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1. Week 1: Covering Compactness

Table of Contents | 2. A Problem with Tubes ≫

Welcome to Math 131BH! We’re going to kick off the quarter by discussing compact sets. Recall the definition of covering compactness:

In addition to describing a metric space as “compact”, we’ll also describe subsets of metric spaces as being “compact”. Namely,

These two definitions have some overlap, as one can treat \(\left( S, d \right)\) as a metric space by restricting the metric. This turns out to not matter at all — the two definitions coincide in this case.

You should think of this as saying that \(X\) has limited space for open sets to “spread out”, and that compactness is a topological notion of what it means to be “small”.

What “spreading out” means is kind of hazy, so take this intuition with a grain of salt — the open interval \((0, 1)\) is not compact with respect to the standard metric, for instance. Perhaps one way to think about this is that \((0, 1)\) extends endlessly towards its “missing boundary points”. In fact, we can formalise this: if \(S\subseteq X\) is a subset of a larger metric space, and if \(S\) is compact, then \(S\) must be closed. (Of course, the converse is far from true.)

In the case of \((0, 1)\), one can quickly construct explicit open covers with no finite subcovers: for instance, \((0, 1) = \bigcup _{n=1}^{\infty} \left( \frac{1}{n+2}, \frac{1}{n} \right)\). One can reproduce this idea more generally in an abstract metric space (exercise!).

You’ve seen hopefully in the first lecture that the closed interval \([0, 1]\) is compact, and that more broadly any bounded closed interval is compact. But this takes a nontrivial amount of effort to prove, and we may be interested in proving that other sets are compact without all the sweat. So, our goal for today is as follows: can we identify a systematic way to quickly prove that certain sets are compact?

We’ll show three big ideas:

1. Finite unions of compact subsets of a metric space are compact. This makes sense — putting a few small things together will still make something small.
2. Closed subsets of compact metric spaces are compact. The “closed” condition is to ensure we don’t miss any boundary points, as described before.
3. Finite products of compact metric spaces are compact.

These three properties of compact metric spaces together will allow us to construct a very wide variety of compact sets. Knowing that \([-1, 1]\) is compact, for instance, the third property says the unit square \([-1, 1]^2\) is compact in \(\mathbb{R}^2\). The second property then tells us that the closed unit disc in \(\mathbb{R}^2\) is also compact. Let’s prove these one at a time, and hopefully this will showcase some helpful techniques for the future.

Note this implies by induction that any finite union of compact sets in a metric space is compact.

Proof

Let \(\left\lbrace U_ \alpha \right\rbrace\) be a collection of open subsets of \(X\) such that \(S_1\cup S_2\subseteq \bigcup U_ \alpha \).

Since \(S_1\subseteq S_1\cup S_2\subseteq \bigcup U_ \alpha \), there exist finitely many indices \(\alpha _1,\ldots, \alpha _n\) such that \(S_1\subseteq U _{\alpha _1} \cup \cdots U _{\alpha _n}\). (This is the definition of compactness.)

Likewise, since \(S_2\subseteq \bigcup U_ \alpha \), there exist finitely many indices \(\beta _1, \ldots, \beta _m\) such that \(S_2\subseteq U _{\beta _1}\cup \cdots \cup U _{\beta _m}\). Then, \[S_1\cup S_2 \subseteq U _{\alpha _1} \cup \cdots \cup U _{\alpha _n}\cup U _{\beta _1}\cup\cdots \cup U _{\beta _m}.\] This is a finite subcover of \(S_1\cup S_2\), hence it’s compact. \(\square\)

Using this idea, we can prove that \([0, 1]\cup [2, 3]\) is compact immediately. However, beware that infinite unions of compact sets may not be compact; \(\mathbb{R} = \bigcup _{n\in \mathbb{Z}} \left[ n, n+1 \right]\) is one such example.

Proof

Let’s begin by considering \(\left\lbrace U_ \alpha \right\rbrace\) a collection of open subsets of \(X\) such that \(S\subseteq \bigcup U_ \alpha \). I want to use the compactness of \(X\) somehow, but I don’t know that \(\bigcup U_ \alpha = X\).

This is where the closed-ness of \(S\) comes in: \(S^\complement\) is an open set, hence \[X = S^\complement\cup S\subseteq S^\complement \cup \left(\bigcup U_ \alpha \right).\] This rightmost quantity is an open cover of \(X\), hence there exist finitely many indices \(\alpha _1, \ldots, \alpha _n\) such that \[X = S^\complement \cup U _{\alpha _1} \cup U _{\alpha _2} \cup \cdots \cup U _{\alpha _n}.\] While the finite subcover may or may not include \(S^\complement\), putting it in won’t hurt. But since \(S\subseteq X\) and \(S\cap S ^\complement = \varnothing\), it immediately follows that \[S\subseteq U _{\alpha _1} \cup \cdots \cup U _{\alpha _n},\] hence \(S\) is compact. \(\square\)

Knowing that every closed interval in \(\mathbb{R}\) is compact, this immediately tells us that every closed and bounded subset of \(\mathbb{R}\) is compact! If \(S\subseteq \mathbb{R}\) is closed and bounded, then \(S\subseteq \left[ -M, M \right]\) for some very large number \(M\). But \(\left[ -M, M \right]\) is compact, hence by the preceeding proposition, so is \(S\). Thus, for instance, we know immediately that the Cantor set is compact, without having to sweat about open covers at all.

Finally, we arrive at perhaps the most difficult of the three claims to prove.

