More generalized: Do uncountable subspaces of separable metric spaces always touch a point of their dense subspaces? (If $X$ is a separable metric space, $S \subseteq X$ countable with $\mathrm{cl}(S)=X$, and $A \subseteq X$ uncountable, is $S \cap \mathrm{cl}(A) \neq \emptyset$?)
I just tried to proof this.
This is what I thought: Assume $S = \bigcap_{i\in \mathbb{N}} O_i$, an intersection of open sets. So $S^c = \bigcup_{i \in \mathbb{N}} O_i^c$, a countable union of closed sets. For any $i \in \mathbb{N}$, the closed set $O_i^c$ can't be uncountable, for otherwise it would touch a rational point in $(0,1)$, this can't be: $O_i^c$ is closed and doesn't meet $S$. Therefore, $S^c$ is countable, a countable union of countable sets. But $S$ surely is uncountable, so the assumption was false.
Now, this relies on the assumption that any uncountable set in $(0,1)$ touches a rational number. I believe it's true, but I'm having a hard time proving it. Is it even true?
Edit: A point touches a set if it's in its closure.
Best Answer
This is not true. Choose an enumeration of rationals $r_1,r_2,r_3,\ldots$. Then pick $A = \mathbb R \setminus \bigcup B(r_n, 2^{-n})$. $A$ is closed, doesn't meet $\mathbb Q$, and has infinite measure (we at worst removed a set of measure $2$), so it is uncountable.