TITLE: Is $[ \sqrt 2, \sqrt 3] \cap \mathbb{Q}$ an open subset of $\mathbb{Q}$?
QUESTION [7 upvotes]: Consider  the  set  of rational number $\mathbb{Q}$ as a subset  of  $\mathbb{R}$  with  the usual metric. Let  $K  = [ \sqrt 2, \sqrt 3] \cap \mathbb{Q}$.
I  have  some  confusion in  my mind that is 
Is  $K$ is an open subset  of $\mathbb{Q}$ ?
My attempt :  my answer  is No,
$K=[\sqrt 2, \sqrt 3]\cap \Bbb{Q}=\{q \in \Bbb{Q}|\sqrt 2< q< \sqrt 3\}$ where$[\sqrt 2, \sqrt 3]$ is closed in $\Bbb{R}$.
From this  I can conclude  that  K  is not  open  subset  of $\mathbb{Q}$ 
Is  it True  ?

REPLY [4 votes]: $K=[\sqrt 2,\sqrt 3]∩\mathbb{Q}=\{q\in \mathbb{Q}|\sqrt 2<q<\sqrt 3\}$  where$[\sqrt 2,\sqrt 3]$ is closed in R.
From this I can conclude that K is not open subset of Q

You're really not making it clear what your reasoning is. You seem to mostly just be restating the problem statement. Reading between the lines, your argument seems to be:

K is an intersection between a closed set and a closed set.

K is therefore closed.

Therefore K is not open.


The third statement is wrong; that a set is closed doesn't mean it's open. Presenting your argument explicitly helps others, and hopefully yourself, see what's wrong with it.
If we have the open ball topology, then since $\sqrt2 <q$, we know that there is "space" between $\sqrt 2$ and $q$, and similarly for $\sqrt3$. So given any $q$, we can take $\epsilon_1$ to be half the distance between $\sqrt2$ and $q$, $\epsilon_2$ to be half the distance between $\sqrt3$ and $q$, and $\epsilon$ to be the minimum of $\epsilon_1$ and $\epsilon_2$. Then everything withing $\epsilon$ of $q$ is in K, so $q$ is interior, and since $q$ is arbitrary, K is open.