left\{ |Tf|_{\uptheta} , |Tg|_{\uptheta}\right\} \leq \uptheta^{4} \max\left\{ |f|_{\uptheta} , |g|_{\uptheta}\right\} .\] ◻

Proof. Applying the infratriangle inequality to \(\require{upgreek} (\upalpha - \upbeta)+\upbeta\) gives \[\require{upgreek} |\upalpha|_{\uptheta} \leq \uptheta^{2}\max\{ |\upalpha-\upbeta|_{\uptheta}, |\upbeta|_{\uptheta} \} =\uptheta^{2} |\upalpha-\upbeta|_{\uptheta},\] the last equality being a consequence of the hypothesis \(\require{upgreek} \uptheta^{2} |\upbeta|_{\uptheta}<|\upalpha|_{\uptheta}\) (for if the maximum were \(\require{upgreek} \uptheta^{2}|\upbeta|_{\uptheta}\), we would obtain \(\require{upgreek} |\upalpha|_{\uptheta} <|\upalpha|_{\uptheta}\)). ◻

Proof. Suppose not: we may assume, after passing to a subsequence, that the sequence of infranorms is strictly monotonically increasing: \[\require{upgreek} |\upalpha_{1}|_{\uptheta} < |\upalpha_{2}|_{\uptheta} < \cdots .\] Then for all \(i\) and \(k\geq 3\), \(\require{upgreek} \uptheta^{2} |\upalpha_{i}|_{\uptheta} < |\upalpha_{i+k}|_{\uptheta}\). By the Cauchy condition, for fixed \(\require{upgreek} \upvarepsilon>0\), there is \(N\) such that \(\require{upgreek} |\upalpha_{j}-\upalpha_{i}|_{\uptheta}<\upvarepsilon\) for all \(j>i\geq N\). By Proposition 8, for fixed \(i\geq N\) and all \(k\geq 3\), \[\require{upgreek} |\upalpha_{i+k}|_{\uptheta} \leq \uptheta^{2} |\upalpha_{i+k}-\upalpha_{i}|_{\uptheta} \leq \uptheta^{2}\upvarepsilon,\] contradiction. ◻

Proof. Suppose that \(\{ f_{n}\}\) is a Cauchy sequence for which \(\lim | f_{n}|\) is not well-defined. By Lemma 2, the sequence \(\require{upgreek} \{ |f_{n}|_{\uptheta}\}\) is uniformly bounded from above, so we must have that \(\lim | f_{n}|\) produces multivalues that are finite. We claim that, by the Reverse Infratriangle Inequality (Proposition 8), these multivalues must lie in a set of three consecutive powers: \(\require{upgreek} \uptheta^{m}, \uptheta^{m+1}\) and \(\require{upgreek} \uptheta^{m+2}\). For suppose not i.e.there exist subsequences \(\{ f_{i_{l}}\}\) and \(\{ f_{j_{l}}\}\) with constant infranorms which are distinct, such that for all \(l\), \[\require{upgreek} c = |f_{i_{l}}|_{\uptheta} >\uptheta^{2} d,\quad d= |f_{j_{l}}|_{\uptheta} .\] Thus, for each \(l\), the hypothesis of the Reverse Infratriangle Inequality is satisfied. But then, since \(\{ f_{n}\}\) is Cauchy, \[\require{upgreek} 0\not=c-d\leq \uptheta^{2} |f_{i_{l}}- f_{j_{l}}|_{\uptheta} \longrightarrow 0,\] contradiction. ◻

Proof. Let \(\require{upgreek} \{ \upalpha_{i}\}, \{ \upbeta_{i}\} \in \mathcal{R}_{\uptheta}\). Then by the infratriangle inequality, inframultiplicativity and Lemma 2, \[\require{upgreek} \begin{align} |\upalpha_{i}\upbeta_{i}-\upalpha_{j}\upbeta_{j}|_{\uptheta} & = |\upalpha_{i}\upbeta_{i}-\upalpha_{i}\upbeta_{j} +\upalpha_{i}\upbeta_{j}- \upalpha_{j}\upbeta_{j}|_{\uptheta} \\ & \leq \uptheta^{4} \max\big\{ |\upalpha_{i}|_{\uptheta} |\upbeta_{i}-\upbeta_{j}|_{\uptheta} , |\upbeta_{i}|_{\uptheta} |\upalpha_{i}-\upalpha_{j}|_{\uptheta} \big\} \\ & \longrightarrow 0, \end{align}\] hence \(\require{upgreek} \mathcal{R}_{\uptheta}\) is a ring. Likewise, \(\require{upgreek} \mathcal{M}_{\uptheta}\) is seen to be an ideal (since, by definition, \(\require{upgreek} \{ \upalpha_{i}\} \in \mathcal{M}_{\uptheta}\) implies \(\require{upgreek} |\upalpha_{i}|_{\uptheta}\rightarrow 0\)). Suppose that \(\require{upgreek} \{ f_{i}\}, \{ g_{i}\}\not\in \mathcal{M}_{\uptheta}\), but \(\require{upgreek} \{ f_{i} g_{i}\}\in\mathcal{M}_{\uptheta}\). Then by inframultiplicativity, \[\require{upgreek} \uptheta^{-2} |f_{i} |_{\uptheta} |g_{i} |_{\uptheta} \leq |f_{i}g_{i} |_{\uptheta}\longrightarrow 0.\] By Theorem 9, it follows that either \(\require{upgreek} |f_{i} |_{\uptheta}\) or \(\require{upgreek} |g_{i} |_{\uptheta}\) \(\rightarrow 0\), contradiction. Thus \(\require{upgreek} \mathcal{M}_{\uptheta}\) is prime and \(\require{upgreek} \widehat{\mathcal{O}}_{\uptheta}\) is an integral domain. ◻

Proof. Let \(\{ f_{n}\}\) be a non null Cauchy sequence. After passing to a subsequence, we may assume that the \(f_{n}\) have constant sign, say positive, and by Infraoscillation (Theorem 9), we may also assume they have constant infranorm, say \(\require{upgreek} \uptheta^{m}\). We may then further assume that for all \(n\) and all \(k>0\), \(f_{n+k}\) and \(f_{n}\) share the same initial sum of \(n\) terms: \[\require{upgreek} g_{n}: = c_{m}\uptheta^{m} + \cdots + c_{m+n}\uptheta^{m+n} .\] So, for example, the first term of all the \(f_{n}\) is \(\require{upgreek} g_{1}=c_{m}\uptheta^{m}\), the sum of the first two terms of \(f_{2},f_{3},\dots\) is \(\require{upgreek} g_{2}=c_{m}\uptheta^{m} + c_{m+1}\uptheta^{m+1}\), and so on. The sequence \(\{ g_{n}\}\) evidently defines the sequence of partial sums of a greedy Laurent series which defines the same Cauchy class as \(\{ f_{n}\}\) (since \(f_{n}-g_{n}\) is already in greedy form and has infranorm \(\rightarrow 0\)). ◻

Proof. Suppose that \(k>2\). Then for \(i\) large, the hypothesis of the Reverse Infratriangle Inequality (Proposition 8) is satisfied for \(\require{upgreek} \upalpha= f'_{n'_{i}}\) and \(\require{upgreek} \upbeta = f_{n_{i}}\). But then \(\require{upgreek} |f'_{n'_{i}}|_{\uptheta}=\uptheta^{m+k}\leq \uptheta^{2} |f'_{n'_{i}}- f_{n_{i}}|_{\uptheta}\rightarrow 0\), contradiction. ◻

Proof. We assume that \(\hat{f}-\hat{g}\not=0\), since otherwise the statement is trivial. Let \(\{ f_{i}-g_{i}\}\) be a Cauchy sequence in the class \(\hat{f} -\hat{g}\), where \(\{ f_{i}\}\), \(\{ g_{i}\}\) represent \(\hat{f}\), \(\hat{g}\), and which realizes \(\require{upgreek} \max |\hat{f} -\hat{g}|_{\uptheta}\): that is, \(\require{upgreek} |f_{i}-g_{i}|_{\uptheta}\) is constant and equal to \(\require{upgreek} \max |\hat{f} -\hat{g}|_{\uptheta}\) for all \(i\) . Then, \[\require{upgreek} \max |\hat{f} -\hat{g}|_{\uptheta} =|f_{i}-g_{i}|_{\uptheta} \leq \uptheta^{2}\max\left\{ | f_{i}|_{\uptheta}, | g_{i}|_{\uptheta} \right\}\leq \uptheta^{2}\max \left\{ \max | \hat{f}|_{\uptheta}, \max | \hat{g}|_{\uptheta} \right\} .\] ◻

Proof. Same proof as Proposition 9, except that we must choose the subsequence of the form \(\require{upgreek} T^{\upepsilon}g^{0}_{n}\), \(g^{0}_{n}\in \mathcal{O}_{K}^{0}\) and \(\require{upgreek} {\upepsilon}\in \{ -1,0\}\) constant, where we are using the representations of elements of \(\mathcal{O}^{1}_{K}\) in (18 ). ◻

Proof. First, suppose \(\require{upgreek} N(\uptheta )=-1\). For \(\require{upgreek} \hat{x}\in\widehat{\mathcal{O}}_{\uptheta}\), choose a representative Cauchy sequence \(\{ x_{i}\}\). We claim that \(\{ x_{i}'\} \subset\mathbb{R}\) defines a real Cauchy sequence independent of the choice of \(\{ x_{i}\}\) in the class of \(\hat{x}\). Indeed, given \(\require{upgreek} \upvarepsilon = \uptheta^{-r} >0\) choose \(N\) such that \(\require{upgreek} |x_{i} -x_{j}|_{\uptheta} \leq \upvarepsilon\) for all \(i,j\geq N\). Then \[\require{upgreek} |x_{i}' - x_{j}' | \leq \uptheta^{-r}a (1+ \uptheta^{-1} + \uptheta^{-2} + \cdots ) = a\uptheta^{-r}\frac{\uptheta}{\uptheta -1} < a\uptheta^{-r}\frac{\uptheta}{\uptheta -a} =a \uptheta^{-r+2}\] giving the Cauchy condition for the conjugates. A similar argument shows that if \(\{ y_{i}\}\in \hat{x}\), then \(\{ y_{i}'\}\) defines the same Cauchy class as \(\{ x'_{i}\}\). We denote the limit of \(\{ x_{i}'\}\) by \(\hat{x}'\).

Now consider the case \(\require{upgreek} N(\uptheta )=1\) and take \(\{ x_{i}\}\in \hat{x}\) as before a representative Cauchy sequence. Choose \(\require{upgreek} \upvarepsilon = \uptheta^{-r} >0\) and \(N\) such that \(\require{upgreek} |x_{i} -x_{j}|_{\uptheta} \leq \upvarepsilon\) for all \(i,j\geq N\). If \(x_{i}-x_{j}\in \mathcal{O}^{0}_{K}\) then we have \[\require{upgreek} |x_{i}'-x_{j}'| \leq \uptheta^{-r}| (a-1) + (a-2)\uptheta^{-1} +\cdots | = \uptheta^{-r+1} ,\] where the final equality above follows from (11 ). Otherwise, if \(x_{i}-x_{j}\in \mathcal{O}^{1}_{K}\) then, using (18 ), we have \[\require{upgreek} |x_{i}'-x_{j}'| \leq |T'|^{-1} \uptheta^{-r}| (a-1) + (a-2)\uptheta^{-1} +\cdots | = \frac{\uptheta^{-r+2}}{\uptheta-1}.\] This shows that the corresponding sequence \(\{ x'_{i}\}\) is Cauchy in \(\mathbb{R}\) and again we can write \(\hat{x}'\) unambiguously. Thus Galois conjugation defines a well-defined function \(\require{upgreek} \widehat{\mathcal{O}}_{\uptheta}\rightarrow \mathbb{R}\). Since the ring structure of \(\require{upgreek} \widehat{\mathcal{O}}_{\uptheta}\) is defined, at the level of Cauchy sequences, component-by-component on entries in \(\mathcal{O}_{K}\), it follows that Galois conjugation extends to a ring homomorphism of \(\require{upgreek} \widehat{\mathcal{O}}_{\uptheta}\). ◻

Proof. Assume first that \(\require{upgreek} N(\uptheta )=-1\). Let \(\hat{x}'\) be defined as in the previous paragraph. Choose a representative sequence \(\{ x_{i}\}\) of \(\hat{x}\) such that \(\require{upgreek} |x_{i}|_{\uptheta} = |\hat{x}|_{\uptheta}\) for all \(i\). In fact, we may assume that the \(x_{i}\) are the partial sums of a Laurent series \[\require{upgreek} \begin{align} \label{defnconjofx} \sum_{i=M}^{\infty} c_{i}\uptheta^{i} \end{align}\tag{26}\] in the class of \(\hat{x}\) (this follows from the proof of Proposition 9). Then, \[\require{upgreek} |\hat{x}'| = \uptheta^{-M} |c_{M}\pm c_{M+1}\uptheta^{-1} \pm \cdots |= |\hat{x}|_{\uptheta} |c_{M}\pm c_{M+1}\uptheta^{-1} \pm \cdots | .\] Since the \(c_{i}\in \big\{ 0,\dots ,a\}\) and satisfy the greedy condition, and the signs alternate according to parity of the index, the expression \(\require{upgreek} |c_{M}\pm c_{M+1}\uptheta^{-1} \pm \cdots |\) is unformly bounded away from 0 and infinity. Indeed, by (11 ), \[\require{upgreek} \uptheta^{-1}= 1-(a-1)\uptheta^{-1} - a\uptheta^{-3}-a\uptheta^{-5}-\cdots \leq |c_{M}\pm c_{M+1}\uptheta^{-1} \pm \cdots | \leq a(1+\uptheta^{-2} +\uptheta^{-4}\cdots ) < \infty .\] This proves (?? ) for \(\require{upgreek} N(\uptheta )=-1\). Now suppose \(\require{upgreek} N(\uptheta )=1\). Here \(\hat{x}\) may be represented by \(\require{upgreek} T^{\upepsilon}\times\) a Laurent series of the shape (26 ), where \(\require{upgreek} \upepsilon \in \{0,-1\}\). Without loss of generality we may assume \(\require{upgreek} \upepsilon =0\). Then we have the bounds \[\require{upgreek} |\hat{x}|_{\uptheta} \leq |\hat{x}'| \leq \uptheta^{-M}\left( (a-1) + (a-2)\uptheta^{-1} + \cdots \right) = |\hat{x}|_{\uptheta} \uptheta ,\] which give again (?? ). ◻

Proof. . Suppose first that \(\require{upgreek} N(\uptheta )=-1\). We recall that uniform continuity (c.f.[19], §II.2) means in our setting that for every \(\require{upgreek} \upvarepsilon >0\), there exists \(\require{upgreek} \updelta >0\) such that for all \(\require{upgreek} (\hat{x},\hat{y})\in U_{\updelta}\), \(\require{upgreek} |\hat{x}'-\hat{y}'|<\upvarepsilon\). Choose \(\require{upgreek} \updelta =D^{-1}\upvarepsilon\) and let \(\{ x_{k}\}\in \hat{x}\), \(\{ y_{k}\}\in \hat{y}\) be representative sequences of elements of \(\mathcal{O}_{K}\), which, since \(\require{upgreek} N(\uptheta )=-1\), may be taken to be \(\pm\) greedy polynomials in \(\require{upgreek} \uptheta\). Then \(z_{k}=x_{k}-y_{k}\) satisfies that eventually \[\require{upgreek} |z_{k}|_{\uptheta}\in \bigg\{ |z|_{\uptheta}, \uptheta^{-1}|z|_{\uptheta}, \uptheta^{-2}|z|_{\uptheta}\bigg\}\] by Infraoscillation (Theorem 9). Therefore, by (?? ), \(\require{upgreek} |z'_{k}| <\upvarepsilon\) and we are done. The proof for \(\require{upgreek} N(\uptheta )=1\) is essentially the same, where we use Infraoscillation appropriate to this case (Theorem 13). ◻

Proof. The product uniform structure on \(\require{upgreek} \widehat{\mathcal{O}}_{\uptheta}\times \widehat{\mathcal{O}}_{\uptheta}\) is generated by sets of the form \(\require{upgreek} U_{\upvarepsilon_{1}}\times U_{\upvarepsilon_{2}}\). When \(\require{upgreek} N(\uptheta )=-1\), the infratriangle inequality shows that the image of such a set by the sum is in \(\require{upgreek} U_{\updelta}\) for \(\require{upgreek} \updelta = \uptheta^{2}\max (\upvarepsilon_{1}, \upvarepsilon_{2})\), hence the sum is a uniformly continuous map and so is continuous. The statement for the product is proved similarly using inframultiplicativity. The case \(\require{upgreek} N(\uptheta )=1\) is proved analogously. ◻

