# Is $[ \sqrt 2, \sqrt 3] \cap \mathbb{Q}$ an open subset of $\mathbb{Q}$?

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 ?

Yes, $$K$$ is an open subset of $$\mathbb Q$$, since $$K=\left(\sqrt2,\sqrt3\right)\cap\mathbb Q$$ and $$\left(\sqrt2,\sqrt3\right)$$ is an open subset of $$\mathbb R$$.

No, that's wrong. The fact that a set is closed doesn't mean it is not open!

In fact $$K$$ is also open because it equals to $$(\sqrt{2},\sqrt{3})\cap \mathbb{Q}$$.

Side note: The space $$\mathbb{Q}$$ with the topology induced by $$\mathbb{R}$$ is "totally disconnected" this means that it has "many" sets which are both closed and open.

• A set is not a door. – Arno Dec 14 '18 at 21:30
• The first line of this answer can't be emphasized enough. It even made it onto the MO list of most common false beliefs mathoverflow.net/a/23580 – Carmeister Dec 15 '18 at 3:58

$$K=[\sqrt 2,\sqrt 3]∩\mathbb{Q}=\{q\in \mathbb{Q}|\sqrt 2 where$$[\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:

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

2. K is therefore closed.

3. 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 $$\sqrt2 , we know that there is "space" between $$\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.

A set can be both open and closed at the same time (such sets are called clopen), and just because you've shown that $$[\sqrt2, \sqrt3]\cap \Bbb Q$$ is closed in $$\Bbb Q$$, that doen't mean it isn't open.

Look at the definition of open in the subspace topology, and se whether $$[\sqrt2, \sqrt3]\cap \Bbb Q$$ is such a set or not.

With the metric $$d(x,y)=|x-y|$$ on $$\Bbb Q$$ and the topology on $$\Bbb Q$$ generated by $$d$$: For $$q\in K$$ let $$r(q)=\min (q-\sqrt 2\,,\sqrt 3 -q\,).$$ Let $$K(q)=\{q'\in \Bbb Q: d(q',q)

Then $$K(q)$$ is open in $$\Bbb Q$$ and $$q\in K(q)\subset K.$$

So $$\cup_{q\in K}K(q)$$ is open in $$\Bbb Q.$$ And we have $$K=\cup_{q\in K} \,\{q\}\subset \cup_{q\in K}\,K(q)\subset \cup_{q\in K}\,K=K,$$ so $$K=\cup_{q\in K}K(q)$$ is open in $$\Bbb Q.$$

(i). Let $$T$$ be a topology on a set $$X$$ and let $$Y \subset X.$$ The subspace topology $$T|Y$$ on $$Y$$ is defined as $$T|Y=\{t\cap Y:t\in T\}.$$ If $$B$$ is a base (basis) for $$T$$ then $$B|Y=\{b\cap Y: b\in B\}$$ is a base for $$T|Y.$$
(ii).Suppose $$T$$ is generated by a metric $$d$$ on $$X,$$ so that the set $$B$$ of open $$d$$-balls of $$X$$ is a base for $$T$$. So $$B|Y$$ is a base for $$T|Y.$$
BUT in general $$B|Y$$ may NOT be the set of open $$d$$-balls of $$Y.$$ An open $$d$$-ball of $$Y$$ is $$B_d^Y(y,r)=\{y'\in Y: d(y',y) for some $$y\in Y,$$ which does belong to $$B|Y,$$ but there may be other members of $$B|Y.$$
For example in your Q, with $$X=\Bbb R$$ and $$Y=\Bbb Q$$ and $$d(u,v)=|u-v|,$$ the set $$K$$ belongs to $$B|Y$$ but is not an open ball of $$\Bbb Q$$ because $$(\sqrt 2 +\sqrt 3)/2\not \in \Bbb Q.$$
(iii). In the general case, for metric spacess it is a useful, widely used result that every member of $$B|Y$$ is a union of open $$d$$-balls of $$Y,$$ so the subspace topology $$T|Y,$$ as a subspace of $$X,$$ co-incides with the topology on $$Y$$ generated by the metric $$d|_{Y\times Y}.$$