Does every uncountable real set touch a rational number? 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.
 A: 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.
A: To expand Brian's comment:
Every uncountable Borel subset of the real numbers contain a perfect subset, i.e. a Cantor set: closed, nowhere-dense, uncountable.
In particular the irrational numbers are a Borel subset of the real line, and therefore contain such closed subset homeomorphic to a Cantor set. Being closed it does not touch any point outside itself, being a subset of the irrational numbers it does not touch any rational point.
(Note that compact subsets of the irrational numbers are also compact subsets of the real numbers since the inclusion map is a continuous embedding and there must map compact sets to compact sets.)
This can be transfered into any uncountable Polish space (completely metrizable, separable), by the same argument, fix a countable dense subset and find a Cantor set in its complement.
