Consider the metric space $Q$ of rational numbers with the Euclidean metric of $R$. Let $S$ consist of all rational numbers in the open interval ($a, b$), where $a$ and $b$ are irrational. Then $S$ is a closed and bounded subset of $Q$ which is not compact.
I am trying to prove this statement. It's easy to show that $S$ is closed and bounded but I'm having trouble showing that it is not compact. I think I need to find an example of an open covering of $S$ which doesn't have any finite subcover, but I can't think of such an example. Any help? Thanks.
Best Answer
HINT: Pick a sequence $a_n$ converging to $a$ from above, and consider $(a_n,b)$ as the open cover.
Another approach would be to prove that $(a,b)\cap\Bbb Q$ is not complete and that a compact metric space is always complete.