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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.

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What do you mean by "touches"? –  Asaf Karagila Oct 3 '12 at 14:30
    
No, it’s not: the irrationals contain many subsets that are homeomorphic to the middle-thirds Cantor set and are therefore closed and uncountable. –  Brian M. Scott Oct 3 '12 at 14:30
    
@Asaf: It’s clear from the parenthesis: $A$ touches $S$ if the closure of $A$ meets $D$. –  Brian M. Scott Oct 3 '12 at 14:32
    
@Brian I would love to see that in a bit more detail. Can you elaborate this? –  k.stm Oct 3 '12 at 14:45

2 Answers 2

up vote 4 down vote accepted

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.

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I actually thought of that and had forgotten all about it while doing something else! Anyway, thanks a lot! –  k.stm Oct 3 '12 at 14:41

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.

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I think you need compactness of the Cantor set to ensure that it is not only closed in the irrational numbers but also closed as a subset of $\mathbb{R}$. In your last paragraph you need a further hypothesis ensuring that the Polish space is uncountable. –  commenter Oct 3 '12 at 15:02
    
@commenter: You are correct about the uncountability of the Polish space in the last paragraph. I think that compact subsets of the irrationals are automatically compact in $\mathbb R$, but I don't see it immediately right now. I'll think it for a bit more and then I will edit the answer to complete the arguments. Thank you for pointing out these things! –  Asaf Karagila Oct 3 '12 at 15:33
    
No problem... The inclusion of the irrationals into the reals is continuous, hence it sends compact sets to compact sets :-) –  commenter Oct 3 '12 at 15:37
    
@commenter: Ah yes. I thought about that at first, but for some reason I backed down from this idea... thanks again! –  Asaf Karagila Oct 3 '12 at 16:20

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