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?

  • $\begingroup$ Pick any countable set, and prove that it has empty interior? Or did you mean that any countable set that you pick will have empty interior? The ambiguity would not be there if you just wrote "every" instead of "any". $\endgroup$ – Michael Hardy Dec 6 '11 at 3:08
  • $\begingroup$ I was under the impression that "every" and "any" meant the same thing in these types of situations (after all, doesn't \forall mean "for all", "for every" or "for any"?), but if it's ambiguous then I'll definitely change it! $\endgroup$ – user20682 Dec 6 '11 at 12:31
  • $\begingroup$ 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." $\endgroup$ – Michael Hardy Dec 7 '11 at 0:55

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 $\mathbb R$?

  • $\begingroup$ Thanks so much for the help. It seems so obvious now; looks like I have some studying to do before the final! ;) $\endgroup$ – user20682 Dec 6 '11 at 1:36

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