Proof. The argument is essentially the same as that of Proposition 9. Assume first that \(\require{upgreek} x,y\in\mathcal{O}^{+}_{\uptheta}\). The sequence \(\{ f_{n} \} =\{ x_{n}+y_{n}\}\) is Cauchy by Theorem 10. If \(f_{n}\rightarrow 0\) then we may take \(w=0\). So assume now that \(f_{n}\not\rightarrow 0\). After passing to a subsequence, we may assume that the \(\require{upgreek} \uptheta\)-norms are constant, equal say to \(\require{upgreek} \uptheta^{-m}\), and moreover, the coefficients of \(\require{upgreek} \uptheta^{m}\) are all equal. Thus \[\require{upgreek} f_{n} =_{\rm gr} c_{m} \uptheta^{m} + c_{m+1}(n) \uptheta^{m+1}+\cdots ,\] where the notation for the coefficients indicates that \(c_{m}\) is independent of the index \(n\), whereas the remaining coefficients will in general depend on \(n\). Inductively, for all \(k\) there exists an infinite set of indices \(I^{(k)}\subset \mathbb{N}\) so that for all \(n\in I^{(k)}\), \[\require{upgreek} f_{n}=_{\rm gr} c_{m} \uptheta^{m} +\cdots + c_{m+k} \uptheta^{m+k} +c_{m+k+1}(n) \uptheta^{m+k+1} +\cdots,\] i.e., the first \(k\) terms are the same for all \(f_{n}\) with \(n\in I^{(k)}\). Then if we take \[I=\{ n_{1}< n_{2}<\cdots \}, \quad n_{k}\in I^{(k)},\] we obtain a sequence of partial sums of a Laurent series \(\require{upgreek} \sum_{i=m}^{\infty}c_{i}\uptheta^{i}\) defining a class \(w\) with \(w\in x+y\) and \(\{ w_{n} \} =\{ x_{n} + y_{n}\} \in w\). If \(\require{upgreek} -x,-y\in -\mathcal{O}_{\uptheta}^{+}\) then \(-w\in (-x)+(-y)\) for \(w\in x+y\) as above. If \(\require{upgreek} x\in \mathcal{O}_{\uptheta}^{+}\) and \(\require{upgreek} -y\in -\mathcal{O}_{\uptheta}^{+}\), then after passing to subsequences if necessary, we may find \(\{ x_{n}\}\in x\) and \(\{ -y_{n}\}\in -y\) so that \(\{ x_{n}-y_{n}\}\) is either always non-negative or always non-positive. Then once again, an argument in the style of Proposition 9 gives \(w\in x+(-y)\). The analogous statement for the product is proved in exactly the same way. ◻

Proof. Let \(t\in (x+y)+z\). Then, there exists \(\require{upgreek} w\in \mathcal{O}_{\uptheta}\) with \(w\in x+y\) along with representative sequences of partial sums \(\{ x_{n}\}\in x\), \(\{ y_{n}\}\in y\) and \(\{ z_{n}\}\in z\) such that \[\{ w_{n}\} = \{ x_{n}+y_{n}\} \in w,\quad \{ w_{n} + z_{n}\} =\{ x_{n}+y_{n}+z_{n}\} \in t.\] Now, it may not be the case that there exists \(r\in y+z\) with \(\{ r_{n}\} =\{ y_{n} +z_{n}\} \in r\), however, by Lemma 4, we may pass to a subsequence for which such an \(r\) exists with \(\{ y_{n} +z_{n}\}_{n\in I} \in r\). But then, by associativity in \(\mathbb{R}\), we have \[\{ t_{n}\}_{n\in I} = \{x_{n}+ y_{n} +z_{n}\}_{n\in I} = \{x_{n}+ r_{n}\}_{n\in I} \in t ,\] and since \(\{ r_{n}\} \in r\), it follows that \(t\in x+(y+z)\). Since the argument is symmetric, this proves associativity. The associativity of the product is shown in the same way. The argument for distributivity is similar: choose \(t\in x\cdot (y+z)\) represented by the sequence of partial sums \(\{ x_{n}(y_{n}+z_{n})\}\), where \(\{ x_{n}\}\in x\), \(\{ y_{n}\}\in y\), \(\{ z_{n}\}\in z\) and the sum \(\{y_{n}+z_{n}\}\) defines a sequence of partial sums in \(y+z\). By ordinary distributivity, \[x_{n}(y_{n}+z_{n} ) = x_{n}y_{n} + x_{n}z_{n} ,\] and by Lemma 4, after passing to a subsequence if necessary, we may assume \(x_{n}y_{n} + x_{n}z_{n}\) defines a sequence of partial sums in \(xy + xz\). Hence \(t\in xy + xz\). ◻

Proof. By Theorem 15, both multioperations \(+\) and \(\cdot\) are associative and \(\cdot\) distributes weakly over \(+\). Thus, the only thing left to verify is the existence of two-sided additive inverses, which is guaranteed by the inclusion of \(\require{upgreek} -\mathcal{O}_{\uptheta}\). ◻

Proof. Let \(x,y\in \mathcal{O}_{K}\). Then any sequence of partial sums of \(x\) resp.\(y\) have the form \(\require{upgreek} \{ x+ \upvarepsilon_{n}\}\) resp. \(\require{upgreek} \{ y+ \updelta_{n}\}\) where \(\require{upgreek} \upvarepsilon_{n}, \updelta_{n}\rightarrow 0\) with respect to \(\require{upgreek} |\cdot |_{\uptheta}\). By the infratriangle inequality, \(\require{upgreek} \upvarepsilon_{n}+ \updelta_{n}\rightarrow 0\). Thus, if \(\require{upgreek} ( x+ \upvarepsilon_{n}) + ( y+ \updelta_{n})\) defines a sequence of partial sums of a Laurent series, this series must be in the class of the element \(x+y\in\mathcal{O}_{K}\). In particular, the restriction of the sum to \(\mathcal{O}_{K}\) is single valued and coincides with the usual sum in \(\mathcal{O}_{K}\). The product \[\require{upgreek} ( x+ \upvarepsilon_{n}) \cdot ( y+ \updelta_{n}) = xy + x\updelta_{n}+y\upvarepsilon_{n} + \upvarepsilon_{n}\updelta_{n}\] is treated similarly, where one uses inframultiplicativity and the infratriangle inequality to assert that the final three terms \(\rightarrow 0\). ◻

Proof. Note that since the range of either map is a ring, the defining condition (30 ) of a multihomomorphism is replaced by \(f(x\cdot y)=f(x)\cdot f(y)\). The statement of the Theorem now follows immediately from the definition of the multioperations. Indeed, for example, in the case of the sum, the definition starts with the sum of partial sums \(\{ x_{n} + y_{n}\}\) and then extracts subsequences which form coherent Laurent series. But all of these series belong to the same Cauchy class. Thus \(\require{upgreek} \uppi (x+y) = \uppi (x) + \uppi (y)\). The conjugation map on \(\require{upgreek} \mathcal{O}_{\uptheta}\) is simply the composition of \(\require{upgreek} \uppi\) with conjugation on \(\require{upgreek} \widehat{\mathcal{O}}_{\uptheta}\), hence it is also an epimorphism by Proposition 11. ◻

Proof. In \(\require{upgreek} \widehat{\mathcal{O}}_{\uptheta}\) the sum \(\hat{x}+ \hat{y}\) is well-defined and by Proposition 10, it contains a sequence of partial sums of a Laurent series. ◻

Proof. Same as the proof of Theorem 15, using Lemma 5 above. ◻

Proof. The proof follows the same structure as its counterpart Theorem 16. ◻

Proof. Clearly \(\iota (f+g) \in \iota (f) + \iota (g)\) and \(\iota (f\cdot g) \in \iota (f) \cdot \iota (g)\), so \(\iota\) is an homomorphism of Marty multirings. What remains to show is injectivity. Suppose on the contrary that there exist \(f,g\in \mathcal{O}_{K}\) with \(\iota (f) = \iota (g)\). Then there exist partial sums of each which are equal: \[\require{upgreek} f+ \updelta_{n} =g + \upvarepsilon_{n}.\] Taking conjugates we have \[\require{upgreek} f'- g'=\upvarepsilon_{n}' -\updelta_{n} '.\] The right-hand side tends to 0 in the euclidean norm but the left hand side is constant, thus the right-hand side is \(0\) and \(f'= g'\) which implies \(f= g\). ◻

Proof. The proof is identical to that of Theorem 18. ◻

Proof. Multiplication by a power \(\require{upgreek} \uptheta^{k}\) takes Cauchy classes to Cauchy classes and Laurent series to (shifted) Laurent series, hence it will be enough to consider \(\require{upgreek} \upalpha\) with \(\require{upgreek} |\upalpha|_{\uptheta}=1\) i.e. \(\require{upgreek} \upalpha = c_{n}\uptheta^{n} + \cdots +c_{0}\). We may also assume that \(n\) is even: if not, we add the term \(\require{upgreek} 0\cdot \uptheta^{n+1}\). Suppose there exists a series \(\require{upgreek} \upgamma=\sum_{i=0}^{\infty} b_{i}\uptheta^{i}\) for which \(\require{upgreek} \upgamma - \upalpha\) defines the \(0\) Cauchy class. Then by Proposition 11, we must have \[\require{upgreek} \upgamma'-\upalpha' =0.\] Notice that \(\require{upgreek} \upgamma\) is necessarily an infinite series, since Galois conjugation is injective on \(\mathcal{O}_{K}\). Then this implies \[\require{upgreek} \sum_{i=0}^{n/2} c_{2i} \uptheta^{-2i} + \sum_{i=0}^{\infty} b_{2i-1}\uptheta^{-2i+1} = \sum_{i=0}^{n/2} c_{2i-1}\uptheta^{-2i+1} + \sum_{i=0}^{\infty} b_{2i} \uptheta^{-2i} .\] After amalgamating terms of the same power – grouping on either the left or right-hand side in such a way that the common coefficient is positive – we obtain \[\require{upgreek} \sum_{i=0}^{n/2} \tilde{c}_{2i} \uptheta^{-2i} + \sum_{i=0}^{\infty} \tilde{b}_{2i-1}\uptheta^{-2i+1} = \sum_{i=0}^{n/2} \tilde{c}_{2i-1}\uptheta^{-2i+1} + \sum_{i=0}^{\infty} \tilde{b}_{2i} \uptheta^{-2i} ,\] in which all coefficients appearing are in the range \(\{ 0,\dots ,a\}\) and satisfy \[\tilde{b}_{i} = 0 \text{ or } \leq b_{i} \text{ and is eventually }=b_{i} ,\] as well as \[\tilde{c}_{i} = 0 \text{ or } \leq c_{i} .\] Each of the series \[\require{upgreek} \sum_{i=n/2+2}^{\infty} \tilde{b}_{2i-1}\uptheta^{-2i+1}, \quad \sum_{i=n/2+1}^{\infty} \tilde{b}_{2i} \uptheta^{-2i}\] is clearly greedy: since their coefficients \(\leq a\) and since they skip powers, there can be no Fibonacci relations. Let us analyze the remaining terms. On the left hand side we have \[\require{upgreek} \begin{align} \label{LHS32Poly} \tilde{c}_{0} + \tilde{b}_{1}\uptheta^{-1} + \tilde{c}_{2}\uptheta^{-2} + \tilde{b}_{3}\uptheta^{-3} +\cdots + \tilde{c}_{n} +\tilde{b}_{n+1} \uptheta^{-n} =: d_{0} + \cdots + d_{n+1}\uptheta^{-n-1} \end{align}\tag{34}\] In what follows, we analyze an algorithm which puts (34 ) in greedy form. Start with the first pair sum \(\require{upgreek} d_{0} + d_{1}\uptheta^{-1}\). Then either

1. There is no Fibonacci relation. Then either

   1. \(d_{0}<a\). Here, we proceed to the next pair sum \(\require{upgreek} d_{1}\uptheta^{-1} + d_{2}\uptheta^{-2}\) and decide whether it gives or does not give a Fibonacci relation.

   2. \(d_{0}=a\) but \(d_{1}=0\). Then, we pass to pair sum \(\require{upgreek} d_{2}\uptheta^{-2} + d_{3}\uptheta^{-3}\) and again determine whether there is a Fibonacci relation.

2. There is a Fibonacci relation. Thus, we suppose \(d_{0}=a\) and \(d_{1}\not=0\). This gives \[\require{upgreek} \uptheta + (d_{1}-1)\uptheta^{-1} + d_{2}\uptheta^{-2} + \cdots .\] Note then that the following pair \(\require{upgreek} (d_{1}-1)\uptheta^{-1} + d_{2}\uptheta^{-2}\) cannot provide a Fibonacci relation since \(d_{1}-1<a\). Thus for this pair, one proceeds to item Ai.above, which instructs us to move on to \(\require{upgreek} d_{2}\uptheta^{-2} + d_{3}\uptheta^{-3}\).

The intuitive idea is that Fibonacci relations imply changes upwards to higher powers, never lower powers. If A.always holds at each step, then the expression (34 ) is already in greedy form. Now suppose there are Fibonacci relations, the first one occuring at the index \(i\), i.e., we have \[\require{upgreek} \begin{align} \label{1stFibRes} \cdots + d_{i-1}\uptheta^{-i+1} +a\uptheta^{-i} +d_{i+1}\uptheta^{-i-1} \cdots =\cdots + (d_{i-1}+1)\uptheta^{-i+1} +(d_{i+1}-1)\uptheta^{-i-1} \cdots \end{align}\tag{35}\] where we assume \(d_{i+1}\not=0\). Since this was the first such Fibonacci relation, \(d_{i-1}<a\), hence the new coefficient at \(i-1\) satisfies \((d_{i-1}+1)\leq a\). After this Fibonacci resolution, the only way that the sum of elements with indices \(\geq i-1\) is not greedy would be if \(d_{i-1}=0\) and \(d_{i-2}=a\): producing a new Fibonacci relation at the index \(i-2\). Resolving the latter would either produce a sum for indices \(\geq i-1\) that is greedy, or else, \(d_{i-3}=0\) and \(d_{i-4}=a\): producing another Fibonacci relation at the index \(i-4\). Thus, inductively, while the indices \(\geq i-1\) may change due to the occurrence of new Fibonacci relations provoked by (35 ), the remaining sum in (35 ) for indices \(>i-1\), \[\require{upgreek} (d_{i+1}-1)\uptheta^{-i-1} \cdots,\] is unchanged. Inductively, this stability persists to the end, and the lowest order exponent satisfies \(\geq -n\). \(\quad\diamond\)

By the Claim, (35 ) has greedy form \[\require{upgreek} \cdots + \tilde{d}_{n} \uptheta^{-n}+ \tilde{d}_{n+1}\uptheta^{-n-1},\quad \tilde{d}_{n+1}\in \{0,\dots, a\} .\] When this is summed with \(\require{upgreek} \sum_{i=n/2+2}^{\infty} \tilde{b}_{2i-1}\uptheta^{-2i+1}\), the result is automatically greedy (since there is no \(\require{upgreek} \uptheta^{-n-2}\) term in either sum) and we obtain an expansion whose powers are eventually all odd. A similar analysis of the right-hand side gives a greedy expansion in which the powers are eventually all even. This contradicts the unicity of greedy expansions of elements of \(\mathbb{R}\). ◻

