On small bases which admit countably many expansions

Abstract: Let $q\in(1,2)$ and $x\in[0,\frac1{q-1}]$. We say that a sequence $(\epsilon_i){i=1}^{\infty}\in{0,1}^{\mathbb{N}}$ is an expansion of $x$ in base $q$ (or a $q$-expansion) if x=\sum{i=1}^{\infty}\epsilon_iq^{-i}. Let $\mathcal{B}{\aleph{0}}$ denote the set of $q$ for which there exists $x$ with exactly $\aleph_{0}$ expansions in base $q$. In \cite{EHJ} it was shown that $\min\mathcal{B}{\aleph{0}}=\frac{1+\sqrt{5}}{2}.$ In this paper we show that the smallest element of $\mathcal{B}{\aleph{0}}$ strictly greater than $\frac{1+\sqrt{5}}{2}$ is $q_{\aleph_{0}}\approx1.64541$, the appropriate root of $x^6=x^4+x^3+2x^2+x+1$. This leads to a full dichotomy for the number of possible $q$-expansions for $q\in (\frac{1+\sqrt{5}}{2},q_{\aleph_{0}})$. We also prove some general results regarding $\mathcal{B}{\aleph{0}}\cap[\frac{1+\sqrt{5}}{2},q_{f}],$ where $q_{f}\approx 1.75488$ is the appropriate root of $x^{3}=2x^{2}-x+1.$ Moreover, the techniques developed in this paper imply that if $x\in [0,\frac{1}{q-1}]$ has uncountably many $q$-expansions then the set of $q$-expansions for $x$ has cardinality equal to that of the continuum, this proves that the continuum hypothesis holds when restricted to this specific case.