# Gracious Living

Compactness and the Heine-Borel Theorem
November 29, 2010, 20:57
Filed under: Math, Topology | Tags: , ,

Most of the topological work we’ve been doing has been in the area of constructing new topologies. We’re now ready to move on and look at their properties. We’re looking in particular for properties that are intrinsic to the topology: we want them to be preserved under homeomorphism (and arbitrary continuous maps, if possible) and not depend on other structures like a metric, a vector space structure, or a specific basis. Of course, it’s always nice to apply topological properties to such specific structures.

We’ve seen a couple examples, importantly connectedness and path-connectedness (and their local versions) and metrizability. Boundedness was a non-example — the interval $(0,1)$ is homeomorphic to $\mathbb{R}$ — and in fact we found a bounded metric for any given metric that generated the same topology. But the idea of a space being “small” is nevertheless compelling. In $\mathbb{R}$, for example, the Extreme Value Theorem states that continuous functions on a closed interval $[a,b]$ attain both a minimum and a maximum, as opposed to approaching either asymptotically. This and other related theorems make doing calculus on closed intervals really nice. All these properties of closed intervals derive from a single topological one: compactness.

(Be warned: this is a pretty long post!  Mostly because I give a very careful proof of a theorem I consider important.)

A cover of a space $X$ is a set of sets $\{U_\alpha\}_{\alpha\in A}$ such that $\bigcup_{\alpha\in A}U_\alpha=X$. An open cover is a cover all of whose elements are open sets. Occasionally we’ll instead deal with a subspace $A\subset X$ and say that an open cover of $A$ is a set of open sets $\{U_\alpha\}_{\alpha\in A}$ in $X$ such that $\bigcup_{\alpha\in A}U_\alpha$ includes $A$. Since $\{U_\alpha\cap A\}$ is then an open cover of $A$ by the first definition, and we can likewise go backwards from open sets in the subspace topology of $A$ to open sets in $X$, the distinction is immaterial.

A subcover of a given cover $\{U_\alpha\}$ is a subset that also covers $X$. We can now formulate our first definition:

• Compactness. A space $X$ is compact if every open cover has a finite subcover.

As easy examples, finite sets are compact, and sets with the cofinite topology (but not the cocountable topology) are compact. The set $\left\{1,\frac{1}{2},\frac{1}{3},\dotsc,0\right\}$ is compact as a subspace of $\mathbb{R}$ because any open set containing $0$ also has to contain all but finitely many of the positive elements, and we finish the subcover off by adding in (finitely many) sets that cover the remaining elements. $\mathbb{R}$ is not compact because $\{(n-1,n+1):n\in\mathbb{Z}\}$ is an open cover with no finite subcover. Likewise, $(0,1)$ is not compact because the set $\{(1/n,1):n\in\mathbb{N}\}$ is an open cover with no finite subcover.

Subspaces of compact spaces aren’t necessarily compact, but closed subspaces are. As a proof, let $C\subset X$, with $X$ compact, and let $\{U_\alpha\}$ be an open cover of $C$ (with $\{U_\alpha\}$ open in $X$). Then $\{U_\alpha\}\cup\{X-C\}$ is an open cover of $X$, with $X-C$ open since $C$ is closed. This has a finite subcover; throwing out $X-C$ gives a finite subset of the $U_\alpha$ covering $C$.

Likewise, compactness is preserved by a continuous function. This is probably my favorite theorem in topology. Let $f:X\rightarrow Y$ be continuous and let $\{U_\alpha\}$ be an open cover of $Y$. Then $\{f^{-1}(U_\alpha)\}$ is an open cover of $X$. If $X$ is compact, then we can find a finite subcover, and applying $f$ again gives a finite cover of $Y$. This is simple, but so insanely useful. For instance, the Extreme Value Theorem is a corollary: any continuous function $\mathbb{R}\rightarrow\mathbb{R}$ maps the compact set $[a,b]$ onto a compact subset of $\mathbb{R}$, which has a maximum and minimum by what we’re about to prove.

Theorem (Heine-Borel). A subspace of $\mathbb{R}^n$ is compact iff it is closed and bounded.

Proof. The set of cubes $[-a,a]^n$ for $a\in (0,\infty)$ covers every set, so if a set is compact, it must be covered by a finite set of such cubes. The largest of these gives a bound for the set.