Proof. We will use the fact that \(\require{upgreek} \uppi (\upalpha_{1})= \uppi (\upalpha_{2})\) implies \(\require{upgreek} \upalpha_{1}'= \upalpha_{2}'\): thus \[\require{upgreek} \upalpha_{1} = \sum_{i=m}^{\infty} b_{i}\uptheta^{i},\quad \upalpha_{2} = \sum_{i=n}^{\infty} c_{i}\uptheta^{i},\] produce the same conjugates \[\require{upgreek} \sum_{i=m}^{\infty} b_{i}(- \uptheta)^{-i} = \sum_{i=n}^{\infty} c_{i}(-\uptheta)^{-i}.\] First suppose that \(m=n=0\) and that the first coefficient in which they differ is \(i=0\). Suppose \(b_{0}>c_{0}\). Then we have \[\require{upgreek} (b_{0}-c_{0}) + c_{1}\uptheta^{-1} + b_{2}\uptheta^{-2} + c_{3}\uptheta^{-3} + b_{4} \uptheta^{-4} +\cdots = b_{1}\uptheta^{-1} + c_{2}\uptheta^{-2} + b_{3}\uptheta^{-3} + c_{4} \uptheta^{-4} +\cdots .\] By the greedy condition \(b_{1}\leq a-1\). Then, the right-hand side of the above satisfies \[\require{upgreek} \begin{align} \text{\rm RHS} &\leq (a-1)\uptheta^{-1} + a(\uptheta^{-2} + \uptheta^{-3} +\cdots ) = -\uptheta^{-1} + \frac{a}{\uptheta} \frac{1}{1-\uptheta^{-1}} = -\uptheta^{-1} + \frac{a}{\uptheta-1} \\ & =\frac{ -\uptheta +1 +a\uptheta}{\uptheta^{2}-\uptheta} = \frac{ -\uptheta +\uptheta^{2}}{\uptheta^{2}-\uptheta} =1. \end{align}\] From this we conclude that \(b_{0}-c_{0}=1\) and for this to happen we must have additionally \(b_{1} =a-1\), \(c_{2}=b_{3}=c_{4}=b_{5}=\cdots =a\) and \(c_{1}=b_{2}=c_{3}=b_{4}=\cdots =0\). We conclude in this case that \(\require{upgreek} \upalpha_{1}, \upalpha_{2}\) are of the form claimed in the statement of the Proposition: \[\require{upgreek} \upalpha_{1} =(c_{0}+1) + (a-1)\uptheta + a (\uptheta^{3} + \uptheta^{5}+\cdots ),\quad \upalpha_{2} =c_{0} + a (\uptheta^{2} + \uptheta^{4}+\cdots )\] Notice that this argument works even when \(c_{0}=0\), so we can always just assume that our series start from \(i=0\), possibly allowing zero coefficients in the beginning. More generally, allowing that \(\require{upgreek} \upalpha_{1}\) and \(\require{upgreek} \upalpha_{2}\) coincide in an initial Laurent polynomial \(\require{upgreek} \upbeta\), the above argument shows that if \(\require{upgreek} \upalpha_{1}'=\upalpha_{2}'\), then we have, say, \[\require{upgreek} \begin{align} \upalpha_{1}=_{\rm gr} \upgamma + (c_{i}+1)\uptheta^{i} + (a-1)\uptheta^{i+1} + a\left(\uptheta^{i+3}+ \uptheta^{i+5} +\cdots \right) \end{align}\] and \[\require{upgreek} \begin{align} \upalpha_{2}=_{\rm gr} \upgamma + c_{i}\uptheta^{i} + a\left(\uptheta^{i+2}+ \uptheta^{i+4} +\cdots \right), \end{align}\] where \[\require{upgreek} \upgamma = \sum g_{k}\uptheta^{k}\] is of degree \(\leq i-1\). Notice the dichotomy in the “tails” of \(\require{upgreek} \upalpha_{1}\) and \(\require{upgreek} \upalpha_{2}\), which are of opposite power parity. Now suppose we have a third series \[\require{upgreek} \updelta = \sum_{j=r}^{\infty} d_{j}\uptheta^{j}\] with \(\require{upgreek} \updelta'=\upalpha_{1}'=\upalpha_{2}'\). If \(\require{upgreek} \updelta\) first differs from either \(\require{upgreek} \upalpha_{1}\) or \(\require{upgreek} \upalpha_{2}\) in an index \(j<i\), then by the above arguments \[d_{j}= g_{j} \pm 1.\] But this implies that \(\require{upgreek} \upalpha_{1}\) and \(\require{upgreek} \upalpha_{2}\) would have to have the same tail parity, opposite to that of \(\require{upgreek} \updelta\), which contradicts the fact that \(\require{upgreek} \upalpha_{1}\) and \(\require{upgreek} \upalpha_{2}\) have opposite tail parity. Thus \(\require{upgreek} \updelta =\upgamma + d_{i}\uptheta^{i}+\cdots\). But then \(d_{i}\) must differ from one of \(c_{i}\), \(c_{i}+1\) , and must be equal to one of the two, which would imply that \(\require{upgreek} \updelta\) is equal to one of either \(\require{upgreek} \upalpha_{1}\) or \(\require{upgreek} \upalpha_{2}\). That \(\require{upgreek} \upalpha_{1}'=\upalpha_{2}'\in\mathcal{O}_{K}\) follows immediately from the calculation \[\require{upgreek} a(1+\uptheta^{-2}+\uptheta^{-4}+\cdots )=\uptheta.\] ◻

Proof. If \(\require{upgreek} \upalpha =-\uptheta^{N}\) then we may take \(\require{upgreek} \upalpha_{1}=(-1)_{1}\uptheta^{N}\) and \(\require{upgreek} \upalpha_{2}=(-1)_{2}\uptheta^{N}\). where \((-1)_{1}, (-1)_{2}\) are as in Note 6. Thus we assume now that \(\require{upgreek} \upalpha \not= -\uptheta^{N}\) for any \(N\). Let \(\require{upgreek} \uptheta^{N+1}\) be the smallest power so that \(\require{upgreek} \upalpha+ \uptheta^{N+1}>0\). Then we may write \[\require{upgreek} \upalpha = \sum_{i=n}^{N} b_{i}\uptheta^{i} - \uptheta^{N+1} ,\] where the sum \(\require{upgreek} \sum_{i=n}^{N} b_{i}\uptheta^{i} >0\) is greedy and by the minimality of \(N+1\), \(b_{N}>0\). Then we have \[\require{upgreek} \upalpha_{1}=\sum_{i=n}^{N} b_{i}\uptheta^{i} + (-1)_{1} \uptheta^{N+1}=_{\rm gr} \sum_{i=n}^{N} b_{i}\uptheta^{i} + a(\uptheta^{N+2}+\uptheta^{N+4} +\cdots )\] defines an element of \(\require{upgreek} \mathcal{O}_{\uptheta}^{+}\) in the Cauchy class of \(\require{upgreek} \upalpha\). On the other hand, consider \[\require{upgreek} \begin{align} \label{alpha2original} \upalpha_{2} & =\sum_{i=n}^{N} b_{i}\uptheta^{i} + (-1)_{2} \uptheta^{N+1} \nonumber \\ & = \sum_{i=n}^{N-1} b_{i}\uptheta^{i} + (b_{N}+1)\uptheta^{N}+ (a-1)\uptheta^{N+1} + a\left(\uptheta^{N+3}+\uptheta^{N+5}+\cdots \right). \end{align}\tag{36}\] If \(b_{N}\leq a-2\), this expression is greedy. Otherwise, if \(b_{N}=a-1\) and \(b_{N-1}>0\), then resolving the implied Fibonacci relation gives \[\require{upgreek} \upalpha_{2}= \sum_{i=n}^{N-2} b_{i}\uptheta^{i} + (b_{N-1}-1)\uptheta^{N-1}+ a\left(\uptheta^{N+1}+\uptheta^{N+3}+\cdots \right) ,\] which is greedy. If \(b_{N}=a\) (so that \(b_{N-1}=0\)), we have a carry at the term \(\require{upgreek} \uptheta^{N}\), which resolves to \[\require{upgreek} \upalpha_{2}= \sum_{i=n}^{N-3} b_{i}\uptheta^{i} + (b_{N-2}+1)\uptheta^{N-2} + (a-1)\uptheta^{N-1}+ a\left(\uptheta^{N+1}+\uptheta^{N+3}+\cdots \right)\] Note that this is of the same shape as the original expression (36 ): thus inductively we conclude that, put into greedy form, \(\require{upgreek} \upalpha_{2}\) gives an element in \(\require{upgreek} \mathcal{O}_{\uptheta}^{+}\) distinct from \(\require{upgreek} \upalpha_{1}\) and defining the same Cauchy class as \(\require{upgreek} \upalpha\). Now suppose that there was a third element \(\require{upgreek} \updelta\in \mathcal{O}^{+}_{\uptheta}\) in the Cauchy class of \(\require{upgreek} \upalpha\) distinct from \(\require{upgreek} \upalpha_{1}\) and \(\require{upgreek} \upalpha_{2}\). But then \(\require{upgreek} \updelta'=\upalpha'= \upalpha_{1}'=\upalpha_{2}'\), and by Proposition 13, this is not possible. ◻

Proof. First note that every \(\require{upgreek} \upalpha\in\mathcal{O}_{K}\) has exactly three pre-images. If \(\require{upgreek} \upalpha >0\), by Theorem 23, it has a unique representative in \(\require{upgreek} \mathcal{O}_{\uptheta}^{+}\) (itself), and, since by Theorem 24, \(\require{upgreek} -\upalpha\) has exactly two distinct representatives \(\require{upgreek} \upalpha_{1}, \upalpha_{2}\in \mathcal{O}_{\uptheta}^{+}\), it follows that \(\require{upgreek} \upalpha\) is also represented by \(\require{upgreek} -\upalpha_{1}\) and \(\require{upgreek} -\upalpha_{2}\). The same applies for \(\require{upgreek} \upalpha<0\). Finally, by Proposition 13, there can be no other elements of \(\require{upgreek} \widehat{\mathcal{O}}_{\uptheta}\) with multiple pre-images. ◻

Proof. By Theorem 18, \(\require{upgreek} \uppi\) is a homomorphism of multirings, and since \(\require{upgreek} \widehat{\mathcal{O}}_{\uptheta}\) is an actual ring, \(\require{upgreek} \uppi (\upalpha + \upbeta) = \uppi (\upalpha)+ \uppi ( \upbeta) = \hat{\upalpha}+\hat{\upbeta}\), hence \(\require{upgreek} \upalpha+\upbeta\) is in the pre-image by \(\require{upgreek} \uppi\) of the unique element \(\require{upgreek} \hat{\upalpha}+\hat{\upbeta}\), hence \(\require{upgreek} \upalpha+\upbeta\) consists of no more than three elements. The same applies to the product. ◻

Proof. First suppose that we have two distinct Laurent series in \(\require{upgreek} \uptheta\), neither having a factor of \(T^{-1}\) or a sign \(-\), having the same conjugate: \[\require{upgreek} \upalpha = \sum_{i=m}^{\infty} b_{i}\uptheta^{i},\quad \upbeta = \sum_{i=n}^{\infty} c_{i}\uptheta^{i}, \quad \upalpha'=\upbeta'.\] Thus we have \[\require{upgreek} \begin{align} \label{equalconj} \sum_{i=m}^{\infty} b_{i}\uptheta^{-i} = \sum_{i=n}^{\infty} c_{i}\uptheta^{-i} . \end{align}\tag{37}\] If there are no infinite forbidden blocks in either expression in (37 ) (as defined in (12 )), then the conjugates, as beta expansions of elements of \(\mathbb{R}\), are greedy. By unicity of the greedy expansion in \(\mathbb{R}\), their coefficients are equal, hence \(\require{upgreek} \upalpha=\upbeta\). If one of them, say \(\require{upgreek} \upalpha'\), contains an infinite forbidden block, it has the form \[\require{upgreek} \begin{align} \label{N611ConjEqual} \upalpha' & = \cdots + (a-2)\uptheta^{-k-2} + (a-2)\uptheta^{-k-1} + (a-1)\uptheta^{-k} + b_{k-1} \uptheta^{-k+1} + \cdots +b_{m}\uptheta^{-m} \nonumber \\ & = (b_{k-1} +1) \uptheta^{-k+1} + b_{k-2} \uptheta^{-k+2}+ \cdots +b_{m}\uptheta^{-m} , \end{align}\tag{38}\] where \(k\geq m\) and the second equality follows from (11 ). We claim that the Laurent polynomial in (38 ) is greedy. Indeed, first we note that necessarily \(b_{k-1}\leq a-2\) (otherwise we would have a consecutive pair of \(a-1\) coefficients, which forms a forbidden block). Hence \(b_{k-1}+1\leq a-1\). If \(b_{k-1}+1 = a-1\), and if it were to form in addition a forbidden block in (38 ), i.e., there exists \(l\leq k-2\) with \(b_{l}=a-1\), it means that we already had a forbidden block in \(\require{upgreek} \upalpha\), of the shape \[\require{upgreek} \cdots (a-1)\uptheta^{k} + (a-2) \uptheta^{k-1} + \cdots + (a-2)\uptheta^{l-1} + (a-1)\uptheta^{l}+\cdots .\] So in greedy form \(\require{upgreek} \upalpha'\) reduces to a polynomial. Therefore the same is true of \(\require{upgreek} \upbeta'\): either

1. \(\require{upgreek} \upbeta\) is a polynomial with \(c_{i}=0\) for all \(i\geq k\), \(c_{k-1}= b_{k-1}+1\) and \(c_{l}=b_{l}\) for \(l\leq k\) in which case we have \(\require{upgreek} \upalpha\not=\upbeta\), or else,

2. \(\require{upgreek} \upbeta\) like \(\require{upgreek} \upalpha\) contains an infinite forbidden block. In this case we claim that \(\require{upgreek} \upalpha=\upbeta\). Indeed, suppose we have \[\require{upgreek} \upalpha' =\cdots + (a-2)\uptheta^{-2} + (a-1) + \upxi = \cdots + (a-2)\uptheta^{-k-1} + (a-1)\uptheta^{-k} +\upeta =\upbeta' ,\] where we suppose (after possibly multiplying \(\require{upgreek} \upalpha, \upbeta\) by a suitable common power of \(\require{upgreek} \uptheta\)) that the infinite forbidden block of \(\require{upgreek} \upalpha'\) begins in the power \(0\), and where \(\require{upgreek} \upxi , \upeta\) are greedy polynomials whose lowest order powers are \(\require{upgreek} \geq \uptheta, \uptheta^{-k+1}\), respectively. Upon resolving the infinite forbidden blocks, we obtain \[\require{upgreek} \upalpha' =_{\rm gr} \uptheta +\upxi = \uptheta^{-k+1} + \upeta =_{\rm gr} \upbeta' .\] This implies that \(k=0\), and thus \(\require{upgreek} \upxi = \upeta\), hence \(\require{upgreek} \upalpha = \upbeta\).

In particular, if there is a third element \(\require{upgreek} \upgamma\), also positive and with no \(T^{-1}\) factor, with \(\require{upgreek} \upgamma'=\upalpha'=\upbeta'\), then \(\require{upgreek} \upgamma\) must be equal to either \(\require{upgreek} \upalpha\) or \(\require{upgreek} \upbeta\). The cases where the signs are negative or each series contains a factor of \(T^{-1}\) are proved in the same way. We now indicate how the statement in the Lemma follows from the above arguments. First, if \(\require{upgreek} r=\upalpha' = \upbeta'\) but \(\require{upgreek} \upalpha\not=\upbeta\), then one of the two must have an infinite forbidden block, hence \(r\in \mathcal{O}_{K}\). In this case, we cannot have a third element \(\require{upgreek} \upgamma\not=\upalpha, \upbeta\) of the same type with with \(\require{upgreek} r=\upgamma'\). ◻

Proof. Consider \(r \in \mathcal{O}_{K}^{0}\); without loss of generality we may assume \(r>0\). Let \(\require{upgreek} \upalpha \in \mathcal{O}_{K}^{0}\) be such that \(\require{upgreek} \upalpha' =r\). Thus \(\require{upgreek} \upalpha >0\). Then the product \[\require{upgreek} \upalpha\cdot 1_{3}= - T^{-1} \upalpha \cdot (a-2)(\uptheta +\uptheta^{2} +\cdots ) \subset \mathcal{O}_{\uptheta}\] (which possibly is multivalued) has conjugate equal to \(r\). Since the product \(\require{upgreek} \upalpha \cdot (a-2)(\uptheta +\uptheta^{2} +\cdots )\) is nonempty, there exists an element of \(\require{upgreek} \mathcal{O}_{\uptheta}\) \[\require{upgreek} \upgamma =- T^{-1} \sum_{i=m}^{\infty} b_{i} \uptheta^{i} \;\; \in \;\; - T^{-1} \upalpha \cdot (a-2)(\uptheta +\uptheta^{2} +\cdots ) .\] We claim that \(\require{upgreek} \upgamma\) is the unique element possessing the factor \(T^{-1}\) such that \(\require{upgreek} \upgamma'=r\). (We note that such an element must carry the sign \(-\) to conjugate to a positive real number.) For otherwise, if we have \(\require{upgreek} \upeta =- T^{-1} \sum_{i=n}^{\infty} c_{i} \uptheta^{i}\) with \(\require{upgreek} \upeta'=r\), then passing to conjugates, we have \[\require{upgreek} \sum_{i=m}^{\infty} b_{i} \uptheta^{-i} =\sum_{i=n}^{\infty} c_{i} \uptheta^{-i} = (-T)'r= -(T\upalpha)'.\] However, \(\require{upgreek} 0<-(T\upalpha)'\) has conjugate \(\require{upgreek} -T\upalpha<0\), hence \(\require{upgreek} -(T\upalpha)' \not\in \mathcal{O}_{K}^{0}\). It follows that neither of the conjugates \(\require{upgreek} \upgamma'\), \(\require{upgreek} \upeta'\) contain an infinite forbidden block (otherwise they would be in \(\mathcal{O}_{K}^{0}\)), hence they are both greedy. Thus, by unicity of greedy expansions in \(\mathbb{R}\), \(\require{upgreek} \upgamma=\upeta\). For \(r= T^{-1}h \in \mathcal{O}_{K, +}^{1}\), the argument is essentially the same: there exists a series \(\require{upgreek} \upalpha\), with no \(T^{-1}\) and with sign \(-\), producing \(r\) as conjugate, and the argument needed to show that this element is unique is the same as that used in the previous case. ◻