By the way, recall that \(d _{X\times Y} \left( \left( x_1, y_1 \right), \left( x_2, y_2 \right) \right) = d_X \left( x_1, x_2 \right) + d_Y \left( y_1, y_2 \right)\) is the metric on the Cartesian product of two metric spaces. To prove this, we’ll need a few lemmas about open sets first:

In \(\mathbb{R}^2\), this can be visualised. Draw an open subset of \(U\subseteq \mathbb{R}^2\). The intersection of this open set with any vertical line looks like an open subset of \(\mathbb{R}\), and if you “squish” \(U\) down to one axis, then it also looks like an open subset of \(\mathbb{R} \). This intuition generalises completely to the abstract setting. These can be proved within finite time, and it’s a good refresher of how open sets work if you need it.

Let’s now prove the earlier proposition.

Proof

Let \(\left\lbrace U_ \alpha \right\rbrace\) be a collection of open subsets of \(X\times Y\) such that \(X\times Y = \bigcup U_ \alpha \). We want to find a finite subcover. The outline of the proof is as follows:

1. For each \(x\in X\), the set \(\left\lbrace x \right\rbrace\times Y\) is compact. Thus, there is a set \(V_x\), a finite union of elements of \(\left\lbrace U_ \alpha \right\rbrace\), with \(\left\lbrace x \right\rbrace\times Y \subseteq V_x\). (This uses the fact that slices of open sets are open.)
2. For a fixed \(x\), let’s write \(V_x = U _{\alpha _1} \cup \cdots U _{\alpha _n}\). Each projection of the individual cover elements is open, so \(W_x = \pi _X U _{\alpha _1} \cap \cdots \cap \pi _X U _{\alpha _n}\) is an open neighbourhood of \(x\). (This uses the fact that projections of open sets are open.)
3. \(\bigcup _{x\in X} W_x\) is an open cover of \(X\), hence admits a finite subcover \(W _{x_1}\cup \cdots W _{x_n} = X\).
4. \(X\times Y\subseteq V _{x_1} \cup \cdots \cup V _{x_n}\), and each of these \(V\)’s is a union of finitely elements of \(\left\lbrace U_ \alpha \right\rbrace\).

Step 1 says that each vertical slice of \(X\times Y\), denoted \(\left\lbrace x \right\rbrace\times Y\), has a finite subcover \(V_x\), and step 2 says that this finite subcover contains some “open tube”, denoted \(W_x\times Y\). Step 3 says that we can cover \(X\times Y\) lengthwise using the tubes, and step 4 lets us turn this back into a finite subcover of \(X\times Y\). Draw a picture to illustrate what the argument is!

Let’s more rigorously prove each step, which will consequently finish the proof.

1. For any \(x\in X\), we have \(Y\subseteq \bigcup \sigma _x U _ \alpha \). That is, if you take all the \(x\)-slices of the \(U_ \alpha \)’s, they cover all of \(Y\). Thus, by compactness of \(Y\), there exist a finite set of indices \(\alpha _1, \ldots, \alpha _n\) such that \(Y = \sigma _x U _{\alpha _1} \cup \cdots \cup \sigma _x U _{\alpha _n}\). In particular, we have \(\left\lbrace x \right\rbrace \times Y\subseteq U _{\alpha _1}\cup \cdots cup U _{\alpha _n} =: V_x\). This \(V_x\) is a finite union of elements of the open cover we started with.
2. With \(x\) and \(\alpha _1, \ldots, \alpha _n\) as in the previous step, we have \(\pi _X U _{\alpha _j}\) is an open set for each \(j=1,\ldots, n\). Moreover, we may assume without loss of generality that \(x\in \pi _X U _{ \alpha _j} \) for each such \(j\). Thus, if \(W_x = \bigcap _{j=1}^{n} \pi _X U _{\alpha _j}\), \(W_x\) is an open subset of \(X\) containing \(x\). Going back up, we have \(W_x \times Y \subseteq V_x\).
3. Clearly \(X = \bigcup _{x\in X} W_x\). Since \(X\) is compact, there exist finitely many indices \(x_1,\ldots, x_n\) such that \(X = W _{x_1}\cup \cdots \cup W _{x_n}\).
4. Finally, we have \[X\times Y \subseteq \bigcup _{j=1}^{n} \left(W _{x_j} \times Y\right)\subseteq \bigcup _{j=1}^{n} V _{x_j}. \] Each \(V \_{x_j}\) is a finite union of \(\left\lbrace U_ \alpha \right\rbrace\)’s, and it follows that \(X\times Y\) is also a finite union of the \(\left\lbrace U_ \alpha \right\rbrace\)’s. \(\square\)

We now know something very powerful: every closed and bounded subset of \(\mathbb{R}^n\) is a compact set! We abstracted away all the open cover business in the above arguments. Note that this is only true in \(\mathbb{R}^n\) (with the standard metric of course); do not make the mistake of saying that closed and bounded subsets of arbitrary metric spaces are compact. We’re secretly using the fact that closed bounded intervals are compact when we make this assertion.

By the way, we used a very cheeky and thematic technique in the above proof. To each point \(x\in X\), we constructed some open set \(W_x\) containing \(x\), then said \(X=\bigcup _{x\in X} W_x\) and used compactness. Keep this trick in your back pocket! This is really useful for taking a property that is true locally (that is, near each point) and turning it into a property that’s true globally (i.e. for every point). We will almost surely see this when we talk about continuous functions.