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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.

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    $\begingroup$ An alternate approach: Observe that when $a$ and $b$ are irrational, $\mathbb{Q}\cap[a,b]=\mathbb{Q}\cap(a,b)$. $\endgroup$ – Michael Burr Dec 20 '18 at 15:53
  • $\begingroup$ Can you please check the proof :( $\endgroup$ – Subhasis Biswas Dec 20 '18 at 15:54
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    $\begingroup$ The converse also holds: for any $a,b\in\Bbb R$ such that $a<b$, $\Bbb{Q}\cap[a,b]$ is an open subset of $\Bbb Q$ if and only if both $a$ and $b$ are irrational numbers. $\endgroup$ – user593746 Dec 20 '18 at 15:54
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    $\begingroup$ Why is every subset of $\mathbb{Q}$ both open and closed? You haven't shown that. Is $\{0\}$ open? $\endgroup$ – Michael Burr Dec 20 '18 at 15:54
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    $\begingroup$ Your formula for $\epsilon$ should depend on $a$ and $b$. Something like $\min\{\frac{t-a}2,...\}$. $\endgroup$ – Michael Burr Dec 20 '18 at 16:03

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