Proof. First suppose that \(r\in \mathcal{O}_{K}\). By Lemmas 6 and 7, there are exactly three elements of \(\require{upgreek} \mathcal{O}_{\uptheta}\) whose conjugate is \(r\). For \(r\not\in\mathcal{O}_{K}\), we have the infinite greedy expansion \[\require{upgreek} r=_{\rm gr} \sum_{i=m}^{\infty} b_{i}\uptheta^{-i} .\] Then \(\require{upgreek} \upalpha =_{\rm gr} \sum_{i=m}^{\infty} b_{i}\uptheta^{i}\) satisfies \(\require{upgreek} \upalpha'=r\), and by Lemma 6, it is the unique element of \(\require{upgreek} \mathcal{O}_{\uptheta}\) with this property and not having a factor of \(T^{-1}\). On the other hand, \(\require{upgreek} \upbeta = \upalpha\cdot 1_{3}\) has a factor of \(T^{-1}\) and, again by Lemma 6, it is the unique such element with conjugate \(r\). ◻

Proof. If \(\require{upgreek} \{ \upalpha_{i} \} \subset \mathcal{O}_{\uptheta, N}\) and \(\require{upgreek} \upalpha_{i}\rightarrow \upalpha\) in \(C_{N}\) in the Tychonoff (weak) topology, then the coordinates of the limit point \(\require{upgreek} \upalpha\) must also satisfy the greedy condition, hence \(\require{upgreek} \upalpha\in \mathcal{O}_{\uptheta, N}\). It is trivial to see that every point \(\require{upgreek} \upalpha\in \mathcal{O}_{\uptheta, N}\) is a limit point of points of \(\require{upgreek} \mathcal{O}_{\uptheta, N}\): any sequence of partial sums gives a sequence of elements of \(\require{upgreek} \mathcal{O}_{\uptheta, N}\) converging to \(\require{upgreek} \upalpha\). ◻

Proof. If \(\require{upgreek} \upalpha_{i}\rightarrow \upalpha\) in the \(\require{upgreek} \uptheta\)-adic topology, then the associated Cauchy classes satisfy \(\require{upgreek} | \hat{\upalpha}_{i} -\hat{\upalpha}|_{\uptheta}\rightarrow 0\) by the infratriangle inequality, which implies convergence. ◻

Proof. It is evident that \(\require{upgreek} \upalpha'\) defines a convergent Laurent series in \(\require{upgreek} \uptheta^{-1}\) since its coefficients are bounded in absolute value by \(a\). If \(\require{upgreek} \upalpha\) and \(\require{upgreek} \upbeta\) are close in the \(\require{upgreek} \uptheta\)-adic topology, it means that they agree up to some large index \(N\), but then the conjugate of \(\require{upgreek} \upalpha-\upbeta\) must be small. In particular, if \(\require{upgreek} \upalpha_{i}\rightarrow \upbeta\) then \(\require{upgreek} \upalpha_{i}'\rightarrow \upbeta'\). Since \(\require{upgreek} \mathcal{O}_{\uptheta}\) has a first countable topology, this implies continuity. Now suppose that \(\require{upgreek} B\subset \mathcal{O}_{\uptheta}\) is closed but non-compact, we claim that \(B'\) is also closed. If \(\require{upgreek} \{ \upalpha_{i}\} \subset B\) is a sequence \(\rightarrow \infty\) in the \(\require{upgreek} \uptheta\)-adic topology, then in particular \(\require{upgreek} |\upalpha_{i}|_{\uptheta}\rightarrow \infty\) and hence \(\require{upgreek} \upalpha'_{i}\rightarrow\pm\infty\) in \(\mathbb{R}\). Indeed, \[\require{upgreek} |\upalpha_{i}'| \geq |\upalpha_{i}|_{\uptheta} \cdot \big(1- (a-1)\uptheta^{-1} - a\uptheta^{-3} -a\uptheta^{-5} -\cdots \big) = |\upalpha_{i}|_{\uptheta} \uptheta ^{-1}\longrightarrow \infty ,\] where the last equality is that found in (16 ). Thus, if \(x\) is a limit point of \(B'\), so that there exists \(\require{upgreek} \upbeta_{i}'\rightarrow x\) for \(\require{upgreek} \upbeta_{i}\in B\), then, after passing to a subsequence, we may assume \(\require{upgreek} \upbeta_{i}\rightarrow \upbeta\in B\). By continuity of conjugation, \(\require{upgreek} \upbeta' = x\), hence \(B'\) is closed. ◻

Proof. See the proof of Theorem 17.16 of [26]. ◻

Proof. We prove the multicontinuity of \(+\) ; that of \(\cdot\) is analogous. Suppose that \(\require{upgreek} x_{\upalpha}\rightarrow x\), \(\require{upgreek} y_{\upalpha}\rightarrow y\) in \(\require{upgreek} \mathcal{O}_{\uptheta}\), and let \(\require{upgreek} w_{\upalpha}\in x_{\upalpha} +y_{\upalpha}\). By Theorem 26, it suffices to show that there exists \(w\in x+y\) and a subsequence of \(\require{upgreek} \{ w_{\upalpha}\}\) converging to \(w\). Since pre-images with respect to conjugation have bounded cardinality, and since \(\require{upgreek} w_{\upalpha}' = x_{\upalpha}' + y_{\upalpha}' \rightarrow x'+y'\), we may find some \(w\) satisfying \(w ' = x' +y'\) for which \(\require{upgreek} w_{\upalpha}\rightarrow w\) (after passing to a subsequence, if necessary). Note that \(w ' = x' +y'\) is a necessary condition for \(w\in x+y\). To show that the latter is indeed the case we must find sequences of partial sums \(\{x_{n} \} \in x\), \(\{y_{n} \} \in y\) so that \(\{ w_{n} = x_{n} + y_{n}\}\in w\). On the other hand, since \(\require{upgreek} w_{\upalpha}\in x_{\upalpha} + y_{\upalpha}\) for each \(\require{upgreek} \upalpha\), we may find sequences of partial sums of all parties so that \[\require{upgreek} \sum_{i=m}^{M_{\upalpha, n}} w_{\upalpha, i}\uptheta^{i} +\updelta_{\upalpha, n} = \sum_{i=m'}^{M'_{\upalpha, n}} x_{\upalpha,i}\uptheta^{i} + \upxi_{\upalpha, n} + \sum_{i=m''}^{M''_{\upalpha, n}} y_{\upalpha,i} \uptheta^{i} +\upeta_{\upalpha, n},\] where for each \(\require{upgreek} \upalpha\), the sequences \(\require{upgreek} M_{\upalpha, n}\), \(\require{upgreek} M_{\upalpha, n}'\) and \(\require{upgreek} M_{\upalpha, n}''\) are monotone increasing and \(\require{upgreek} \updelta_{\upalpha, n}\), \(\require{upgreek} \upxi_{\upalpha, n}\) and \(\require{upgreek} \upeta_{\upalpha, n}\) are the defects. By definition of convergence in the \(\require{upgreek} \uptheta\)-topology, given \(M > m,m',m''\), for \(\require{upgreek} \upalpha\) sufficiently large and \(i\leq M\), \[\require{upgreek} \begin{align} \label{convequ} x_{\upalpha, i}= x_{i},\;\;, y_{\upalpha, i}= y_{i} ,\;\;w_{\upalpha, i}= w_{i}. \end{align}\tag{39}\] Now take a diagonal subsequence \(\left\{ (\widetilde{M}_{n}, \widetilde{M}_{n}' ,\widetilde{M}_{n} '') \right\}\) of the bi-indexed sequence of triples \(\require{upgreek} \left\{ (M_{\upalpha, n}, M'_{\upalpha, n},M''_{\upalpha, n}) \right\}\), so that for \(M=\widetilde{M}_{n}\), resp., \(M=\widetilde{M}'_{n}\), resp., \(M=\widetilde{M}_{n}''\) we have the respective equalities shown in (39 ). Denote the corresponding defects using the same notation. Then \[\require{upgreek} \begin{align} \sum_{i=m}^{\widetilde{M}_{n}} w_{i}\uptheta^{i} +\widetilde{\updelta}_{n} = \sum_{i=m'}^{\widetilde{M}_{n}} x_{i}\uptheta^{i} + \widetilde{\upxi}_{n} + \sum_{i=m''}^{\widetilde{M}_{n}} y_{i} \uptheta^{i} +\widetilde{\upeta}_{n} , \end{align}\] which gives the relation \(w\in x+y\). ◻

Proof. Rescaling by a power of \(\require{upgreek} \uptheta\) we may assume that \[\require{upgreek} g = b_{0} + b_{1}\uptheta + \cdots + b_{m}\uptheta^{m},\] so that \(\require{upgreek} |g|_{\uptheta}=1\). In what follows, we abbreviate \(b=b_{0}\). Represent \(h\) as a Cuachy sequence \(h= \{ h_{n}\}\) so that \[\begin{align} \label{InvCondForG} gh_{n}=_{\rm gr} 1+ e_{n}, \end{align}\tag{40}\] where \(e_{n}\rightarrow 0\) in \(\require{upgreek} |\cdot |_{\uptheta}\). Thus we wish to show that eventually \(\require{upgreek} |h_{n}|_{\uptheta}=1\). Since \(g\) starts with the constant term \(b\), the conjugate \(g'\) is positive: indeed, we have the strict lower bound \[\require{upgreek} \begin{align} \label{g39lower} g' > 1 - (a-1)\uptheta^{-1} - a\uptheta^{-3} -a\uptheta^{-5}-\cdots = \uptheta^{-1} +1 - \frac{a}{\uptheta-\uptheta^{-1}} =\uptheta^{-1} >0, \end{align}\tag{41}\] where the lower bound involves only odd powers of \(\require{upgreek} \uptheta\) so that all the signs of conjugates of terms following \(1\) are negative. We have also the strict upper bound \[\require{upgreek} \begin{align} \label{g39upper} g'< b+ a\uptheta^{-2} +a\uptheta^{-4} +\cdots = b+\uptheta^{-1} . \end{align}\tag{42}\] The condition (40 ) implies after conjugation that \(g'h_{n}'\rightarrow 1\), thus \(h_{n}'\rightarrow g'^{-1}\) in the euclidean topology. Moreover, by the uniform bounds (41 ), (42 ) on \(g'\), for \(n\) large we have \[\require{upgreek} \begin{align} \label{boundonhn} \frac{1}{b+\uptheta^{-1}} < h'_{n}< \uptheta , \end{align}\tag{43}\] which we emphasize are strict inequalities. The positivity condition on \(h_{n}'\) implies that the lowest power of \(\require{upgreek} \uptheta\) in \(h_{n}\) is even (e.g., if \(\require{upgreek} h_{n} =_{\rm gr} c_{1} \uptheta +\text{higher powers of \uptheta}\), then \(\require{upgreek} h_{n}'< -\uptheta^{-1} +(a-1)\uptheta^{-2} + a\uptheta^{-4} + \cdots =-\uptheta^{-2}<0\)). Suppose the lowest power is negative, e.g., \[\require{upgreek} h_{n} = c_{-2}\uptheta^{-2}+\cdots;\] then we have \[\require{upgreek} h_{n} ' > \uptheta^{2} - (a-1)\uptheta -a(\uptheta^{-1} + \uptheta^{-3}+\cdots ) = \uptheta .\] This contradicts the strict upper bound of (43 ). For a more general lowest negative power, \(\require{upgreek} h_{n}=c_{-2k}\uptheta^{-2k}+\cdots\), \(k\geq 1\), we obtain \(\require{upgreek} h_{n} '>\uptheta^{2k-1}\geq \uptheta\) which again contradicts (43 ). Now suppose the lowest power is positive and \(\geq 2\), e.g., \[\require{upgreek} h_{n} = c_{2}\uptheta^{2}+ \cdots .\] Then we have \[\require{upgreek} h'_{n} < a(\uptheta^{-2} + \uptheta^{-4} +\cdots ) =\uptheta^{-1}\leq \frac{1}{b+\uptheta^{-1}},\] since \[\require{upgreek} \uptheta^{-1}(b+\uptheta^{-1}) \leq \uptheta^{-1} ( a+ \uptheta^{-1}) = 1.\] This contradicts the strict lower bound in (43 ). And for a general lowest positive power \(\require{upgreek} h_{n} = c_{2k}\uptheta^{2k}+ \cdots\) we obtain \(\require{upgreek} h'_{n}<\uptheta^{-2k+1}\leq\uptheta^{-1}\) which continues to contradict (43 ). Therefore \(h_{n}\) must start with a constant term, implying that its norm is \(1\). ◻

Proof. Since \(n\) is prime and \(c<n\), \(n\nmid ac\), so \(ac/n = m+r/n\) for \(r\in \{1,\dots , n-1\}\). Thus \[n \left\lfloor \frac{ac}{n}\right\rfloor = nm = ac -r = ac +d -n ,\quad d=n-r\in \{1,\dots , n-1\}.\] Moreover, as \(ac/n\not\in \mathbb{Z}\), \[n\left\lceil\frac{ac}{n}\right\rceil = n \left\lfloor \frac{ac}{n}\right\rfloor +n = ac +d.\] ◻

Proof. We claim that \[\begin{align} \label{littleclaim} c_{i}\equiv c_{i-2} +d_{i-1}-n\equiv c_{i-2} +d_{i-1}\mod n. \end{align}\tag{48}\] Indeed, by definition of the sequence \(\{ c_{i}\}\), \(c_{i} \equiv c_{i-2} - bc_{i-1} \mod n\). On the other hand, by (47 ), \(d_{i-1}\equiv -ac_{i-1} \mod n\) and since \(a\equiv b\mod n\), \(d_{i-1}\equiv -bc_{i-1} \mod n\), which gives (48 ). We now prove (?? ). Either \(c_{i-2} +d_{i-1}-n\) or \(c_{i-2} +d_{i-1} \in \{ 0,\dots , n-1\}\), since \(c_{i-2} \in \{ 0,\dots , n-1\}\) and \(d_{i-1} \in \{ 1,\dots , n-1\}\). If \(c_{i-2} +d_{i-1}-n\geq 0\), then it belongs to the set \(\{ 0,\dots , n-2\}\) and, therefore, by (48 ), it is equal to \(c_{i}\). Otherwise, if \(c_{i-2} +d_{i-1}-n< 0\), then \(c_{i-2} +d_{i-1}\in \{ 1,\dots , n-1\}\), and again by (48 ), it is equal to \(c_{i}\). To show (?? ), first assume \(c_{i-2} +d_{i-1}-n\geq 0\): then by (47 ), \(e_{i}\) is calculated using the floor, and by (?? ), \[ne_{i}=ac_{i-1} + d_{i-1}-n =ac_{i-1} + c_{i} - c_{i-2}.\] Otherwise, if \(c_{i-2} +d_{i-1}-n<0\), by (47 ) \(e_{i}\) is calculated using the ceiling, and by (?? ), \[ne_{i}=ac_{i-1} + d_{i-1} =ac_{i-1} + c_{i} - c_{i-2}.\] ◻

