Compactness (1)

Since I haven't posted in a million years, I figured a good way to jump back into it would be to digress back to some good, old fashioned point-set topology. Somehow I never wrote about compactness, so I think it's about time I fixed that.

Unlike the other topological invariants we've discussed, such as connectedness, compactness does not have an immediately obvious geometric description. In fact, the concept is notorious for throwing students for a loop and seeming completely unintuitive. However, it is a concept that is impossible to avoid, as it shows up everywhere and in every field of mathematics. I would encourage you to adopt the following mentality if you have never been introduced to compactness before:

Don't get discouraged by the weird, seemingly unmotivated definition. Years of research yielded this definition to unify several more concrete ideas, so the consequences of compactness are more important than its definition. The implications of compactness in Euclidean spaces, which are arguably the spaces we care the most about, are particularly nice, and so the concept is worth studying even if it is not intuitive.

With that being said, I'll stop beating around the bush and just do the thing. We'll need the preliminary notions of covers and subcovers, which I may have mentioned before at some point but it bears repeating.

Definition. If $X$ is a set, then a cover of $X$ is a collection of sets whose union contains $X$.

If $\mathcal{O}$ is a cover of a set $X$, a subcover of $\mathcal{O}$ is a subset of $\mathcal{O}$ which still covers $X$.

If $X$ is a topological space, an open cover is a cover of $X$ comprised solely of open sets. Trivially, any subcover of an open cover is still an open cover.

Of course, every set is a cover of itself, and every cover is a subcover of itself. More interesting examples of covers that we've seen before include topologies and partitions. We've seen examples of subcovers before too, although we didn't call them that. For example, a topological basis is a subcover of the topology it induces!

Now here's the definition of compactness:

Definition. A topological space $X$ is compact if every open cover of $X$ has a finite subcover.

A subset $A$ of $X$ is compact if it is a compact space when viewed as a subspace of $X$.

As I said earlier, the best way to get a feel for why compactness is important is to examine its implications. First of all, it's a topological property (meaning it is preserved by continuous maps). That alone gives us an additional tool in our never-ending journey to characterize topological spaces. We need to prove it first, though!

The following proof requires the fact that, for any set function $f:X\to Y$ and any subsets $A\subseteq X$ and $B\subseteq Y$, we always have that $A\subseteq f^{-1}\big[f[A]\big]$ and $f\big[f^{-1}[B]\big]\subseteq B$. I have not explicitly proved this, but it is easy enough to do and I will use these facts freely.

Theorem. Let $X$ and $Y$ be topological spaces, let $X$ be compact, and let $f:X\to Y$ be continuous. Then $f[X]\subseteq Y$ is compact.

Proof. Choose any open cover $\{U_i\}_{i\in I}$ of $f[X]$. Then from the definition of a cover, we know that $f[X]\subseteq\bigcup_{i\in I} U_i$. It follows then that

$$\begin{align}
X &\subseteq f^{-1}\big[f[X]\big] \\[.5em]
&\subseteq f^{-1}\Big[\bigcup_{i\in I} U_i\Big] \\
&=\bigcup_{i\in I} f^{-1}[U_i].
\end{align}$$

Since $f$ is continuous, $f^{-1}[U_i]$ is open in $X$ for each $i\in I$. This means that $\{f^{-1}[U_i]\}_{i\in I}$ is an open cover of $X$. But $X$ is compact, so this cover of $X$ has a finite subcover $\{f^{-1}[U_{i_j}]\}_{j=1}^n$. This means that $X\subseteq \bigcup_{j=1}^n f^{-1}[U_{i_j}]$, so

$$\begin{align}
f[X] &\subseteq f\Big[\bigcup_{j=1}^n f^{-1}[U_{i_j}]\Big] \\
&= \bigcup_{j=1}^n f\big[f^{-1}[U_{i_j}]\big] \\
&\subseteq \bigcup_{j=1}^n U_{i_j}.
\end{align}$$

Thus, $\{U_{i_j}\}_{j=1}^n$ covers $f[X]$. It is a subcover of our original cover for $f[X]$, so it follows that $f[X]$ is compact.

So at the very least, compactness is a topological invariant. That makes it useful enough in its own right. Before we move on to some of the other nice properties of compact spaces, it would probably be a good idea to look at a few spaces that are and aren't compact.

Example. The empty set is compact in any topological space $X$.

Proof. Take any open cover of the empty set (literally any collection of open sets is a cover). We can use zero of the sets in this cover (i.e., the empty set) as a subcover. Zero is certainly finite, so the empty set is compact.

That kind of feels like cheating, though. If the only compact set was the empty set, compactness wouldn't be a very interesting property at all. Luckily, there are many more nontrivial examples.

Example. In any non-empty topological space $X$, every finite subset of $X$ is compact.

