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Prove that any finite set in a metric space is compact

It is obvious that any finite set is bounded. I don't understand if any finite set is closed? And is it ok to say any finite set is compact because they are bounded and closed?

ViktorStein
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user378456
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    Metric spaces need not have the Heine-Borel property. So you cannot use closed and boundedness. Instead use the traditional definition. If you have an open cover of this set, then can't you find a finite subcover? – Sarvesh Ravichandran Iyer Oct 20 '16 at 06:06
  • @астонвіллаолофмэллбэрг right, some do not have the Bolzano-Weierstrass property. And the definition is a set K is compact if every sequence in K has a subsequence that converges to a limit that is also in K. How do i prove every sequence in finite set has a subsequence that converges to a limit that is also in the finite set? – user378456 Oct 20 '16 at 06:20

4 Answers4

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There are two ways to look at this.

One is this: suppose that we have a sequence from this finite set. Then, at least one of the values must appear infinitely many times, because if each value appears only finitely many times, then the sequence itself would be a finite number of values appearing finitely many times, which would make it a finite sequence, a contradiction. Hence, one of the values appears infinitely many times. Then, take the subsequence formed by this value appearing infinitely many times. Of course, this converges. Hence, the finite set is sequentially compact, hence compact.

The other way is even simpler: suppose we have an open cover. Then, each point is contained in some open set from the cover depending upon that point. This means there is a finite subcover (infact, the size of the subcover is at most the size of the set). Hence, the set is compact.

Both methods would do.

Sarvesh Ravichandran Iyer
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In metric spaces in general, being closed and bounded is not equivalent to being compact. A set $S$ is compact iff whenver $F$ is an open family (a family of open sets) such that $S\subset \cup F,$ there is a finite $G\subset F$ such that $S\subset \cup G.$

A finite set is compact.Proof by induction.

(1). If $S=\emptyset$ and $F$ is any open family then $S\subset \cup F$ and we may let $G=\emptyset.$

(2). Suppose that $n\geq 0$ and that every open cover of any $n$-member set has a finite subcover. Then if $S=\{x\}\cup T$ has $n+1$ members with $x\not \in T,$ and $F$ is an open cover of $S,$ then $F$ is also an open cover of $T,$ and $T$ has $n$ members. So there exists a finite $G^*\subset F$ such that $\cup G^*\supset T.$

Now there exists $f\in F$ with $x\in f$ because $F$ covers S. And for any such $f,$ the set $G=\{f\}\cup G^*$ is a finite subset of $F ,$ and $G$ covers $S.$

DanielWainfleet
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You can prove it directly takin $\{U_{\alpha}\}$ any open cover of your finite set, we name $F$. Now for each point $x \in F$, as $\bigcup U_{\alpha}$ covers $F$ take $\alpha_{x} $ such that $x \in U_{\alpha_{x}} $ so $\{U_{\alpha_x} : x \in F\}$ is a finite subcover. Hence F is compact

Iñaki Mendieta
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In order to prove that a set is compact, you must show that it is bounded and closed.

To show that it is bounded, let F be a finite set, then since it is finite, by the arch median principle of natural numbers, there exists an infimum and supremum of the set therefore you can find an M > 0 s.t for some point x contained in F, abs(x) <= M and I will leave it to you to determine the other side of this. So F is bounded.

To show that F is closed, there are multiple ways to accomplish this. The easiest would be notice that F can contain no limit points, if F = {1}, then F would have no limit points and thus the method of showing F contains all pf its limit points would be futile. But rather, consider the complement of a finite set it open, therefore F must be closed

Hence, F is compact

Sam Lerner
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