Infinite closed subset of $S^1$ such that the squaring map is a bijection?

Question

Is there an infinite closed subset $X$ of the unit circle in $\mathbb C$ such that the squaring map induces a bijection from $X$ to itself?

Answer 1

Let us think of $S^1$ as $\mathbb{R}/\mathbb{Z}$, so we want an infinite closed subset $X$ on which multiplication by $2$ is a bijection. Suppose you have such an $X$; write $T:X\to X$ for the multiplication by $2$ map. Each element $x\in X$ determines a biinfinite binary expansion $f_x:\mathbb{Z}\to\{0,1\}$, such that $T^n(x)=\sum_{k=1}^\infty f_x(k+n)2^{-k}$ for each $n\in \mathbb{Z}$. Say that a finite string of $0$s and $1$s is admissible if it appears as a sequence of consecutive values of some $f_x$ (i.e., if it appears as a sequence of consecutive digits in the binary expansion of some element of $X$). For each $n$, let $A_n\subseteq\{0,1\}^n$ consist of those sequences $s$ such that both $0^\frown s$ and $1^\frown s$ are admissible (where $^\frown$ is string concatenation). If $A_n$ is empty for some $n$, that means that given a sequence of $n$ consecutive digits in any $f_x$, all of the preceding digits are uniquely determined. By pigeonhole, for each $x$, some sequence of $n$ digits must appear infinitely often in the restriction of $f_x$ to $\mathbb{N}$, and it follows that every $f_x$ is periodic. Furthermore, there is a uniform bound on the periods of all the $f_x$ (because if some particular $s\in\{0,1\}^n$ appears infinitely often in $f_x|_\mathbb{N}$, that determines the period of $f_x$, and there are only finitely many different such $s$). So there are only finitely many different $f_x$, so $X$ is finite. This is a contradiction.

Thus each $A_n$ is nonempty. By König's lemma, it follows that there exists an infinite string $s:\mathbb{N}\to\{0,1\}$ such that every initial segment of $s$ is in the appropriate $A_n$. But then since $X$ is closed, the numbers $y_0$ and $y_1$ whose binary expansions are $0^\frown s$ and $1^\frown s$, respectively, are in $X$. Since $T(y_0)=T(y_1)$, this is a contradiction.

Thus no such $X$ exists.

Answer 2

For a small $\theta$, close the set $\{e^{ix} : x \in (-\theta, \theta)\}$ under square roots to get an open set $U_{\theta}$. If $\theta$ is small, the measure of $U_{\theta}$ is small. So we can take $A$ to be the complement of $U_{\theta}$.

Sorry, this only gives that $A$ is closed under squaring. Anyway, if $A$ is also required to be closed under square roots, then there is no such proper subset. This is because $(1, 0)$ is not an accumulation point of $A$ and therefore for some $\theta >0$, $\{e^{ix} : x \in (-\theta, \theta)\}$ is disjoint with $A$ and closing this under squaring gives the whole circle.