Proof. As indicated above, we may assume without loss of generality that \(n\) is prime. We first show that the series (44 ) converges. We note that \(e_{i}\leq a\) for all \(i\) since \(ac_{i-1}/n < a\): thus, if the series is not greedy, it is because there exist unresolved Fibonacci relations. If we denote \(\require{upgreek} n^{-1}_{j}=1 + e_{1}\uptheta + \cdots + e_{j}\uptheta^{j}\), we must show that as \(i<j\rightarrow\infty\), \[\require{upgreek} n^{-1}_{j} -n^{-1}_{i} = e_{i+1}\uptheta^{i+1} + \cdots + e_{j}\uptheta^{j}\longrightarrow 0\] with respect to the \(\require{upgreek} \uptheta\)-adic infranorm. We may write \[\require{upgreek} n^{-1}_{j} -n^{-1}_{i} = \left( a \sum_{\substack{k=i+1 \\ e_{k}>0}}^{j} \uptheta^{k} \right) -g\] where \[\require{upgreek} g = g_{i+1}\uptheta^{i+1} + \cdots + g_{j}\uptheta^{j},\quad 0\leq g_{k} = a-e_{k}\leq a-1.\] Now \(g\) is greedy and has infranorm \(\require{upgreek} \leq \uptheta^{-i-1}\), and by infra multiplicativity \[\require{upgreek} \left|a \sum_{\substack{k=i+1 \\ e_{k}>0}}^{j} \uptheta^{k} \right|_{\uptheta} \leq \uptheta^{2} |a|_{\uptheta}\left|\uptheta^{i+1} + \cdots + \uptheta^{j} \right|_{\uptheta} = \uptheta^{1-i} .\] Then, by the infratriangle inequality, \[\require{upgreek} | n^{-1}_{j} -n^{-1}_{i} |_{\uptheta} =\bigg|a \sum_{\substack{k=i+1 \\ e_{k}>0}}^{j} \uptheta^{k} -g\bigg|_{\uptheta}\leq \uptheta^{3-i}\longrightarrow 0.\] Thus the expression for \(n^{-1}\) is convergent. We now consider \[\require{upgreek} n\cdot n^{-1} = 1 + (n-1) + ne_{1}\uptheta + \cdots =: 1+x .\] We claim first that \[ne_{1}=n \left\lceil \frac{a(n-1)}{n}\right\rceil = (n-1)a + b.\] Indeed, for \(a=mn+b\), \[n \left\lceil \frac{(n-1)a}{n}\right\rceil = n\left( (n-1)m + \left\lceil \frac{(n-1)b}{n}\right\rceil \right) = (n-1)a -(n-1)b+ n \left\lceil \frac{(n-1)b}{n}\right\rceil ;\] by Hermite’s identity [27], [28] (see page 85, (3.24)) \[\left\lceil \frac{(n-1)b}{n}\right\rceil = \left\lceil \frac{n-1}{n}\right\rceil + \left\lceil \frac{n-1}{n} - \frac{1}{b}\right\rceil + \cdots + \left\lceil \frac{n-1}{n} - \frac{b-1}{b}\right\rceil = b.\] This proves the claim. Thus, the first two terms of \(x\) are \[\require{upgreek} (n-1) + n \left\lceil \frac{(n-1)a}{n}\right\rceil \uptheta = b\uptheta + (n-1)\uptheta^{2} = c_{1}\uptheta + c_{0}\uptheta^{2}.\] Now add \(ne_{2}\); we have \(c_{0} + d_{1}-n = d_{1}-1\geq 0\) since \(d_{1}>0\). Thus, \[e_{2} = \left\lfloor \frac{ac_{1}}{n}\right\rfloor=\left\lfloor\frac{ab}{n}\right\rfloor\] and \[n \left\lfloor \frac{ac_{1}}{n}\right\rfloor = ac_{1}+d_{1}-n .\] By Lemma 10, \[c_{2}= c_{0}+d_{1}-n\] so that the first three terms of \(x\) are \[\require{upgreek} c_{1}\uptheta + c_{0}\uptheta^{2}+ (ac_{1}+d_{1}-n)\uptheta^{2} =c_{2}\uptheta^{2}+ c_{1}\uptheta^{3}.\] Inductively, if the first \(k\) terms of \(x\) resolve to \[\require{upgreek} c_{k-1}\uptheta^{k-1}+ c_{k-2}\uptheta^{k} ,\] then adding to this \(\require{upgreek} ne_{k} \uptheta^{k}\) and using (?? ), we obtain \[\require{upgreek} c_{k-1}\uptheta^{k-1}+ c_{k-2}\uptheta^{k} + \big(ac_{k-1}+c_{k} - c_{k-2}\big)\uptheta^{k} = c_{k}\uptheta^{k}+ c_{k-1}\uptheta^{k+1} .\] We conclude that \(x=0\) in \(\require{upgreek} \widehat{\mathcal{O}}_{\uptheta}\). ◻

Proof. By Theorem 28, \(\require{upgreek} \mathbb{Q}\subset\widehat{\mathcal{O}}_{\uptheta}\), and since \(\require{upgreek} \uptheta\) is a unit, \(\require{upgreek} K\subset\widehat{\mathcal{O}}_{\uptheta}\). Now let \(\require{upgreek} x = \{ \upalpha_{i}\}\in \widehat{\mathcal{O}}_{\uptheta}\), \(\require{upgreek} \upalpha_{i}\in\mathcal{O}_{K}\). Each \(\require{upgreek} \upalpha_{i}\) has an inverse (as an element of \(\require{upgreek} \widehat{\mathcal{O}}_{\uptheta}\)) \(\require{upgreek} \upalpha^{-1}_{i} = \{ \upalpha^{-1}_{ij}\}\); choose a diagonal subsequence \[\require{upgreek} x_{i}^{-1}:= \upalpha^{-1}_{ij}\] so that \(\require{upgreek} x_{i}^{-1}\upalpha_{i} =1 + \updelta_{i}\rightarrow 1\) and \(\require{upgreek} |x_{i}^{-1}|_{\uptheta} = 1/ |\upalpha_{i}|_{\uptheta}\) (the latter choice is possible due to Lemma 8). Write \(x^{-1}= \{ x_{i}^{-1}\}\); then the product of sequences \(\require{upgreek} x\cdot x^{-1} = \{ 1+ \updelta_{i} \}\) converges to \(1\) in \(\require{upgreek} \widehat{\mathcal{O}}_{\uptheta}\). What remains is to show that \(x^{-1}\) is Cauchy. Suppose otherwise, that there is a constant \(C>0\) such that for all \(N\in\mathbb{N}\), there exists \(i,j>N\) with \[\require{upgreek} \begin{align} \label{notCauchy1} |x_{i}^{-1} - x_{j}^{-1}|_{\uptheta}\geq C. \end{align}\tag{49}\] By Theorem 9, either \(\require{upgreek} |x|_{\uptheta} = \lim |\upalpha_{i}|_{\uptheta}\) exists or \(\require{upgreek} |\upalpha_{i}|_{\uptheta}\) oscillates between at most two values, so for \(x\not=0\), the limit points of the sequence \(\require{upgreek} \{ |\upalpha_{i}|_{\uptheta}\}\) cannot be zero. Thus (49 ) and inframultiplicativity imply that \[\require{upgreek} |\upalpha_{i}x_{i}^{-1} - \upalpha_{i}x_{j}^{-1}|_{\uptheta}\geq C'>0\] for some new constant \(C'\). On the other hand, \[\require{upgreek} \begin{align} |\upalpha_{i}x_{i}^{-1} - \upalpha_{i}x_{j}^{-1}|_{\uptheta} & = |\upalpha_{i}x_{i}^{-1} - \upalpha_{j}x_{j}^{-1} + \upalpha_{j}x_{j}^{-1} -\upalpha_{i}x_{j}^{-1}|_{\uptheta} \\ &\leq \uptheta^{2} \left( |\upalpha_{i}x_{i}^{-1} - \upalpha_{j}x_{j}^{-1} |_{\uptheta} + | \upalpha_{j}x_{j}^{-1} - \upalpha_{i}x_{j}^{-1} |_{\uptheta} \right) . \end{align}\] However, \(\require{upgreek} |\upalpha_{i}x_{i}^{-1} - \upalpha_{j}x_{j}^{-1} |_{\uptheta} = |\updelta_{i}- \updelta_{j} |_{\uptheta} \rightarrow 0\) and \[\require{upgreek} | \upalpha_{j}x_{j}^{-1} - \upalpha_{i}x_{j}^{-1} |_{\uptheta} \leq \frac{\uptheta^{2}}{|\upalpha_{j}|_{\uptheta}} | \upalpha_{j} -\upalpha_{i}|_{\uptheta} \longrightarrow 0,\] since \(\require{upgreek} |\upalpha_{i}|_{\uptheta} \rightarrow |x|_{\uptheta}\) and the latter assumes at most three (non zero) values. Thus the sequence defining \(x^{-1}\) must be Cauchy. ◻

Proof. By Proposition 11 and Corollary 4, conjugation defines a continuous homomorphism of rings, and since \(\require{upgreek} \widehat{\mathcal{O}}_{\uptheta}\) is a field by Theorem 29, is a monomorphism. By Lemma 3, conjugation is bicontinuous, hence the image is complete; it is all of \(\mathbb{R}\) since it contains the dense subfield \(K\). ◻

Proof. Since we have already shown that \(\require{upgreek} \mathcal{O}_{\uptheta}\) is a Marty multiring, it will be enough to show that every element \(\require{upgreek} x\in{\mathcal{O}}_{\uptheta}\) has an inverse (not necessarily unique). Denote by \(\hat{x}\) the corresponding class in \(\require{upgreek} \widehat{\mathcal{O}}_{\uptheta}\), which by Theorem 29 has an inverse \(\hat{y}\). Let \(\require{upgreek} y\in \mathcal{O}_{\uptheta}\) be any greedy Laurent series in the class of \(\hat{y}\), represented by a sequence of partial sums \(\{ y_{n}\}\). Then if \(\{ x_{n}\}\) is a sequence of partial sums for \(x\), we have \[\require{upgreek} x_{n}y_{n} = 1 + \upepsilon_{n}\] where \(\require{upgreek} \upepsilon_{n}\rightarrow 0\) in the \(\require{upgreek} \uptheta\)-adic infranorm. But \(\require{upgreek} \{ 1 + \upepsilon_{n}\}\) is a sequence of partial sums in the class of \(\require{upgreek} 1 \in \mathcal{O}_{\uptheta}\), hence \(1=xy\). ◻

Proof. First suppose \(g\in\mathcal{O}^{0}_{K}\); without loss of generality, we assume \(g>0\) and \(g\not=1\). Rescaling by a power of \(\require{upgreek} \uptheta\) we may assume that \[\require{upgreek} g = b_{0} + b_{1}\uptheta + \cdots + b_{m}\uptheta^{m}\] so that \(\require{upgreek} g\in\mathbb{Z}_{\uptheta}\) and \(\require{upgreek} |g|_{\uptheta}=1\). Note that this implies that \(\require{upgreek} 1<g' <\uptheta\).

\(00\): \(h_{n} \in \mathcal{O}^{0}_{K}\) eventually.

Then, \(h_{n}g = 1+e_{n}\) where \(\require{upgreek} |e_{n}|_{\uptheta}\rightarrow 0\). Note that \(e_{n} =h_{n}g - 1\) can be in either \(\mathcal{O}_{K}^{0}\) or \(\mathcal{O}_{K}^{1}\). The latter condition implies that \(e'_{n}\rightarrow 0\) in the euclidean norm: indeed, for those elements \(e_{n}\in \mathcal{O}_{K}^{0}\), then \(\require{upgreek} |e_{n}| =_{\rm gr} \sum_{i=N_{n}}^{R_{n}} b_{n,i}\uptheta^{i}\), hence \(\require{upgreek} |e_{n}'| =_{\rm gr} \sum_{i=N_{n}}^{R_{n}} b_{n,i}\uptheta^{-i}\rightarrow 0\) as \(N_{n}\rightarrow \infty\) (by the greedy property of the expansion). On the other hand, if \(e_{n}\in \mathcal{O}_{K}^{1}\), then \(Te_{n}\in \mathcal{O}_{K}^{0}\) ; by definition, \(\require{upgreek} |e_{n}|_{\uptheta}=|Te_{n}|_{\uptheta}\), hence \(T'e_{n}'\rightarrow 0\), which implies \(e_{n}'\rightarrow 0\) in the euclidean norm. Thus, \(h_{n}'g'\rightarrow 1\) implying \(h_{n}'>0\) eventually, hence \(h_{n}>0\) eventually (as \(h_{n}\in \mathcal{O}_{K}^{0}\)). Moreover, \(h_{n}' \rightarrow 1/g'\) and thus eventually \(\require{upgreek} \uptheta^{-1}< h_{n}'<1\), which forces \(\require{upgreek} |h_{n}|_{\uptheta}=\uptheta^{-1}\).

\(01\): \(h_{n}\in \mathcal{O}^{1}_{K}\) for infinitely many \(n\).

Let us pass to a subsequence so that \(h_{n}\in \mathcal{O}_{K}^{1}\) eventually. In this case, the condition \(h_{n}g = 1+e_{n}\) with \(\require{upgreek} |e_{n}|_{\uptheta}\rightarrow 0\) forces \(h_{n}'>0\) eventually, hence \(h_{n}<0\) eventually. As in the previous case, we have \[\require{upgreek} \begin{align} \label{h39ineq01} \uptheta^{-1}<h_{n}'<1 \end{align}\tag{50}\] eventually. Now by the inequality (?? ) of Lemma 1 (applied to \(f=-Th_{n}>0\)), \[\require{upgreek} |h_{n}|_{\uptheta} := |T h_{n}|_{\uptheta} \leq -T'h_{n}' \leq \uptheta |T h_{n}|_{\uptheta}=: \uptheta | h_{n}|_{\uptheta},\] and since \(\require{upgreek} -T'=1-\uptheta^{-1}\), (50 ) gives \[\require{upgreek} \uptheta^{-1} (1-\uptheta^{-1}) <-T'h_{n}'<1-\uptheta^{-1}.\] From the above two inequalities, we deduce \[\require{upgreek} \uptheta^{-2} (1-\uptheta^{-1})< | h_{n}|_{\uptheta} <1-\uptheta^{-1} .\] Since \(\require{upgreek} N(\uptheta )=1\), \(\require{upgreek} \uptheta>2\), so the only values of \(\require{upgreek} | h_{n}|_{\uptheta}\) which are consistent with the above inequality are \(\require{upgreek} \uptheta^{-1}\) and \(\require{upgreek} \uptheta^{-2}\).

Now assume \(g\in\mathcal{O}^{1}_{K}\), \(g>0\). After rescaling by a power of \(\require{upgreek} \uptheta\), we may assume \(\require{upgreek} |g|_{\uptheta}=|Tg|_{\uptheta}=1\). Thus, since for \(\require{upgreek} N(\uptheta )=1\) the greedy expansion of a polynomial in \(\require{upgreek} \uptheta\) is also greedy, \[\require{upgreek} Tg =_{\rm gr} b_{0} + b_{1}\uptheta + \cdots +b_{n}\uptheta^{n}, \quad (Tg)' =_{\rm gr} b_{0} + b_{1}\uptheta^{-1} + \cdots +b_{n}\uptheta^{-n},\] which give the bounds \[\require{upgreek} \begin{align} \label{boundsonTG39} 1\leq (Tg)' <\uptheta . \end{align}\tag{51}\] The lower bound \(1\) can be realized only when \(a=3\): \((Tg)' = 1\) implies \(g= T^{-1}\), and since \(\require{upgreek} N(T)= (\uptheta -1)(\uptheta^{-1} -1)= 2-a\), \(T^{-1}\in \mathcal{O}_{K}\) precisely when \(a=3\). Otherwise, \(T^{-1}\not\in \mathcal{O}_{K}\) and the value \(g= T^{-1}\) is not possible.

\(10\): \(h_{n}\in \mathcal{O}^{0}_{K}\) eventually.

Since \(g'<0\), the condition \(h_{n}g=1+e_{n}\), \(e_{n}\rightarrow 0\) in the \(\require{upgreek} \uptheta\)-adic norm, forces \(h_{n}'<0\) eventually, i.e., \(h_{n}<0\) and \(h_{n}'\rightarrow 1/g'<0\) eventually. By (51 ) we have \[\require{upgreek} \frac{\uptheta}{\uptheta -1}\leq -g'< \frac{\uptheta^{2}}{\uptheta -1}\] or \[\require{upgreek} \begin{align} \label{ineqforhn39-} \uptheta^{-1} -\uptheta^{-2}< -h_{n}' \leq 1-\uptheta^{-1} <1. \end{align}\tag{52}\] The right hand inequality in (52 ) implies immediately that \(\require{upgreek} |h_{n}|_{\uptheta}<1\), but is consistent with \(\require{upgreek} |h_{n}|_{\uptheta}\leq \uptheta^{-1}\). Indeed, \(\require{upgreek} |h_{n}|_{\uptheta}= \uptheta^{-1}\) implies that \(\require{upgreek} -h_{n}' =_{\rm gr} b_{1}\uptheta^{-1} +\cdots\) and if \(\require{upgreek} (b_{1}+1)\uptheta^{-1} +\cdots\) is greedy, it is \(<1\). On the other hand, the left hand inequality is consistent with \(\require{upgreek} |h_{n}|_{\uptheta}\geq \uptheta^{-2}\), but not with lower values. Indeed, if \(\require{upgreek} |h_{n}|_{\uptheta}= \uptheta^{-1}\), then \[\require{upgreek} -h_{n}' =_{\rm gr} b_{1}\uptheta^{-1} +\cdots > \uptheta^{-1} -\uptheta^{-2} .\] In the case \(\require{upgreek} |h_{n}|_{\uptheta}= \uptheta^{-2}\), we may have \[\require{upgreek} \uptheta^{-1} -\uptheta^{-2}< -h_{n}' =_{\rm gr} b_{2}\uptheta^{-2} +\cdots,\] provided that \(\require{upgreek} (b_{2}+1)\uptheta^{-2} +\cdots\) is not greedy (either because \(b_{2}+1=a-1\) and a forbidden block is formed, or \(b_{2}+1=a\) and a carry is required to put it into greedy form).

