Krdytkyu

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 $K=left(sqrt2,sqrt3right)capmathbb Q$ and $left(sqrt2,sqrt3right)$ 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 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 $qin K$ 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 $qin K(q)subset K.$

So $cup_{qin K}K(q)$ is open in $Bbb Q.$ And we have $K=cup_{qin K}
,{q}subset cup_{qin K},K(q)subset cup_{qin K},K=K,$ so $K=cup_{qin K}K(q)$ 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={tcap Y:tin T}. $ If $B$ is a base (basis) for $T$ then $B|Y={bcap Y: bin B}$ 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 $yin 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 $(sqrt 2 +sqrt 3)/2not in Bbb Q.$

(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 $d|_{Ytimes Y}.$

$K=[sqrt 2,sqrt 3]∩mathbb{Q}={qin mathbb{Q}|sqrt 2<q<sqrt 3}$ 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.