Proof. Suppose $A=\{x_j\}_{j=1}^n$ is a finite subset of $X$. Choose any open cover $\{U_i\}_{i\in I}$ of $A$. Then by definition, $A \subseteq \bigcup_{i\in I} U_i$, so for any point $x_j\in A$, there exists $i_j\in I$ for which $x_j\in U_{i_j}$. It follows then that $A \subseteq \bigcup_{j=1}^n U_{i_j}$, and so $\{U_{i_j}\}_{j=1}^n$ is a finite subcover. Thus, $A$ is compact, as desired.

You might start to get the impression that closed and compact sets are related. And while this is (partially) correct in metric spaces, I'll take this opportunity to remind you that finite sets aren't even guaranteed to be closed unless you are working in a Hausdorff space (take the trivial topology as a counterexample). This means that compactness is different than just being closed. In fact, in metric spaces, it's a strictly stronger condition. The next example demonstrates that even closed sets in metric spaces, the "nicest" topological spaces, are not necessarily compact.

Example. The real line $\R$ with the standard topology is not compact.

Proof. It suffices to demonstrate an open cover of $\R$ which does not have a finite subcover. We claim that $\{(n, n+2)\}_{n=-\infty}^\infty$ is such a cover. This is an infinite collection of overlapping open intervals extending infinitely in each direction, so certainly it is an open cover of $\R$. However, if we remove even one set $(i, i+2)$ from the cover, then suddenly for any $x\in (i, i+1]$, $x$ is not in any remaining set. Thus, we cannot remove a single set from our cover and still have a cover of $\R$. It follows that our open cover has no proper subcovers, and so it certainly does not have any finite ones. Thus, $\R$ is not compact.

So even though $\R$ is closed, it is not compact. It turns out that the only thing preventing its compactness is that $\R$ is unbounded. We shall see in the next continuation of this post that a subset of $\R$ is compact if and only if it is closed and bounded. For general metric spaces, though, even this is not sufficient. However, it is necessary, and we will show this shortly.

Before proceeding, I need to tell you what "bounded" actually means, and we will also need a lemma which simplifies the proof of the theorem I just promised you.

Definition. A subset $A$ of a metric space $X$ is bounded if there is an open ball $B\subseteq X$ which contains $A$.

That should seem like a reasonable definition. Basically a set is bounded if it fits inside some ball of finite radius. If it is impossible to find such a ball, then the set is unbounded.

Lemma. Every subset $A$ of a metric space $X$ has an open cover comprised of bounded sets.

Proof. For each $x\in A$, define $B_x=B(x, 1)$, the open ball of radius $1$ centered at $x$. Then, by the union lemma,

$$A=\bigcup_{x\in A}B_x,$$

and so $\{B_x\}_{x\in A}$ is an open cover of $A$.

Now we are equipped to prove the theorem I mentioned above.

Theorem. If $A$ is a compact subset of a metric space $X$ then $A$ is closed and bounded.

Proof. For the first part, we will actually prove something stronger. We will assume only that $X$ is a Hausdorff space (of which metric spaces are a special case) and show that if $A\subseteq X$ is compact then it is closed. We proceed by showing that $X-A$ is open. Choose any point $x\in X-A$. Then since $X$ is Hausdorff, for any point $y\in A$ we have that $x\neq y$ and so there exist disjoint neighborhoods $U_y\ni y$ and $V_y\ni x$. Clearly $A\subseteq\bigcup_{y\in A}U_y$ by a weakened form of the union lemma, and so $\{U_y\}_{y\in A}$ is an open cover of $A$. Since $A$ is compact, there is a finite subcover $\{U_{y_i}\}_{i=1}^n$ of $A$.

Define $U=\bigcup_{i=1}^n U_{y_i}$ and $V=\bigcap_{i=1}^n V_{y_i}$. Then $U$ is open since it is the union of open sets and $V$ is open because it is a finite intersection of open sets. Furthermore,

$$\begin{align}
U\cap V &= \Big(\bigcup_{i=1}^n U_{y_i}\Big) \cap \Big(\bigcap_{i=1}^n V_{y_i}\Big) \\
&= \bigcup_{i=1}^n \Big(U_{y_i}\cap\bigcap_{i=1}^n V_{y_i}\Big) \\
&= \bigcup_{i=1}^n\bigcap_{i=1}^n\big(U_{y_i}\cap V_{y_i}\big) \\
&= \varnothing,
\end{align}$$

since $U_{y_i}$ and $V_{y_i}$ are disjoint for each $i$ by construction. It follows that $x\in V\subseteq X-A$, and so $X-A$ is open. Thus, $A$ is closed.

For the second part, we need to work in the context of metric spaces since that is the only setting where boundedness makes any sense. Suppose then that $A$ is a compact subset of a metric space $X$. By the lemma above, $A$ has an open cover $\{U_i\}_{i\in I}$ of bounded sets. Since $A$ is compact, this cover has a finite subcover $\{U_{i_j}\}_{j=1}^n$. Since each set in this subcover is bounded and there are finitely many of them, their union is bounded. It follows that $A$ is bounded since it is contained in this union.

There's a lot more to be said about compactness, and so I will be writing another post on the topic soon (and likely more than one). Stay tuned!