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Let $(M,d)$ be a metric space. If $K\subset M$ is compact, then it is closed (and bounded).

Proof Let's see that $M\setminus K$ is open. Let $x\in K$ $$\exists \varepsilon_1 (x), \varepsilon_2(x) \text{ so that } B(x, \varepsilon_1(x))\cap B(y,\varepsilon_2(x)) = \emptyset$$ then $K \subset \cup_{x\in K} B(x, \varepsilon_1(x))$ and, since $K$ is compact $\exists N \in \mathbb N \; \exists x_1,...,x_N \in K$ s.t. $$ K\subset \bigcup_{i=1}^N B(x_i, \varepsilon_1(x_i)) $$ let $r = \min\{\varepsilon_2(x), i = 1,...,N \} > 0$, then $$ B(y,r)\cap B(x_i, \varepsilon_1(x_i)) = \emptyset \quad\forall i = 1,...,N $$ therefore $B(y,r)\subset M\setminus K$ and $K$ is closed.

Question Why uses $\varepsilon_1(x)$ and $\varepsilon_2(x)$? If we consider, in $\mathbb R$ the interval $[0,1]$ it can't be covered using open balls without covering elements of $\mathbb{R}\setminus[0,1]$, so what happens when choosing $r$? Am I missing something?

Thanks in advance.

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$y$ should be fixed at the start. Then for each $x\in K$, you choose disjoint open balls containing $x$ and $y$. You can't cover $[0,1]$ without covering points of its complement; but you can cover $[0,1]$ without covering a fixed point outside of it. – David Mitra Dec 15 '12 at 20:54
I guess you need to read Hausdorff Separation property to understand what you are missing. – Sudhanshu Srivastava Dec 15 '12 at 20:55
up vote 4 down vote accepted

I think this proof might be a little clearer in its more general setting. That is if $X$ is a Hausdorff space and $Y \subset X$ is compact then $Y$ is closed. We call a space Hausdorff if for any two distinct points $x,y \in X$ there exists open neighborhoods $U,V \subset X$ so that $x \in U$ and $y \in V$ and $U \cap V =\emptyset$. It's clear that metric spaces are Hausdorff because you can separate two points taking balls around each half the distance between the two. Concretely we take $r=d(x,y)/2$ then have $U=B(x,r)$ and $V=B(y,r)$.

Now let $Y \subset X$ be compact. Pick $y \in X \setminus Y$. For each $x \in Y$ by the Hausdorff property we can find open sets $U_x$ and $V_x$ so that $x \in U_x$, $y \in V_x$ and $U_x \cap V_x=\emptyset$. Now it follows that

$$Y \subset \bigcup_{x \in Y} U_x,$$

since $Y$ is compact there exists some $x_1,\dots,x_n$ such that $U_{x_1},\dots,U_{x_n}$ covers $Y$. Then set $V=V_{x_1} \cap \cdots \cap V_{x_n}$ it follows that $V$ is open and contains $y$ but $U_{x_i} \cap V_{x_i}=\emptyset$. Since $Y$ is contained in the union of the $U_{x_i}$ we have that

$$Y \cap V = \emptyset.$$

Thereby around every point $y \in X \setminus Y$ we can find an open set contained in $X \setminus Y$. It follows that $Y$ is closed since $X \setminus Y$ is open.

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I prefer the proof using limit point compactness, which is equivalent to cover compactness in metric spaces:

Definition: limit point compact

Let $X$ be a metric space. A set $A \subset X$ is limit point compact if every infinite subset of $A$ has a limit point in $A$.

How do we use this to show closed? Well, a limit point of $A$ (if one exists) has a sequence in $A$ converging to that point by definition. The set of all points in such a sequence would be an infinite set in $A$. Can you finish?

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Moral: don't use cover compactness unless you have to. In metric spaces, the fact that limit point compactness and sequential compactness are equivalent to cover compactness is one great improvement over arbitrary topological spaces. – Gyu Eun Lee Dec 15 '12 at 20:45
Thanks!, but I'm trying to understand this proof and unfortunately I've to rewrite it. – user50554 Dec 15 '12 at 20:46

I'm guessing from your question that you've mentally switched the order of the quantifiers. Your argument shows that $\epsilon_1(x)$ cannot be chosen independently of $y$, but it doesn't have to be. Given $y$, both $\epsilon_1(x)$ and $\epsilon_2(x)$ can be chosen small enough that there's no overlap; but the union of the balls with radii $\epsilon_1(x)$ will contain other points in $K\setminus M$ closer to $K$ than $y$, for which $\epsilon_1(x)$ will have to be chosen smaller.

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