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?
Best Answer
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 $\mathbb R$?