\(11\): \(h_{n}\in \mathcal{O}_{K}^{1}\) for infinitely many \(n\).

Again, we pass to a subsequence so that \(h_{n}\in \mathcal{O}_{K}^{1}\) eventually. Here, we are forced to have \(h_{n}'<0\) and \(h_{n}>0\), wherein \(-h_{n}'\rightarrow -1/g'\), so as in Case 10, we have (52 ). Multiplying the latter by \(\require{upgreek} -T' = 1-\uptheta^{-1}\) we obtain \[\require{upgreek} \uptheta^{-1}(1-\uptheta^{-1})^{2}<T'h'_{n} \leq (1-\uptheta^{-1})^{2} .\] The right hand inequality is consistent with \(\require{upgreek} |h_{n}|_{\uptheta}:= |Th_{n}|_{\uptheta}\leq \uptheta^{-1}\) but not with any larger values. Indeed, if \(\require{upgreek} |Th_{n}|_{\uptheta} =1\), then we have \[\require{upgreek} T'h_{n}' =_{\rm gr} c_{0} + c_{1}\uptheta^{-1}+\cdots > 1 - 2\uptheta^{-1} + \uptheta^{-2}\] since \(\require{upgreek} -2\uptheta^{-1} + \uptheta^{-2} <0\). On the other hand, if \(\require{upgreek} |Th_{n}|_{\uptheta} =\uptheta^{-1}\), then \[\require{upgreek} T'h_{n}' =_{\rm gr} c_{1}\uptheta^{-1}+c_{2} \uptheta^{-2} +\cdots \leq 1 - 2\uptheta^{-1} + \uptheta^{-2}\] is consistent, e.g., if \(\require{upgreek} (c_{1}+2)\uptheta^{-1}+c_{2} \uptheta^{-2} +\cdots\) is greedy. As for the left hand inequality, it is consistent for \(\require{upgreek} |h_{n}|_{\uptheta}:= |Th_{n}|_{\uptheta}\geq \uptheta^{-2}\) but not for smaller values. Indeed, if \(\require{upgreek} |Th_{n}|_{\uptheta}= \uptheta^{-3}\) we have necessarily, for \(a>3\), \[\require{upgreek} T'h_{n}' =_{\rm gr} c_{3}\uptheta^{-3}+c_{4} \uptheta^{-4} +\cdots < \uptheta^{-1} - 2\uptheta^{-2} + \uptheta^{-3} .\] If \(a=3\), this inequality still holds, even if a forbidden block is formed when adding \(\require{upgreek} 2\uptheta^{-2}\) to the left hand side: for then, the resolution of such a block would be bounded from above by the short block \(\require{upgreek} 2\uptheta^{-2} + 2\uptheta^{-3}= \uptheta^{-1}+\uptheta^{-4}\). One might think that there is the possibility that the new \(\require{upgreek} \uptheta^{-4}\) term might combine with an existing one to give another forbidden block, producing \(\require{upgreek} \uptheta^{-3}\): but for this to occur, we need that \[\require{upgreek} T'h_{n}' = 2\uptheta^{-3} + \uptheta^{-4} +\cdots + \uptheta^{-k} + 2\uptheta^{-k} +\cdots ,\] which is not greedy. On the other hand, if \(\require{upgreek} |Th_{n}|_{\uptheta}= \uptheta^{-2}\) the inequality \[\require{upgreek} T'h_{n}' =_{\rm gr} c_{2}\uptheta^{-2}+c_{3} \uptheta^{-3} +\cdots > \uptheta^{-1} - 2\uptheta^{-2} + \uptheta^{-3}\] is possible, for example, if \(c_{2}+2=a\), since then the term \(\require{upgreek} a\uptheta^{-2}\) resolves to greedy form via a carry, giving \[\require{upgreek} \uptheta^{-1} + \uptheta^{-3} <a\uptheta^{-2}+c_{3} \uptheta^{-3} +\cdots = \uptheta^{-1} + (c_{3}+1) \uptheta^{-3} +\cdots ,\] which holds if \(c_{k}>0\) for any \(k\geq 3\). ◻

Proof. First, \[\require{upgreek} \begin{align} xn & = T + (n-1) \left(T+ (a-2)\uptheta + (a-2)\uptheta^{2} + \cdots \right) \\ & = T+ (n-1) \left(-1+ (a-1)\uptheta + (a-2)\uptheta^{2} + \cdots \right) . \end{align}\] Using the norm = \(1\) Fibonacci identity, \[\require{upgreek} \uptheta^{m+2} = a\uptheta^{m+1} -\uptheta^{m},\] we may write \[\require{upgreek} \begin{align} -1+ (a-1)\uptheta + (a-2)\uptheta^{2} + \cdots & = -1+ a\uptheta -\uptheta + (a-2)\uptheta^{2} + \cdots \\ & = -\uptheta + (a-1)\uptheta^{2} + (a-2)\uptheta^{3} \cdots \\ &\longrightarrow 0. \end{align}\] Thus, \(xn =T\), and therefore it defines an element of \(\require{upgreek} \widehat{\mathcal{O}}_{\uptheta}\). ◻

Proof. First we show that formally \(n\cdot n^{-1}\) gives a Cauchy sequence converging to 1. Thus we study the expression \[\require{upgreek} \begin{align} n\cdot n^{-1} & = n \left\lfloor \frac{a-1}{n} \right\rfloor \uptheta + ne_{2}\uptheta^{2} + \cdots + ne_{i}\uptheta^{i} + \cdots \\ & =mn \uptheta + ne_{2}\uptheta^{2} + \cdots + ne_{i}\uptheta^{i} + \cdots , \end{align}\] where the second line follows upon making the substitution \(a=mn +b\). Using the identity \(\require{upgreek} \uptheta^{2}+1=a\uptheta\), we have \[\require{upgreek} \begin{align} n\cdot n^{-1} & = 1 -b\uptheta + \left( ne_{2}+1\right)\uptheta^{2} +ne_{3}\uptheta^{3} + \cdots .\\ \end{align}\] Note that, \[ne_{2}+1= ne_{2} +c_{0} = ac_{1}-c_{2} = ab-c_{2} ,\] so we may write (using \(\require{upgreek} b\uptheta^{3} +b\uptheta=ab\uptheta^{2}\)), \[\require{upgreek} \begin{align} n\cdot n^{-1} & = 1 -c_{2}\uptheta^{2} + \left( ne_{3}+c_{1}\right)\uptheta^{3} + ne_{4}\uptheta^{4} + \cdots .\\ \end{align}\] Inductively, at the \(i\)th step, we obtain \[\require{upgreek} \begin{align} n\cdot n^{-1} & = 1 -c_{i-1}\uptheta^{i-1} + \left( ne_{i}+c_{i-2}\right)\uptheta^{i} + ne_{i+1}\uptheta^{i+1} + \cdots , \\ \end{align}\] which may be resolved to the next higher power using the identity \(\require{upgreek} c_{i-1}\uptheta^{i+1} + c_{i-1}\uptheta^{i-1} = ac_{i-1}\uptheta^{i}\). This show that \(n^{-1}\) is formally an inverse of \(n\). We end by showing that \(n^{-1}\) indeed defines an element of \(\require{upgreek} \widehat{\mathcal{O}}_{\uptheta}\). It will suffice to show that \(n^{-1}_{+}\) defines an element of \(\require{upgreek} \widehat{\mathcal{O}}_{\uptheta}\), since we have already remarked that \(n^{-1}_{-}\) defines an element of the ring \(\require{upgreek} \widehat{\mathcal{O}}_{\uptheta}\). Thus, without loss of generality, we may assume \(n^{-1}=n^{-1}_{+}\). With this assumption, we have \(0\leq e_{i}\leq a-1\) for \(i\geq 2\). Moreover, clearly \(\left\lfloor \frac{a-1}{n} \right\rfloor\in \{ 0,\dots ,a-1\}\) so all of the coefficients belong to the set \(\{ 0,\dots ,a-1\}\). Write \(n^{-1}= \{ n^{-1}_{m}\}\), where \(n^{-1}_{m}\) consists of the first \(m\) terms of \(n^{-1}\). By definition, \(n^{-1}_{m}\in \mathcal{O}_{K}^{0}\), so \(n^{-1}_{m}\), if not already in greedy form, has a finite greedy expansion. Now for \(\tilde{m}>m\), \[\require{upgreek} \begin{align} | n^{-1}_{\tilde{m}}-n^{-1}_{m}|_{\uptheta} & = \uptheta^{-m-1} \left| e_{m+1} +e_{m+2} \uptheta +\cdots +e_{\tilde{m}}\uptheta^{\tilde{m} -m-1}\right|_{\uptheta} \\ & =: |f+g|_{\uptheta} \end{align}\] where \[\require{upgreek} f = d_{0} + d_{1}\uptheta + \cdots + d_{\tilde{m}-m-1}\uptheta^{\tilde{m}-m-1}\] and where \[d_{i} =\left\{ \begin{array}{ll} e_{m+i+1}-1 & \text{if } e_{m+i+1} \not=0 \\ & \\ 0 & \text{otherwise} \end{array} \right.\] Then \(f\) is greedy since its coefficients are \(\leq a-2\), and so is \(g\), since its coefficients are all either \(0\) or \(1\). By the shape of \(f\) and \(g\), \(\require{upgreek} |f|_{\uptheta}\), \(\require{upgreek} |g|_{\uptheta}\leq 1\), so by the infratriangle inequality (Theorem 8, item 1), we have \[\require{upgreek} | n^{-1}_{\tilde{m}}-n^{-1}_{m}|_{\uptheta} \leq \uptheta^{-m} \longrightarrow 0\] as \(m\rightarrow\infty\). This shows that \(n^{-1}= \{n^{-1}_{m}\}\) defines an element of \(\require{upgreek} \widehat{\mathcal{O}}_{\uptheta}\). ◻

Proof. The proof has the same structure as that of Theorem 29. By Theorem 31, \(\require{upgreek} \mathbb{Q}\subset\widehat{\mathcal{O}}_{\uptheta}\), and since \(\require{upgreek} \uptheta\) is a unit, \(\require{upgreek} K\subset\widehat{\mathcal{O}}_{\uptheta}\). Now let \(\require{upgreek} x = \{ \upalpha_{i}\}\in \widehat{\mathcal{O}}_{\uptheta}\), \(\require{upgreek} \upalpha_{i}\in\mathcal{O}_{K}\). Each \(\require{upgreek} \upalpha_{i}\) has an inverse (as an element of \(\require{upgreek} \widehat{\mathcal{O}}_{\uptheta}\)) \(\require{upgreek} \upalpha^{-1}_{i} = \{ \upalpha^{-1}_{ij}\}\); choose a diagonal subsequence \[\require{upgreek} x_{i}^{-1}:= \upalpha^{-1}_{ij}\] so that \(\require{upgreek} x_{i}^{-1}\upalpha_{i} =1 + \updelta_{i}\rightarrow 1\) and \(\require{upgreek} |x_{i}^{-1}|_{\uptheta} = c/ |\upalpha_{i}|_{\uptheta}\), where \(\require{upgreek} c\in \{ \uptheta^{-2},\uptheta^{-1}\}\) (the latter choice is possible due to Lemma 11). Write \(x^{-1}= \{ x_{i}^{-1}\}\); then the product of sequences \(\require{upgreek} x\cdot x^{-1} = \{ 1+ \updelta_{i} \}\) converges to \(1\) in \(\require{upgreek} \widehat{\mathcal{O}}_{\uptheta}\). What remains is to show that \(x^{-1}\) is Cauchy. Suppose otherwise, that there is a constant \(C>0\) such that for all \(N\in\mathbb{N}\), there exists \(i,j>N\) with \[\require{upgreek} \begin{align} \label{notCauchy} |x_{i}^{-1} - x_{j}^{-1}|_{\uptheta}\geq C. \end{align}\tag{54}\] By Theorem 13, either \(\require{upgreek} |x|_{\uptheta} = \lim |\upalpha_{i}|_{\uptheta}\) exists or \(\require{upgreek} \{ |\upalpha_{i}|_{\uptheta}\}\) oscillates between at most five values, so for \(x\not=0\), the limit points of the sequence \(\require{upgreek} \{ |\upalpha_{i}|_{\uptheta}\}\) cannot be zero. Thus (54 ) and inframultiplicativity imply that \[\require{upgreek} |\upalpha_{i}x_{i}^{-1} - \upalpha_{i}x_{j}^{-1}|_{\uptheta}\geq C'>0\] for some new constant \(C'\). On the other hand, \[\require{upgreek} \begin{align} |\upalpha_{i}x_{i}^{-1} - \upalpha_{i}x_{j}^{-1}|_{\uptheta} & = |\upalpha_{i}x_{i}^{-1} - \upalpha_{j}x_{j}^{-1} + \upalpha_{j}x_{j}^{-1} -\upalpha_{i}x_{j}^{-1}|_{\uptheta} \\ &\leq \uptheta^{4} \left( |\upalpha_{i}x_{i}^{-1} - \upalpha_{j}x_{j}^{-1} |_{\uptheta} + | \upalpha_{j}x_{j}^{-1} - \upalpha_{i}x_{j}^{-1} |_{\uptheta} \right), \end{align}\] where the factor of \(\require{upgreek} \uptheta^{4}\) comes from the infratriangle Inequality for \(\require{upgreek} N(\uptheta )=1\), Theorem 8. However, \(\require{upgreek} |\upalpha_{i}x_{i}^{-1} - \upalpha_{j}x_{j}^{-1} |_{\uptheta} = |\updelta_{i}- \updelta_{j} |_{\uptheta} \rightarrow 0\) and \[\require{upgreek} | \upalpha_{j}x_{j}^{-1} - \upalpha_{i}x_{j}^{-1} |_{\uptheta} \leq \frac{c\uptheta^{2}}{|\upalpha_{j}|_{\uptheta}} | \upalpha_{j} -\upalpha_{i}|_{\uptheta} \longrightarrow 0\] since \(\require{upgreek} |\upalpha_{j}|_{\uptheta}\) and the latter assumes at most five (non zero) values. (The factor \(\require{upgreek} \uptheta^{2}\) comes from Inframultiplicativity in the case \(\require{upgreek} N(\uptheta )=1\), Theorem 7.) Thus the sequence defining \(x^{-1}\) must be Cauchy. ◻

Proof. Since we have already shown that \(\require{upgreek} \mathcal{O}_{\uptheta}\) is a Marty multiring, it will be enough to show that every element \(\require{upgreek} x\in{\mathcal{O}}_{\uptheta}\) has an inverse (not necessarily unique). Denote by \(\hat{x}\) the corresponding class in \(\require{upgreek} \widehat{\mathcal{O}}_{\uptheta}\), which by Theorem 32 has an inverse \(\hat{y}\). Let \(\require{upgreek} y\in \mathcal{O}_{\uptheta}\) be any greedy Laurent series in the class of \(\hat{y}\) (guaranteed by Proposition 10) represented by a sequence of partial sums \(\{ y_{n}\}\). Then if \(\{ x_{n}\}\) is a sequence of partial sums for \(x\), we have \[\require{upgreek} x_{n}y_{n} = 1 + \upepsilon_{n}\] where \(\require{upgreek} \upepsilon_{n}\rightarrow 0\) in the \(\require{upgreek} \uptheta\)-adic infranorm. But \(\require{upgreek} \{ 1 + \upepsilon_{n}\}\) is a sequence of partial sums in the class of \(\require{upgreek} 1 \in \mathcal{O}_{\uptheta}\), hence \(1= xy\). ◻

