# *Was sind und was sollen die* compact spaces?

I'm reading the "Basic Topology" chapter in Rudin's book, and have hit a snag.

Rudin defines compactness of a set $\Sigma$ in a metric space $S$ as every open cover of $\Sigma$ admitting a finite subcover. This much I get, and I followed the pursuing proofs, but there was one cryptic comment of his which I had glossed over: after proving that $\Sigma \subset S \subset T$ means $\Sigma$ is compact relative to $S$ iff it is compact relative to $T$, he says

"By virtue of this theorem we are able, in many situations, to regard compact sets as metric spaces in their own right, without paying any attention to any embedding space.... it does make sense to talk of compact metric spaces."

I realized that there is something here which I am not getting. I do not see how any finite collection of sets open in $S$ could possibly cover any of the compact sets I can think of (by which I mean closed/bounded sets in $\mathbb{R}^k$). That is, I can't see how there could be a compact metric space. At first I thought I might have misunderstood the meaning of open sets, but since Rudin makes it clear that (1) "open" is relative to a given metric space and (2) we do not have any given metric space in mind, I cannot think of what he means.

Anyway, the problem set has forced the issue, and I find myself still confused. Can somebody clarify what compact metric spaces are, preferably with a few examples?

• All he’s saying is that a compact set is compact in any space in which it is embedded: the compactness of the set does not depend on how the set is embedded in some surrounding space. Thus, we can consider it as a compact space in its own right, without regard for any larger space in which it might be situated. Metric isn’t needed here; he just happens to be talking about metric spaces. – Brian M. Scott Apr 8 '13 at 3:51
• @BrianM.Scott The problem I'm looking at is "Prove that every compact metric space $K$ has a countable base, and that $K$ is therefore separable. Hint: For every positive integer $n$, there are finitely many neighborhoods of radius $\frac{1}{n}$ whose union covers $K$." What I'm not getting is: neighborhoods in what? – Chris Apr 8 '13 at 3:56
• Neighborhoods in $K$. Since $K$ is a metric space, the neighborhoods $B_{1/n}(x) := \{y \in K : d(x,y) < 1/n\}$ are open. – Paul Gustafson Apr 8 '13 at 3:57
• In $[0,1]$ the open ball of radius $\frac12$ centred at $0$ is $\left[0,\frac12\right)$; the open ball of radius $\frac12$ centred at $\frac14$ is $\left[0\frac34\right)$. – Brian M. Scott Apr 8 '13 at 4:05
• Naja, so geht es manchmal! – Brian M. Scott Apr 8 '13 at 4:09

Some compact metric spaces: $[0,1] \subset \mathbb R$; more generally, $[a,b]$ for some $a \le b$; $\{|x| \le 1\} \subset {\mathbb R}^n$.

Some metric spaces that are not compact: $(a,b)$ for $a<b$; $\mathbb N$.

A general theorem: A metric space $M$ is compact iff it is complete and totally bounded.

A trivial example is the empty space.

A slightly less trivial is any one-point space (call its point "P"). Any such space is metrizable: in fact, there is a unique choice of metric: the one with $d(P,P)=0$.

As $\emptyset$ and $\{ P \}$ are the only open sets, it's clear that any open cover of the one-point space is already a finite cover.

The metric space $X$ whose points are the interval $[0,1]$ and whose metric is $d(x,y) = |x-y|$ is also a compact metric space. In this space, $[0,1]$ is an open set, so there clearly exist finite covers. Furthermore, you can show every open subset of $X$ is equal to the intersection $U \cap [0,1]$ for some open subset $U$ of $\mathbb{R}$. The fact $[0,1]$ is a compact subset of $\mathbb{R}$ can then be used to show that $X$ is compact. (or, if you will, that $X$ is a compact subset of $X$)