Is a intersection of chain of non enumerable dense sets non enumerable?

Question

Let $M_1\supset M_2\supset \cdots$ sets such that each $M_k$ is a non enumerable set, all of then open and dense in $S^1$ and such that $Leb(M_k)\to 0$ when $k\to \infty$, where $Leb$ denote the lebesgue measure in $S¹$. Is the intersection non enumerable in $S^1$?

An Example:

Let $x_0\in S^1$, and $\epsilon>0$. Consider a irrational rotation $R_{\alpha}$. So call $M_1$ the set $$ M_1=\bigcup_{n=1}^{\infty}\widehat{(-\epsilon,\epsilon)_n} $$ where $\widehat{(-\epsilon,\epsilon)_n}$ means the interval of length $\dfrac{\epsilon}{10^n}$ centered in $R^n_{\alpha}x_0.$ Consider the sequênce $\epsilon_k=\dfrac{1}{2^{k}}$, and put $$ M_k=\bigcup_{n=1}^{\infty}\widehat{(-\epsilon_k,\epsilon_k)_n}. $$ So each $M_k$ is non enumerable, open and dense in $S¹$. Also $M_1\supset M_2\supset \cdots$.

Answer 1

Not necessarily. For example, suppose $M_n$ is the union of an interval of length $1/n$ and a dense countable set.

For a more interesting question, you might want to assume $M_n$ has uncountable intersection with every nonempty open interval.

EDIT: If you assume each $M_n$ is open, then the Baire Category Theorem says the intersection is a dense $G_\delta$ in $S^1$. A dense $G_\delta$ set $G$ in $S^1$ (or any complete metric space without isolated points) is uncountable, because if it was countable you could intersect it with another sequence of dense open sets (namely $S^1 \backslash \{x\}$ for $x \in G$) and get the empty set.

Answer 2

This is a collection of various "standard" facts that add up to solve the question:

• Every $G_\delta$ subspace of a complete separable metric space is homeomorphic to a complete separable metric space (i.e. the subspace is completely metrizable.)

• The intersection of countably many dense open sets in a complete metric space is dense in the original space (Baire's theorem).

• No subset of $S^1$ that is dense in $S^1$ has any isolated points in the subspace topology.

Combining these:

• A dense $G_\delta$ subspace of $S^1$ will be homeomorphic to a complete separable metric space with no isolated points.

• So a dense $G_\delta$ subset of $S^1$ will contain a Cantor set, and in particular its cardinality will be the same as $S^1$.