I came up with this lemma (although not confident enough about it) while solving Baby Rudin. In the chapter "Basic Topology", I attempted to solve question 16, in which $\mathbb{Q}$ is regarded as the metric space with the usual metric, whose subset $E$ contains rationals $p$, such that $2<p^2<3$. The question asks to show that $E$ is not compact.
I now take the family of sets $\mathbb{Q} \cap [- \sqrt{3} +1/(n+4), \sqrt{3} -1/(n+4)]$ where $n$ runs through $1$ to $\infty$. Clearly, for no finite $n=m$, it can cover $E$. Hence, we are done.
But, in this process, I wanted to make sure, that the sets $\mathbb{Q} \cap [- \sqrt{3} +1/(n+4), \sqrt{3} -1/(n+4)]$ are all open in $\mathbb{Q}$.
I just want you to kindly verify the validity of my approach to the problem and the lemma which follows.
LEMMA:
$\mathbb{Q} \cap [a,b]$ is an open set in $\mathbb{Q}$, where $a$, $b$ are irrational.
Proof:
We know, the ordered set of rational numbers doesn't have the supremum property.
Hence the set $P = \{ t \in \mathbb{Q}: a < t < b\}$ , although bounded, doesn't have the supremum ( and infimum) in $\mathbb {Q}$. We now take a very small rational $\epsilon >0$, such that $a< t -\epsilon < t < t + \epsilon < b$ , hence $N_{\epsilon}(t) \subset P$. This is the case for every $t \in P$. Hence $P$ is open.
Additional Question: Can $\mathbb{Q}$ alone be regarded as a discrete metric space, since any subset of it is both open and closed? [If false, please provide counterexample or arguments].
Thank you.