Distance between two sets $X$ and $Y$.

Question

Let $X$ and $Y$ be compact and closed subset of a metric spaces $M$ respectively, then prove that $inf {d(x,y)|xin X, yin Y}=d(x,y)$ for some $xin X$ and $yin Y$.

I am just here to verify my proof. Please help if I am wrong.
Proof:
Let $f:Xrightarrow [0,infty)$ be the distance function from $Y$ defined by $f(x)=inf {d(x,y)| yin Y}$, then $f$ is a continuous function. Also, $X$ is a compact set so $min_{xin X} f(x)=f(x_1)$ for some $x_1in X$. Hence, $inf {d(x,y)|xin X, yin Y}$ $=inf {d(x_1,y)|yin Y}=d(x_1,y_1)$ for some $y_1in Y$ as $Y$ is closed.

Paul Sinclair · Answer

I'm afraid I must disagree. Your proof doesn't work, and the reason is that this "theorem" as stated is false.
Let $X = [0,1], Y = (2,infty)$ and $M = X cup Ysubset Bbb R$. $X$ is compact, and $Y$ is closed in $M$. Then $d(X, Y) = 1$, yet there is no $yin Y$ such that $d(X, y) = 1$.
In addition to the issue that Hagen von Eitzen pointed out (it also fails when $X$ and $Y$ are empty), the problem with your proof is that you simply assumed that $Y$ being closed was enough to claim $y_1$ existed.
(The function you defined earlier is continuous, so that is not a problem with the logic, though the commenters are correct that unless it had been proven elsewhere, you need to prove it here.)
The correct result must assume that $Y$ is compact as well.

Distance between two sets $X$ and $Y$.

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