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Good afternoon all. This question appeared on my real analysis midterm. I got it wrong (very wrong!) and the prof isn't releasing solutions. Out of curiosity, I'd like to know how to attack the question, which I'll reproduce in full here:

Prove that any countable subset of $\mathbb{R}$ has empty interior. Is the converse true? Explain.

Here're a few ideas I had: Any countable subset $A$ of $\mathbb{R}$ is equivalent to $\mathbb{N}$ (or to some subset of $\mathbb{N}$). We can express the interior of $A$ as the union of all open sets contained in $A$. So if we can show that this union is empty, we'll be done.

As for the converse, if $A$ has empty interior, then the union of all open sets contained in $A$ must be empty. What's the best way to proceed from here?

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    Lot's of mathematicians use "any" synonymously with "every" in these situations, and then sometimes get misunderstood when the difference between "any" and "every" in standard English become relevant. Just one example: "If any citizen so wishes, the council will make that information available." means something very different from "If every citizen so wishes, the council will make that information available."2011-12-07

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Suppose that $A$ has non-empty interior. This implies that $A$ contains some non-trivial open interval.

Hence the cardinality of $A$ is at least the cardinality of the interval. Any non-degenerate interval has cardinality $\mathfrak c=2^{\aleph_0}>\aleph_0$.

Thus $\operatorname{card} A>\aleph_0$ and $A$ is not countable.


The converse is not true. The set $\mathbb R\setminus\mathbb Q$ is an example of a set which is uncountable, but it has empty interior. (No open interval is a subset of this set.)


For the fact about cardinality of intervals see e.g. this question: Does any interval of $\mathbb R$ have the same number of elements as $\mathbb R$?

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    Thanks so much for the help. It seems so obvious now; looks like I have some studying to do before the final! ;)2011-12-06