Definition 5.2.1.
An open cover for the set \(A\) is a collection of open sets \(\mathcal{U}=\{U_\alpha\}_{\alpha\in\Lambda}\) so that \(A\subseteq \bigcup \mathcal{U}\text{.}\)
Section 5.2 compact sets
Our next topological notion is that of compactness, which has to do with (in some sense) the size of a set.
Think about the intervals \([1,3]\) and \([1,\infty)\text{.}\) They’re both closed sets. They’re both intervals. But \([1,3]\) has a feeling of finiteness about it. (For example, its length is finite.) But the interval \([1,3]\) is not finite as a set -- it’s got infinitely many elements.
We’d like to articulate a distinction between the "finite-feeling" sets like \([1,3]\) and "infinite-feeling" sets like \([1,\infty)\text{.}\) (If you want to think about notions like "length" of a set, you’ll get there maybe at the end of the next analysis course, or the beginning of a course in measure theory. But we’re not ready for this yet.)
Subsection 5.2.1 open covers
We need a preliminary notion first:
Example 5.2.2.
Set \(U_k=(k-1,k+1)\) and \(V_k=(k,k+1)\text{.}\) Then \(\left\{U_k\middle\vert k\in\mathbb{Z}\right\}\) is an open cover for \(\mathbb{R}\text{,}\) whereas \(\left\{U_k\middle\vert k\in\mathbb{N}\right\}\) and \(\left\{V_k\middle\vert k\in\mathbb{Z}\right\}\) are not.
Example 5.2.3.
For any metric space \(M\text{,}\) consider the collections
\begin{equation*}
\mathcal{All\ Ball}=\left\{B_r(x)\middle\vert x\in M,r\in(0,\infty)\right\}
\end{equation*}
and, for each \(x\in M\text{,}\)
\begin{equation*}
\mathcal{Ball}_x=\left\{B_r(x)\middle\vert r\in(0,\infty)\right\}\ \ .
\end{equation*}
These are open covers for \(M\text{.}\)
Example 5.2.4.
For any metric space \(M\text{,}\) and any subset \(A\subseteq M\text{,}\) the singleton \(\{M\}\) is an open cover for \(A\text{.}\)
Subsection 5.2.2 subcovers
Generally speaking, open covers are highly redundant -- the open sets overlap so much that you might wonder if we could get away with omitting some of them. Consider \(\mathbb{R}^2\) with its standard 2-norm. The open cover \(\mathcal{All\ Ball}\) has three real parameters -- the radius and each coordinate of the center. So, roughly speaking, \(\mathcal{All\ Ball}\) is the same size as \(\mathbb{R}^3\text{.}\) But we can actually throw away most of the balls in \(\mathcal{All\ Ball}\) and still cover all of \(\mathbb{R}^2\text{:}\)
Example 5.2.5.
The collection
\begin{equation*}
\mathcal{Ball}_{\mathbb{Q}}=\left\{B_r((x,y))\middle\vert r\in \mathbb{Q}\cap (0,\infty),x\in\mathbb{Q},y\in\mathbb{Q}\right\}
\end{equation*}
is an open cover for \(\mathbb{R}^2\) (with the standard 2-norm).
Checkpoint 5.2.6.
Definition 5.2.7.
If \(\mathcal{U}=\{U_\alpha\}_{\alpha\in \Lambda}\) is an open cover for \(A\text{,}\) and \(\Omega\subseteq\Lambda\) is a subset of the index set for \(\mathcal{U}\) so that \(\mathcal{V}=\{U_\alpha\}_{\alpha\in \Omega}\) is an open cover for \(A\text{,}\) we call \(\mathcal{V}\) a subcover of \(\mathcal{U}\text{.}\)
Example 5.2.8.
Example 5.2.9.
Consider \(A=\left[\frac{1}{2},\frac{7}{2}\right]\) as a subset of \(\mathbb{R}\text{,}\) and \(\mathcal{U}=\left\{(k-1,k+1)\middle\vert k\in\mathbb{Z}\right\}\text{.}\) Then \(\mathcal{U}\) is an open cover for \(A\) and \(\mathcal{V}=\{(0,2),(1,3),(2,4)\}\) is a subcover of \(\mathcal{U}\text{.}\)
It turns out that we can always get pare an open cover down to a countable subcover, i.e. one indexed by the natural numbers.
Proposition 5.2.10.
Let \(M\) be a metric space and \(A\subset M\text{.}\) Every open cover for \(A\) has a countable subcover.
The proof of Proposition 5.2.10 is a little beside the point of this course (it belongs to a course in general topology), and isn’t particularly enlightening for us. So I’ve omitted it.
Subsection 5.2.3 compact sets
Example 5.2.9 was chosen to illustrate that sometimes, we get a finite subcover. That’s nice, because generally speaking you can get away with just about anything finitely many times.
Definition 5.2.11.
Example 5.2.12.
Consider the open cover \(\mathcal{U}=\left\{(k-1,k+1)\middle\vert k\in \mathbb{Z}\right\}\) for \(\mathbb{R}\text{.}\) If we remove \((k-1,k+1)\text{,}\) we leave out \(k\text{.}\) So each element of \(\mathcal{U}\) is essential. Since \(\mathcal{U}\) is infinite, we see that \(\mathcal{U}\) has no finite subcovers. Therefore, \(\mathbb{R}\) is not compact.
Example 5.2.13.
The empty set is compact.
As we saw with connectedness, proving a set is noncompact is generally easier than proving it’s compact.
Example 5.2.14.
Any nonempty open interval \((a,b)\) is noncompact.
Proof.
