Showing that $\mathbb{Q}\cap [0,2]$ is Not Compact

Question

I am wanting to show that $\mathbb{Q}$ is not compact on $[0,2]$ by describing an open cover for which there is no finite subcover. The cover I thought of is:

$\left(-1, \sqrt{2}-\frac{1}{n}\right) \cup\left(\sqrt{2}+\frac{1}{n} ,3\right)$ for $ n \in \mathbb{N}$

Does this cover work? Because there would be always be a rational number between$\sqrt{2}-\frac{1}{n}$ and $\sqrt{2}$ for any $ n \in \mathbb{N}$

Note I am defining $\mathbb{N}$ as all integers $\ge1$

Answer 1

Yes, your open cover works just fine.

I must say that I don't like the title of your question or the description of the problem. What you want to show is that $\mathbb{Q}\cap[0,2]$ is not compact.

Answer 2

Consider a sequence that "wants to" converge to, say $\sqrt 2$ ( Any irrational in [0,2] will do), like, e.g. {$ 1, 1.4, 1.41.., ....$}. It does not have a convergent subsequence, so it cannot be compact, since in a compact metric space every sequence has a convergent subsequence. Basically this results from the fact that $\mathbb Q$ is not complete; for metric spaces we have : Compact = Complete + Totally - Bounded.