Proof. First note that the constants of relative density and uniform discreteness of the translates \(\require{upgreek} \Upomega-\upalpha_{i}\) are equal to those of \(\require{upgreek} \Upomega\). Since \(\require{upgreek} \Upomega_{\hat{\upalpha}}\) is a nested union of the \(\require{upgreek} (\Upomega-\upalpha_{i})_{R}\), it is also uniformly discrete and relatively dense, having the same constants of relative density and uniform discreteness. By the characterization theorem of quasicrystals (see Theorem 9.1, (vi), of [12]), if \(\require{upgreek} \Uplambda\subset\mathbb{R}^{n}\) is relatively dense, then \(\require{upgreek} \Uplambda\) is a quasicrystal if and only if \(\require{upgreek} \Uplambda-\Uplambda\) is uniformly discrete. Thus, it is enough to show that \(\require{upgreek} \Upomega_{\hat{\upalpha}}- \Upomega_{\hat{\upalpha}}\) is uniformly discrete. But any element \(\require{upgreek} x-y\in \Upomega_{\hat{\upalpha}}- \Upomega_{\hat{\upalpha}}\) is contained in an approximation \[\require{upgreek} (\Upomega-\upalpha_{i})_{R} - (\Upomega-\upalpha_{i})_{R} \subset (\Upomega-\upalpha_{i} ) -(\Upomega-\upalpha_{i} ) = \Upomega-\Upomega.\] From this it follows that \(\require{upgreek} \Upomega_{\hat{\upalpha}}- \Upomega_{\hat{\upalpha}}\subset\Upomega-\Upomega\). Again by Theorem 9.1, (vi), of [12], since \(\require{upgreek} \Upomega\) is a quasicrystal, \(\require{upgreek} \Upomega-\Upomega\) is uniformly discrete, hence so is \(\require{upgreek} \Upomega_{\hat{\upalpha}}- \Upomega_{\hat{\upalpha}}\). We conclude that \(\require{upgreek} \Upomega_{\hat{\upalpha}}\) is a quasicrystal. ◻

Proof. Given \(R\leq S\) and \(\require{upgreek} \Upomega ' \in \mathcal{O}_{R}( \Upomega_{\hat{\upalpha}}) \cap \mathcal{O}_{S}( \Upomega_{\hat{\upbeta}})\), then \(\require{upgreek} \mathcal{O}_{S}( \Upomega')\subset \mathcal{O}_{R}( \Upomega_{\hat{\upalpha}}) \cap \mathcal{O}_{S}( \Upomega_{\hat{\upbeta}})\), since \(\require{upgreek} \Upomega''\in \mathcal{O}_{S}( \Upomega')\) if and only if \(\require{upgreek} \Upomega''_{S}= \Upomega'_{S}\), which implies 1) \(\require{upgreek} \Upomega''_{S}=( \Upomega_{\hat{\upbeta}} )_{S}\), i.e., \(\require{upgreek} \Upomega'' \in \mathcal{O}_{S}( \Upomega_{\hat{\upbeta}})\) and 2) \(\require{upgreek} \Upomega''_{R} = \Upomega'_{R} = ( \Upomega_{\hat{\upalpha}})_{R}\), implying \(\require{upgreek} \Upomega'' \in \mathcal{O}_{R}( \Upomega_{\hat{\upalpha}})\). ◻

Proof. The topology of the uniform structure, restricted to the subset \(\require{upgreek} \{ \upalpha +\Upomega\; :\;\; \upalpha\in\Upomega\}\), gives the quasicrystal topology. Indeed, \(\require{upgreek} \Upomega+\Upomega\) is also a quasicrystal ([12], Corollary 6.8, (ii)) and for \(\require{upgreek} \upalpha, \upbeta \in \Upomega\) we may view each of \(\require{upgreek} \Upomega':= \upalpha + \Upomega\), \(\require{upgreek} \Upomega'': = \upbeta + \Upomega\) in the quasicrystal \(\require{upgreek} \upalpha +\upbeta + \Upomega+\Upomega\). Then, for \(\require{upgreek} \upvarepsilon<\) constant of uniform discreteness of \(\require{upgreek} \Upomega +\Upomega\), \(\require{upgreek} (\Upomega', \Upomega'')\in {\sf U}_{\upvarepsilon,R}\) if and only if \(\require{upgreek} \Upomega_{R}' = \Upomega_{R}''\). In particular, the quasicrystal completion \(\contour{black}{\color{white}\boldsymbol{\Omega}}_{\;{\sf qc}}\) is a closed subset of \(\require{upgreek} \hat{\mathbb{S}}_{\Upomega}\), hence is compact. On the other hand, if \(\require{upgreek} \Upomega'\not\in \mathcal{O}_{R}( \Upomega_{\hat{\upalpha}})\) i.e.\(\require{upgreek} \Upomega_{R}'\not=( \Upomega_{\hat{\upalpha}})_{R}\), then \(\require{upgreek} \mathcal{O}_{R}(\Upomega') \subset \mathcal{O}_{R}( \Upomega_{\hat{\upalpha}})^{\complement}\). Thus \(\require{upgreek} \mathcal{O}_{R}( \Upomega_{\hat{\upalpha}})\) is also closed, hence \(\contour{black}{\color{white}\boldsymbol{\Omega}}_{\;{\sf qc}}\) is totally disconnected. ◻

Proof. Suppose that \(\require{upgreek} \Upomega\) is defined using \(W\) and let \[\require{upgreek} \begin{align} \label{completelistinofOmega} \Upomega_{R}=\pm \{ 0=\upalpha_{0} <\upalpha_{1} < \cdots < \upalpha_{k} \} . \end{align}\tag{56}\] In what follows, to simplify certain formulas we will write \(\require{upgreek} \upalpha_{k+1}:= R\) (abusively, since \(R\) need not belong to \(\require{upgreek} \Upomega\)). In particular,

1. By definition of \(\require{upgreek} \Upomega\) and \(\require{upgreek} \Upomega_{R}\), for all \(i=0,\dots ,k\), \(\require{upgreek} \upalpha'_{i}\in (-\uptheta^{-x},\uptheta^{-x})\) and \(\require{upgreek} \upalpha_{i}\in (0,R)\).

2. Since the ordered list (56 ) gives a complete account of the elements of \(\require{upgreek} \Upomega_{R}\), there is no \(\require{upgreek} 0<\upgamma\in \mathcal{O}_{K}\) such that for some \(i\in \{0,\dots ,k\}\), \[\require{upgreek} \upgamma < \upalpha_{i+1} -\upalpha_{i}\] and \[\require{upgreek} \upalpha_{i}+\upgamma \in \Upomega.\]

Item 2.implies that for all \(0\leq i\leq k\) and \(\require{upgreek} 0<\upgamma< \upalpha_{i+1}-\upalpha_{i}\), \(\require{upgreek} \upalpha_{i}+\upgamma\not\in\Upomega\), i.e., \[\require{upgreek} \upgamma' \not\in W-\upalpha'_{i} = (-\uptheta^{-x} -\upalpha'_{i} , \uptheta^{-x} -\upalpha'_{i} ).\] Consider the set \[\require{upgreek} \Upsilon = \big\{0\not= \upgamma \in\mathcal{O}_{K} \; : \;\; 0<\upgamma< \upalpha_{i+1}-\upalpha_{i}\text{ for some }i\leq k \big\}\subset (0,\upalpha),\] where \[\require{upgreek} \upalpha:= \max_{i\leq k}(\upalpha_{i+1}-\upalpha_{i}) .\] The set of conjugates \(\Upsilon'\) is a subset of the model set in \(\{ 0\}\times\mathbb{R}\) defined using the window \(\require{upgreek} (0,\upalpha)\) in \(\mathbb{R}\times\{0\}\), hence \(\Upsilon'\) is uniformly discrete. Therefore, we may choose \(\require{upgreek} \updelta>0\) small so that that for all \(\require{upgreek} \upbeta\in\mathcal{O}_{K}\) with \(\require{upgreek} |\upbeta'|<\updelta\),

1. \(\require{upgreek} \upbeta'\pm\upalpha_{i}' \in (-\uptheta^{-x},\uptheta^{-x})\): that is to say, \(\require{upgreek} \upbeta \pm \upalpha_{i}\in\Upomega\). This follows since the set of \(\require{upgreek} \pm \upalpha_{i}'\in (-\uptheta^{-x},\uptheta^{-x})\) is a finite set.

2. For \(\require{upgreek} \upgamma\in\Upsilon\) and all \(i\in \{ 0,\dots ,k\}\) for which \(\require{upgreek} 0<\upgamma< \upalpha_{i+1}-\upalpha_{i}\), \[\require{upgreek} \begin{align} \label{Statementb} \upgamma' \not\in (-\uptheta^{-x}\-\upalpha'_{i} -\upbeta', \uptheta^{-x} -\upalpha'_{i} -\upbeta'). \end{align}\tag{57}\] Indeed, we already have, for some \(i\), \[\require{upgreek} \upgamma' \not\in (-\uptheta^{-x}-\upalpha'_{i} , \uptheta^{-x} -\upalpha'_{i} )\] and since \(\Upsilon'\) is uniformly discrete, we can find a uniform lower bound for the distance of \(\require{upgreek} \upgamma'\) to the interval \(\require{upgreek} (-\uptheta^{-x}-\upalpha'_{i} , \uptheta^{-x} -\upalpha'_{i} )\); hence choosing \(\require{upgreek} \updelta<\) the lower bounds for all \(i\) implies (57 ) for any \(\require{upgreek} |\upbeta'|<\updelta\). In particular, \[\require{upgreek} \begin{align} \label{thenoninclusion}\upbeta+ \upalpha_{i}+\upgamma\not\in \Upomega \end{align}\tag{58}\] for \(\require{upgreek} 0<\upgamma < \upalpha_{i+1}-\upalpha_{i}\) and \(i\leq k\).

Condition a.implies that \(\require{upgreek} \pm\upalpha_{i} = (\pm\upalpha_{i} +\upbeta)-\upbeta\in (\Upomega-\upbeta)_{R}\). Moreover, for \(\require{upgreek} \updelta\) sufficiently small, \(\require{upgreek} \upbeta\in \Upomega\), so \(\require{upgreek} 0\in (\Upomega-\upbeta)_{R}\). This gives the inclusion \(\require{upgreek} \Upomega_{R}\subset (\Upomega-\upbeta)_{R}\). Condition b. gives the opposite inclusion: indeed, suppose there exists \[\require{upgreek} x=\upalpha-\upbeta\in (\Upomega-\upbeta)_{R}\setminus \Upomega_{R},\quad \upalpha\in\Upomega.\] We assume without loss of generality that \(x>0\); the case \(x<0\) can be handled using identical arguments. Then, \(x\) must fall within the gaps in \((0,R)\) demarcated by the \(\require{upgreek} \upalpha_{i}\). Thus there exists a \(\require{upgreek} \upgamma>0\) with \(\require{upgreek} \upgamma<\upalpha_{i+1}-\upalpha_{i}\) and for which \[\require{upgreek} x = \upalpha -\upbeta =\upalpha_{i}+\upgamma.\] But then, \[\require{upgreek} \upalpha= \upbeta + \upalpha_{i}+\upgamma\in\Upomega ,\] contradicting (58 ). We conclude that \(\require{upgreek} (\Upomega-\upbeta)_{R} =\Upomega_{R}\), hence \(\require{upgreek} \upbeta\in {\rm Per}_{R}(\Upomega )\). The set of \(\require{upgreek} \upbeta\) with \(\require{upgreek} |\upbeta'|<\updelta\) is a quasicrystal, hence relatively dense, hence \(\require{upgreek} {\rm Per}_{R}(\Upomega )\) is also relatively dense. This proves repetitivity. On the other hand, if \(\require{upgreek} \Upomega\) is defined with \(\overline{W}\) but cannot be defined with \(W\), there exists \(\require{upgreek} \upalpha\in\Upomega\) with \(\require{upgreek} \upalpha'=\uptheta^{-x}\). The set \(\require{upgreek} \Upomega_{R}\) for \(\require{upgreek} R>|\upalpha|\) cannot occur anywhere else in \(\require{upgreek} \Upomega\): that is, there cannot exist \(\require{upgreek} \upbeta\) with \(\require{upgreek} (\Upomega-\upbeta)_{R}= \Upomega_{R}\). Indeed, suppose on the contrary there exists \(\require{upgreek} \upbeta\not=0\) with \(\require{upgreek} (\Upomega-\upbeta)_{R}= \Upomega_{R}\). Since \(\require{upgreek} \upalpha\in \Upomega_{R}\), there exists \(\require{upgreek} \upgamma\in\Upomega\) with \(\require{upgreek} \upgamma' = \upbeta'+\upalpha' = \upbeta'+\uptheta^{-x}\), which can happen only if \(\require{upgreek} \upbeta'<0\). But then, since we also have \(\require{upgreek} -\upalpha\in \Upomega_{R}\), there exists \(\require{upgreek} \upeta\in \Upomega\) with \(\require{upgreek} \upeta' = \upbeta'-\upalpha' = \upbeta'-\uptheta^{-x}\not\in \overline{W}\), contradiction. In particular, for such \(R\), \(\require{upgreek} {\rm Per}_{R}(\Upomega ) = \{0\}\), so \(\require{upgreek} \Upomega\) is not repetitive. ◻

Proof. Since \(\require{upgreek} \Upomega\) is relatively dense, the set of differences of consecutive elements is uniformly bounded. Thus, if there were an infinite set of distinct differences, we may find a non trivial convergent sequence \(\require{upgreek} \updelta_{n}=\upbeta_{n}-\upalpha_{n}\in \Upomega-\Upomega\). But since \(\require{upgreek} \Upomega\) is a quasicrystal, \(\require{upgreek} \Upomega-\Upomega\) is uniformly discrete, which is contradicted by the existence of the sequence \(\require{upgreek} \updelta_{n}\). ◻

Proof. By Lemma 14, the 1-dimensional quasicrystal \(\require{upgreek} \Upomega\) may be described as a bi-infinite word on a finite alphabet, where the letters of the alphabet index consecutive differences of elements of \(\require{upgreek} \Upomega\). For \(\require{upgreek} \Upomega\) defined as a model set using \(W\), by Lemma 13, \(\require{upgreek} \Upomega\) is repetitive, which implies that the bi-infinite word indexing \(\require{upgreek} \Upomega\) is repetitive. The minimality of \(\contour{black}{\color{white}\boldsymbol{\Omega}}_{\;{\sf qc}}\) then follows from Proposition 4.3 on page 78 of [13]. ◻

Proof. Item 1.follows from Corollary 11, since the latter shows that every point is a limit point, hence \(\contour{black}{\color{white}\boldsymbol{\Omega}}_{\;{\sf qc}}\) is perfect. As for item 2., since \(\require{upgreek} \Upomega\) is not repetitive, for any \(\require{upgreek} \upalpha\in\Upomega\), there can be no non-trivial sequence \(\require{upgreek} \upalpha_{i} +\Upomega\) converging to \(\require{upgreek} \upalpha+\Upomega\). Thus, the points of \(\require{upgreek} \Upomega\) are isolated. ◻