Consider the open cover \(\mathcal{U}=\left\{(a,r)\middle\vert r\lt b\right\}\text{.}\) Since any \(x\in(a,b)\) has some \(x\lt r\lt b\text{,}\) this is an open over for \((a,b)\text{.}\) On the other hand, any finite subcollection \((a,r_1),\ldots,(a,r_K)\) of \(\mathcal{U}\) has a maximum such \(r_j\text{,}\) which we might as well assume is \(r_K\text{;}\) then the union \(\bigcup (a,r_j)=(a,r_K)\) does not hit all of \((a,b)\text{.}\)
Checkpoint 5.2.16.
Write a blueprint for a proof of any claim of the form "If such and such and so on, then \(A\) is compact.".
Proposition 5.2.17.
Any closed interval \([a,b]\) is compact.
Proof.
Consider any open cover \(\mathcal{U}=\{U_\alpha\}_{\alpha\in \Lambda}\text{.}\) We’ll construct a finite subcover. First, we may as well assume that each \(U_\alpha\) intersects \([a,b]\) (we can certainly remove any that don’t from \(\mathcal{U}\)). We may further, using Proposition 5.2.10, assume that \(\mathcal{U}\) is countable and write \(\mathcal{U}=\{U_k\}_{k\in\mathbb{N}}\text{.}\) Further, let’s label this countable subcover so that \(a\in U_1\text{.}\)
Now consider the related collection \(\mathcal{V}=\{V_k=U_1\cup\cdots\cup U_k\}_{k\in\mathbb{N}}\text{.}\) Observe that for any \(j\text{,}\) \(\displaystyle\bigcup_{j=1}^k V_j=\bigcup_{j=1}^kU_j\text{.}\) So \(\mathcal{V}\) is an open cover for \(A\text{.}\) We’ll show first that \(\mathcal{V}\) has a finite subcover.
If \(\mathcal{V}\) didn’t have a finite subcover, we’d be able to find, for each \(k\in\mathbb{N}\text{,}\) some \(x\in[a,b]\cap V_k^c\text{.}\) Because the set \([a,b]\cap V_k^c\) is closed, it contains its infimum, which we’ll call \(x_k\text{.}\) Notice that because the \(V_k\) are nested, \((x_k)_{k\in\mathbb{N}}\) is an increasing sequence.
The Monotone Sequence Theorem says that \(x_k\to x\) for some \(x\in[a,b]\text{.}\) Because \(x\in [a,b]\text{,}\) there is some \(J\in\mathbb{N}\) so that \(x\in V_J\text{.}\) The topological characterization of open set says that in fact, there is some \(N\in\mathbb{N}\) so that \(n\gt N\) guarantees \(x_n\in V_J\text{.}\) But this contradicts the choice of the sequence \(x_n\text{.}\) We conclude that \(\mathcal{V}\) has a finite subcover, say \(V_1,\ldots,V_K\) suffice to cover \(A\text{.}\)
Since \(\displaystyle\bigcup_{j=1}^K V_j=\bigcup_{j=1}^KU_j\text{,}\) this means \(\{U_1,\ldots,U_K\}\) is a finite subcover for \(\mathcal{U}\text{.}\)
Remark 5.2.18.
Using compactness, on the other hand, just amounts to picking out which open cover \(\mathcal{U}\) to start with. For example:
Proposition 5.2.19.
Any compact set is bounded.
Proof.
Let \(A\) be a compact set. Pick \(x\in M\) and consider the open cover \(\mathcal{Ball}_x\) of \(A\) by balls centered at \(x\text{.}\) By compactness, there are finitely many such balls; call them \(B_{r_1}(x),\ldots,B_{r_N}(x)\text{.}\) Set \(B=\max\{r_1,\ldots, r_N\}\text{;}\) then \(A\subseteq B_B(x)\text{.}\)
Proposition 5.2.20.
Any compact set is closed.
Proof.
Let \(A\) be a compact set. To show \(A\) is closed, we need to show \(A^c\) is open. To this end, let \(x\in A^c\text{.}\) Consider the open cover \(\mathcal{U}=\left\{D_r(x)^c\right\}\) of complements of closed balls centered at \(x\text{.}\) By compactness, finitely many of these suffice to cover \(A\text{,}\) say \(\{D_{r_1}(x)^c,\ldots,D_{r_N}(x)^c\}\text{.}\) Let \(\epsilon=\min\{r_1,\ldots,r_N\}\text{.}\) Then for any \(z\in B_\epsilon(x)\text{,}\) \(z\notin D_{r_i}(x)^c\text{,}\) hence \(z\notin A\text{,}\) i.e. \(z\in A^c\text{.}\)
Definition 5.2.21.
We’ll call set which is both closed and bounded a Heine-Borel set.
So we have:
Proposition 5.2.22.
Compact sets are Heine-Borel.
Proposition 5.2.23.
Proof.
Consider an open cover \(\mathcal{U}\) for \(C\text{.}\) Because \(C\) is closed, \(C^c\) is open. Therefore \(\mathcal{U}\cup\{C^c\}\) is an open cover for \(K\text{.}\)
Since \(K\) is compact, finitely many, say \(U_1,\ldots, U_N,C^c\in \mathcal{U}\cup\{C^c\}\text{,}\) suffice to cover \(K\text{.}\) Since \(C\subseteq K\text{,}\) these also cover \(C\text{.}\) Clearly we do not need \(C^c\text{,}\) so \(U_1,\ldots,U_N\) cover \(C\) and we have found our finite subcover of \(\mathcal{U}\text{.}\)
Remark 5.2.24.
The essence of any proof which uses compactness is to pick an appropriate open cover. The essence of any proof which establishes compactness is to start with an arbitrary open cover.
