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$S,T\subset X$. $S$ is compact. $d(S,T):=inf\{d(s,t): s\in S, t\in T\}$.

My idea is to look at the set $A:=\{d(s,T), s\in S\}$, where by definition $d(s,T):=inf\{d(s,t),t\in T\}$

I know that $S$ is compacts means $S$ is closed and bounded in $X$. I also know that if $A$ is closed then we are done. So for the sake of contradiction I assumed that $A$ is open. Then for every $a\in A$, there is some $x\in S$ such that $d(x,T)=a$. I want to reach a contradiction so I want to prove that for every $x\in S$ there is some open ball $B(x,r)$ such that $B(x,r)\subset S$. However, I am stuck on the last step. any help?

user162089
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    When $X=\mathbb R^n$, $S$ is compact means $S$ is closed and bounded. However, this is not true for an arbitrary space $X$. – ThePortakal Sep 21 '16 at 10:26
  • You should assume that $A$ is not closed instead of open. These are different. –  Sep 21 '16 at 10:26
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    By http://math.stackexchange.com/questions/8066/is-the-distance-function-in-a-metric-space-uniformly-continuous, the function $x\to d(x,T)$ is continuous. Then we apply http://math.stackexchange.com/questions/109548/x-compact-metric-space-fx-rightarrow-mathbbr-continuous-attains-max-min – Robert Z Sep 21 '16 at 10:37
  • In your notation you want to show that $inf A$ is actually just a minimum. Now you should only really need the hint, to use approximating sequences. – user369397 Sep 21 '16 at 10:39
  • But the course I'm taking has so far made no mention of sequences so I'm curious if there is a way to attack the problem without sequences – user162089 Sep 21 '16 at 10:47

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Assuming that $S$ and $T$ are not empty.

Prove that the function $f(x)=\inf \{d(x,y):y\in T\}$ is continuous from $S$ to $\mathbb R.$ Since $S$ is compact and not empty, there exists $x_0\in S$ such that $f(x_0)=\min \{f(x):x\in S\} .$

If $x\in S, y\in T$ and $d(x,y)<f(x_0)$ then $f(x)\leq d(x,y)<f(x_0),$ a contradiction.

Remark. For metric spaces in general, compact is not equivalent to closed & bounded.

DanielWainfleet
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