Proof. First suppose that \(\require{upgreek} \Upomega\) is defined by \(W\), hence is repetitive. Let \[\require{upgreek} \Upomega_{\hat{\upalpha}} := \lim \left( \upalpha_{i} +\Upomega\right) \in\contour{black}{\color{white}\boldsymbol{\Omega}}_{\;{\sf qc}} .\] Since \(\require{upgreek} \upalpha_{i} +\Upomega\subset\mathcal{O}_{K}\) for all \(i\), \(\require{upgreek} \Upomega_{\hat{\upalpha}}\subset \mathcal{O}_{K}\). By Lemma 4.1 of [29], \[\require{upgreek} \bigcap_{t\in \Upomega_{\hat{\upalpha}}} ( t'-\overline{W}) = \{ c_{\hat{\upalpha}}\}\] for some \(\require{upgreek} c_{\hat{\upalpha}}\in \mathbb{R}\). Note that for \(\require{upgreek} \Upomega_{\upalpha}:=\upalpha +\Upomega\in\contour{black}{\color{white}\boldsymbol{\Omega}}_{\;{\sf qc}}\), since \(\require{upgreek} \upalpha' \in t'- W\) for all \(\require{upgreek} t\in \upalpha +\Upomega\), \(\require{upgreek} c_{\upalpha} = \upalpha'\in W\). Therefore, the association \[\require{upgreek} \contour{black}{\color{white}\boldsymbol{\Omega}}_{\;{\sf qc}}\longrightarrow \mathbb{R},\quad \Upomega_{\hat{\upalpha}}\longmapsto c_{\hat{\upalpha}}\] extends the conjugation map. We must show that this association is continuous; once we have shown continuity it will follow that \(\require{upgreek} c_{\hat{\upalpha}}\in \overline{W}\). Let \(\require{upgreek} V'\ni c_{\hat{\upalpha}}\) be an open neighborhood; then \[\require{upgreek} \bigcap_{t\in \Upomega_{\hat{\upalpha}}} [( t'-\overline{W}) \setminus V' ] =\emptyset .\] The set of conjugates \(t'\) for \(\require{upgreek} t\in \Upomega_{\hat{\upalpha}}\) belongs to the relatively compact set \(\require{upgreek} \bigcup (\upalpha_{i}' +W)\). Thus, there exists a compact set such containing all of the closed sets \(( t'-\overline{W}) \setminus V'\), and we may therefore deduce that there is a finite subset \(\require{upgreek} F\subset\Upomega_{\hat{\upalpha}}\) such that \[\bigcap_{t\in F} [( t'-\overline{W}) \setminus V' ] =\emptyset .\] This shows that there is an \(R>0\) such that \[\require{upgreek} \bigcap_{t\in (\Upomega_{\hat{\upalpha}})_{R}} [( t'-\overline{W}) \setminus V' ] =\emptyset\] This implies \[\require{upgreek} \bigcap_{t\in(\Upomega_{\hat{\upalpha}})_{R}} ( t'-\overline{W}) \subset V' .\] Thus given any \(\require{upgreek} \Upomega_{\hat{\upbeta}}\in\contour{black}{\color{white}\boldsymbol{\Omega}}_{\;{\sf qc}}\) for which \[\require{upgreek} (\Upomega_{\hat{\upbeta}})_{R} = (\Upomega_{\hat{\upalpha}})_{R} ,\] we have then \(\require{upgreek} c_{\hat{\upbeta}} \in V'\). This proves continuity when \(\require{upgreek} \Upomega\) is defined by \(W\). If \(\require{upgreek} \Upomega\) is defined by \(\overline{W}\), it is not repetitive, however the above argument shows that the conjugation map on the Cantor set \[\contour{black}{\color{white}\boldsymbol{\Omega}}_{\;{\sf qc}}\setminus \{\text{i\small solated points}\}\] is continuous. In particular, the conjugation map is thus continuous on all of \(\require{upgreek} \hat{\Upomega}\). ◻

Proof. The result is trivial if \(\require{upgreek} \hat{\upalpha} =\upalpha \in\mathcal{O}_{K}\). Now suppose that \(\require{upgreek} \upbeta:=\hat{\upalpha}'\in \mathcal{O}_{K}\) but the defining sequence \(\require{upgreek} \{ \upalpha_{i}\}\) is not trivial. Then the sequence of conjugates \(\require{upgreek} \{ \upalpha'_{i}\}\) must eventually converge monotonically: otherwise, the sequence of windows \(\require{upgreek} W_{\upalpha_{i}}\) will oscillate between containing or not containing \(\require{upgreek} \upbeta\), which will mean that for \(\require{upgreek} R>|\upbeta'|\), the sets \(\require{upgreek} \Upomega_{\upalpha_{i},R}\) will oscillate between containing or not containing \(\require{upgreek} \upbeta'\). Hence the convergence condition (55 ) cannot be realized for \(\require{upgreek} R>|\upbeta'|\). Depending on whether the convergence is monotone increasing or decreasing, the \(R\)-balls in (55 ) will contain \(\require{upgreek} r+ \hat{\upalpha}'\) but not \(\require{upgreek} s + \hat{\upalpha}'\), or the contrary. Otherwise, if \(\require{upgreek} \hat{\upalpha}' \not\in \mathcal{O}_{K}\), the limit is allowed to be oscillatory on windows, since \(\require{upgreek} \hat{\upalpha}'\) cannot belong to any quasicrystal associated to the lattice \(\mathcal{O}_{K}\). Thus the model set having window \(\require{upgreek} W_{\hat{\upalpha}}\) coincides with \(\require{upgreek} \Upomega_{\hat{\upalpha}}\) . ◻

Proof. We first indicate how to define (?? ) on all \(\require{upgreek} \contour{black}{\color{white}\boldsymbol{\mathfrak{m}}}_{\uptheta}\), extending the association \(\require{upgreek} \upalpha\mapsto \upalpha +\mathfrak{m}\) for \(\require{upgreek} \upalpha\in \mathfrak{m}\). Let \(\require{upgreek} x= \sum_{m}^{\infty} b_{i}\uptheta^{i}\in \contour{black}{\color{white}\boldsymbol{\mathfrak{m}}}_{\uptheta}\). If the conjugate of \(x\) is not in \(\mathcal{O}_{K}\), we associate to \(x\) the limit quasicrystal with window \(W=(-1+x', 1+x')\) obtained from the sequence \(\{ x_{i} + \mathfrak{m}\}\) where \(\{ x_{i}\}\) is a sequence of partial sums. We note that there can be no other possible choice, as including either of the boundary points in the window does not change the model set, since \(\pm 1+x'\not\in \mathcal{O}_{K}\). Moreover, this definition clearly does not depend on the choice of the sequence of partial sums \(\{ x_{i}\}\in x\), since \(x'\) does not depend on this choice. Now suppose that \(x\) is either of \(\require{upgreek} \upalpha_{1}\), \(\require{upgreek} \upalpha_{2}\) as in Proposition 13, in which \(\require{upgreek} \uppi (\upalpha_{1}) =\uppi (\upalpha_{2})=\upalpha \in \mathcal{O}_{K}\subset \widehat{\mathcal{O}}_{K}\). One of the two, say \(\require{upgreek} \upalpha_{1}\), is such that its sequence of partial sums \(\require{upgreek} \{ \upalpha_{1,i}\}\), satisfies \(\require{upgreek} \upalpha_{1,i}'\) is (eventually) monotone decreasing, and then \(\require{upgreek} \upalpha_{2}\), is such that its sequence of partial sums \(\require{upgreek} \{ \upalpha_{2,i}\}\), satisfies \(\require{upgreek} \upalpha_{2,i}'\) is (eventually) monotone increasing. We assign to \(\require{upgreek} \upalpha_{1}\) the limit quasicrystal \(\require{upgreek} \mathfrak{m}_{\upalpha_{1}}\) having window \(\require{upgreek} W_{R}=(-1+\upalpha', 1+\upalpha']\) and to \(\require{upgreek} \upalpha_{2}\) the limit quasicrystal \(\require{upgreek} \mathfrak{m}_{\upalpha_{2}}\) having window \(\require{upgreek} W_{L}=[-1+\upalpha', 1+\upalpha')\). This completes the definition of (?? ).

The topologies on both \(\require{upgreek} \contour{black}{\color{white}\boldsymbol{\mathfrak{m}}}_{\uptheta}\) and \(\contour{black}{\color{white}\boldsymbol{\mathfrak{m}}}_{\;{\sf qc}}\) are the weak topologies: two series in \(\require{upgreek} \contour{black}{\color{white}\boldsymbol{\mathfrak{m}}}_{\uptheta}\) are close if they agree up to a very high power of \(\require{upgreek} \uptheta\) and two quasicrystals \(\mathfrak{m}_{1}, \mathfrak{m}_{2}\) in the completion are close if they agree on a large interval \([-R,R]\).

The two convergence conditions are equivalent:

Let \(x_{i}\rightarrow x\) in \(\require{upgreek} \contour{black}{\color{white}\boldsymbol{\mathfrak{m}}}_{\uptheta}\). We note that If \(\require{upgreek} x=\upalpha\in\mathcal{O}_{K}\) the result is trivial since the \(\require{upgreek} \uptheta\)-adic topology dictates that eventually \(x_{i}=x\). Thus we may assume that \(x\not\in \mathcal{O}_{K}\). We divide the argument according to whether \(x'\in \mathcal{O}_{K}\) or not. Suppose first that \(x'\not\in\mathcal{O}_{K}\). Denote by \(W_{x_{i}}, W_{x}\) the windows defining \(\mathfrak{m}_{x_{i}}\), \(\mathfrak{m}_{x}\). By the hypothesis \(x'\not\in\mathcal{O}_{K}\), we may take \(W_{x} = (-1+x',1+x')\). The points of \((\mathfrak{m}_{x})_{R}\), which are finite in number, have conjugates belonging to a compact set \(K\subset W_{x}\). Therefore, for large \(i\), \((\mathfrak{m}_{x})_{R}\subset (\mathfrak{m}_{x_{i}})_{R}\). Suppose that for arbitrarily large \(i\), there exists \(\require{upgreek} \upalpha_{i}\in (\mathfrak{m}_{x_{i}})_{R}\setminus (\mathfrak{m}_{x})_{R}\). Then, after passing to subsequences if necessary, the sequence of conjugates \(\require{upgreek} \upalpha'_{i}\) converges (to either \(-1+x'\) or \(1+x'\)) and \(\require{upgreek} \upalpha_{i}\) converges to some \(y\in [-R,R]\). But this is impossible: for if we write \(\require{upgreek} \upalpha_{i}= m_{i} + n_{i}\uptheta\), \(\require{upgreek} \upalpha_{i}' = m_{i}-n_{i}\uptheta^{-1}\rightarrow \pm 1+x'\not\in \mathcal{O}_{K}\) implies that both of \(m_{i}\), \(n_{i}\) tend to \(\pm\infty\) with the same sign. But then we could not have \(\require{upgreek} m_{i} + n_{i}\uptheta\) convergent. Therefore, eventually \((\mathfrak{m}_{x})_{R}=(\mathfrak{m}_{x_{i}})_{R}\) which gives convergence in this case.

If \(x'\in \mathcal{O}_{K}\) but \(x\not\in \mathcal{O}_{K}\), then per Proposition 13, \(x\) either has a tail of the form \(\require{upgreek} a(\uptheta^{2m} + \uptheta^{2m+2}+ \cdots)\) or \(\require{upgreek} a(\uptheta^{2m+1} + \uptheta^{2m+3}+ \cdots)\). Suppose the former, so that the window \(W_{x}=[-1+x', 1+x')\). If \(x_{i}\rightarrow x\) then \(x_{i}'\rightarrow x'\) from below. Indeed, the difference \(x'-x_{i}>0\) is positive since it must either start with odd power \(\require{upgreek} -c_{2m_{i}+1}(-\uptheta)^{-(2m_{i}+1)} = c_{m_{i}}\uptheta^{-(2m_{i}+1)}\) or an even (positive) power \(\require{upgreek} (a-c_{2m_{i}})\uptheta^{-2m_{i}} \geq 0\), where the \(c\)’s are the relevant coefficients of \(x_{i}\). It follows then that \(W_{x_{i}}\rightarrow W_{x}\). Once again, in this case we have for large \(i\), \((\mathfrak{m}_{x})_{R}\subset (\mathfrak{m}_{x_{i}})_{R}\). If there were \(\require{upgreek} \upalpha_{i}\in (\mathfrak{m}_{x_{i}})_{R}\setminus (\mathfrak{m}_{x})_{R}\), after passing to a subsequence, \(\require{upgreek} \upalpha_{i}\) converges to some \(r\in [-R,R]\). At the same time, we have \(\require{upgreek} \upalpha'_{i}\rightarrow \pm 1+ x'\), and since \(\require{upgreek} \{\upalpha'_{i} = m_{i}-n_{i}\uptheta^{-1}\}\) is not the constant sequence, once again we must have \(m_{i}, n_{i}\rightarrow \pm\infty\) with the same signs, violating the convergence of \(\require{upgreek} \upalpha_{i}\). The proves that convergence \(\require{upgreek} \uptheta\)-adic convergence implies quasicrystal convergence.

Now suppose that \(\mathfrak{m}_{x_{i}} \rightarrow \mathfrak{m}_{x}\) in \(\contour{black}{\color{white}\boldsymbol{\mathfrak{m}}}_{\;{\sf qc}}\): then we have convergence of the defining windows \(W_{x_{i}}\rightarrow W_{x}\) implying convergence of the conjugates \(x_{i}'\rightarrow x'\). If \(x'\not\in\mathcal{O}_{K}\), we must have \(x_{i}\rightarrow x\) in the \(\require{upgreek} \uptheta\)-adic topology, since the conjugation map \(\require{upgreek} \mathcal{O}_{\uptheta}\rightarrow \mathbb{R}\) is continuous with single pre-images over \(\mathbb{R}\setminus \mathcal{O}_{K}\). On the other hand, if \(\require{upgreek} x' =\upalpha'\in\mathcal{O}_{K}\), there are three pre-images in \(\require{upgreek} \mathcal{O}_{\uptheta}\): \(\require{upgreek} \upalpha,\upalpha_{1}\) and \(\require{upgreek} \upalpha_{2}\). We may suppose that \(\require{upgreek} x\not= \upalpha\) so that \(x\) is say \(\require{upgreek} \upalpha_{1}\) with window \(\require{upgreek} [-1+\upalpha' , 1+\upalpha')\), wherein the tail of \(\require{upgreek} \upalpha_{1}\) consists of even powers, all with coefficient \(a\). Therefore, the windows \(W_{x_{i}}\) must converge to \(W_{x}\) monotonically from below. Therefore \(x'_{i}\nearrow x'\), and for this to be true, on the level of partial sums we must have \(x_{i}\rightarrow x\) in the weak topology. This completes the proof of the Claim. Moreover, the map (?? ) is a bijection: indeed, if \(\require{upgreek} \upalpha, \upbeta\in \contour{black}{\color{white}\boldsymbol{\mathfrak{m}}}_{\uptheta}\) have the same image in \(\contour{black}{\color{white}\boldsymbol{\mathfrak{m}}}_{\;{\sf qc}}\), then \(\require{upgreek} \upalpha'= \upbeta'\). If their common conjugate value is not in \(\mathcal{O}_{K}\), \(\require{upgreek} \upalpha=\upbeta\); otherwise, the definition of (?? ) takes the three possible elements in \(\require{upgreek} \contour{black}{\color{white}\boldsymbol{\mathfrak{m}}}_{\uptheta}\) to three distinct possible limit quasicrystals. Thus the map (?? ) is a homeomorphism.

If the conjugate of \(\require{upgreek} x=\sum b_{i}\uptheta^{i}\) is not in \(\mathcal{O}_{K}\), we associate to \(x\) the limit quasicrystal with window \(W=(-1+x', 1+x')\) obtained from the sequence \(\{ x_{i} + \mathfrak{m}\}\) where \(x_{i}\) is a sequence of partial sums. Note since \(r=x'\not\in \mathcal{O}_{K}\), there will be another series involving \(T^{-1}\) having conjugate \(r\) (by Theorem 25), and this series will produce the same quasicrystal limit. If \(\require{upgreek} \upalpha\in \mathcal{O}_{K,+}^{0}\), we associate to \(\require{upgreek} \upalpha\) the quasicristal \(\require{upgreek} \mathfrak{m}_{\upalpha}\), to the series \(\require{upgreek} \upalpha_{1}=\sum b_{i}\uptheta^{i}\) with \(\require{upgreek} \upalpha_{1}'=\upalpha'\) the quasicrystal with window \(W_{L}\) and to the series \(\require{upgreek} \upalpha_{2}=-T^{-1}\sum c_{i}\uptheta^{i}\) with \(\require{upgreek} \upalpha_{2}'=\upalpha'\) the quasicrystal with window \(W_{R}\). When \(\require{upgreek} \upalpha\in \mathcal{O}_{K,+}^{1}\), the series with \(T^{-1}\) term is mapped to the quasicrystal with window \(W_{L}\) and the series without \(T^{-1}\) is mapped to the quasicrystal with window \(W_{R}\). Note that this map is two-to-one over points whose conjugate is not in \(\mathcal{O}_{K}\) and otherwise is bijective. This defines the map (?? ); its continuity follows from an argument similar to that presented in the case \(\require{upgreek} N(\uptheta )=-1\). ◻

Proof. The proof is essentially an addendum to the proof of Theorem 35, rounded out by the following observations. The \(\require{upgreek} \uptheta\)-adic completion \(\require{upgreek} \contour{black}{\color{white}\boldsymbol{A}}_{\uptheta}\) contains extra points coming from Laurent series whose conjugates are in \(\mathcal{O}_{K}\). Thus the map (?? ) is not onto. Moreover, it is not bicontinuous since \(\contour{black}{\color{white}\boldsymbol{A}}_{\;{\rm qc}}\) is Stone whereas \(\require{upgreek} \contour{black}{\color{white}\boldsymbol{A}}_{\uptheta}\) is Cantor. Similar remarks apply to (?? ). ◻