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

Question

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 ?

Answer 1

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 $\mathbb R$.

Answer 2

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.

Answer 3

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.

Answer 4

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 $K(q)=\{q'\in \Bbb Q: d(q',q)<r(q)\}.$

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 $\Bbb Q.$

An easily overlooked point about this 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 $B_d^Y(y,r)=\{y'\in Y: d(y',y)<r\},$ 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 $d|_{Y\times Y}.$

Answer 5

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

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 $\sqrt2 <q$, 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.