Now suppose that $C$ is compact but not closed. Let $p$ be a limit point of $C$ that is not in $C$. For every $q\in C$, let $U_q$ be a neighborhood of $q$ disjoint from some neighborhood $V_q$ of $p$. Then any finite cover $\{U_{q_1},\dotsc,U_{q_n}\}$ is disjoint from $V_{q_1}\cap\dotsb\cap V_{q_n}$, which is also a neighborhood of $p$, because it is a finite intersection of open sets. But then we have a neighborhood of $p$ disjoint from $C$, contradicting $p$‘s limit point status. Thus, compact subspaces of $\mathbb{R}^n$ are closed and bounded.

Finally, let $D$ be closed and bounded. Then $D\subset [-a,a]^n$ for some $a$. If $[-a,a]^n$ is compact, then $D$, a closed subspace, is as well. So we only have to prove that cubes are compact!

The final step is a bit trickier. Let $K_0=[-a,a]^n$ be covered by $\mathcal{U}=\{U_\alpha\}$, an open cover without a finite subcover. By bisecting the cube in all the cardinal directions, we get $2^n$ smaller cubes, each also covered by elements of the cover. At least one of these cubes can’t be covered by a finite subcover; call it $K_1$. Inductively, let $K_i$ be a subcube of $K_{i-1}$ of side length $2^{-i}(2a)$, such that no finite subcover of $\mathcal{U}$ covers it. Then $K_0\supset K_1\supset K_2\supset\dotsb$, in order of decreasing size.

We want to show that the intersection of all $K_i$ is just one point. If we can do this, we’re done, and here’s why. Choose an element $U_\alpha$ of $\mathcal{U}$ containing this point. Since all elements of $\mathcal{U}$ are open, $U_\alpha$ contains some open cube $\prod_{i=1}^n (c_i,d_i)$ (as these cubes form a basis for the Euclidean topology). This cube has some finite side length $s$, which is greater than $2^{-i}(2a)$ for all but finitely many $i$. So this cube, and hence $U_\alpha$, contains all but finitely many of the $K_i$. This is a contradiction, since we assumed that each $K_i$ needed infinitely many elements of $\mathcal{U}$ to cover it. This proves that closed and bounded subspaces of $\mathbb{R}^n$ are compact. $\square$

For the final step, a lot of people will end up using some disguised form of compactness to prove that there’s a point in $\bigcap_i K_i$. Don’t let them! One insidious example is the finite intersection property, which states that if $\{A_\alpha\}$ are a collection of closed sets such that any intersection of finitely many of them is nonempty, then the intersection of all of them is nonempty. This is exactly equivalent to compactness, just using closed sets. See, $\{X-A_\alpha\}$ is then a collection of open sets such that no finite subset covers $X$, and the conclusion is then that the collection itself does not cover $X$, which is the contrapositive of the compactness property.

Fortunately, we can use properties of $\mathbb{R}^n$ to prove that this property holds, at least in the very specific case where $\alpha\in\mathbb{N}$ and $A_\alpha=[a_\alpha,b_\alpha]$ is a closed interval such that $A_\alpha\subset A_{\alpha-1}$. I prefer to use $\alpha$ for uncountable index sets, so I’m changing to $i$.

Lemma. Let $\{[a_i,b_i]\}$ be a nested sequence of closed intervals. Then $\bigcap_i [a_i,b_i]$ is nonempty. If moreover $\lim_{n\rightarrow\infty} b_i-a_i=0$, then the intersection is just one point.

Proof. For the first part, observe that we have two sequences: $a_1,a_2,\dotsc,$ which is increasing, and $b_1,b_2,\dotsc$, which is decreasing. For any $i$, $a_i\le b_i$, and then for any $i,j$, if $i\le j$, then $a_i\le a_j\le b_j$, and if $j\le i$, then $a_i\le b_i\le b_j$. So for any $i,j,$latex a_i\le b_j$. In particular, $a_i\le\inf_j b_j$ for all $i$ (recall that the infimum of a set is its greatest lower bound, and the supremum of a set is its least upper bound). Then $\sup_i a_i\le\inf_j b_j$. If we let $a=\sup_i a_i,b=\inf_j b_j$, then $[a,b]$ is nonempty. But $a_i\le a\le b\le b_i$ for all $i$, so $[a,b]\subset\bigcap_i [a_i,b_i]$. This proves the first part of the lemma. For the second part, notice also that if $c, then $c for some $i$ by the definition of supremum (otherwise $c$ would be a smaller upper bound). Similarly, if $c>b$, then $c>b_j$ for some $j$. In either case, $c\not\in\bigcap_i [a_i,b_i]$. So $\bigcap_i [a_i,b_i]$ is actually equal to $[a,b]$. Now we have to invoke the definition of sup again. The limit point of the sequence $(a_i)$, which we’ll call $a^\prime$, has some$a_i$in $B_\epsilon(a^\prime)$ for each$\epsilon$. We must have $a^\prime>a_i$ for all $i$, for otherwise, $a_{i-1}a^{\prime\prime}>a_i$ for all$i$, then $(a^{\prime\prime},\infty)$ is a neighborhood of $a^\prime$ not intersecting the sequence. Thus, $\lim a_i=a^\prime=\sup a_i=a$. Similarly, $\lim b_j=\inf b_j=b$. But $\lim(b_i-a_i)=0$, so $\lim b_i=\lim a_i$, meaning $a=b$. Then $\bigcap [a_i,b_i]=[a,a]=\{a\}$. $\square$ In proving this, we used a property of metric spaces that’s very special to $\mathbb{R}^n$ called completeness. This is going to get its own post, but for $\mathbb{R}$, it basically says that any set that’s bounded above has a sup, and any set that’s bounded below has an inf. We can actually use completeness to give a really easy proof of the Heine-Borel theorem for $\mathbb{R}$: given $[a,b]$ with an open cover $\{U_\alpha\}$, let $c$ be the least upper bound of the set $\{d:[a,d]\text{ is covered by finitely many }U_\alpha\}$. But then $c\in U_\alpha$ for some $\alpha$, so the same open cover actually covers $[a,c+\epsilon]$ for some small $\epsilon$. This is a contradiction unless $c=b$. The problem is that this doesn’t extend to $\mathbb{R}^n$ for $n>1$, because these sets don’t have an order giving them the order topology. We could extend it if we could prove that finite products of compact spaces are compact. This is true: in fact, using the axiom of choice, arbitrary products of compact spaces are compact. But it’s difficult to prove, so it’s going to get its own post. Meanwhile, what we proved extends naturally to a complete proof of the Heine-Borel theorem. The final step is as follows: Given our sequence of nested cubes $(K_i)$, projecting onto each coordinate gives $n$ sequences of nested intervals $([a_{ij},b_{ij}])$, where $1\le j\le n$ is the coordinate and $i$ indexes the sequence. By the lemma, each of these sequences has a single point $a_j$ in its intersection. Then $\{(a_1,a_2,\dotsc,a_n)\}$ is the intersection of the cubes. The rest of the proof, listed above, follows. This monumental theorem about $\mathbb{R}^n$ is only one facet of compactness. Soon, we’ll prove the Bolzano-Weierstrass theorem and use it to investigate sequential compactness, the property that all sequences have convergent subsequences. The Bolzano-Weierstrass and Heine-Borel theorems together generalize to the Arzela-Ascoli theorem, which says “the same thing” for topological spaces of functions on $\mathbb{R}^n$. In addition, there are weaker statements that look like compactness and are worth a look: countable compactness says that every countable open cover has a finite subcover, and limit point compactness says that every infinite set has a limit point. Finally, I’d like to prove the very cool Tychonoff’s theorem, that products of compact spaces are compact; to do this, I’ll need to introduce nets, which generalize sequences. Stay tuned! Advertisements 3 Comments so far Leave a comment […] to topology. The interesting thing about compactness, as I see it, is that its definition isn’t very intuitive. We want to talk about what are […] […] axioms tell us in what ways it is possible to break our space apart into pieces. The compactness axioms tell us how bounded the space is. What we’re going to look at today is a set of […] Pingback by Countability Axioms « Gracious Living Thank you for the great post. Now suppose that C is compact but not closed. Let p be a limit point of C that is not in C. … Thus, compact subspaces of \mathbb{R}^n are closed and …. I believe that this argument does not need limit points. Use the fact that$A$is closed iff$\operatorname{cl}(A)\subseteq A$and a fact that$a\in \operatorname{cl}(A)$iff every open neighborhood of$a$meets$A\$. Link in: Hu, Sze-Tsen. “Elements of general topology.” Corollary 2.5.

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