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 is homeomorphic to — 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 , for example, the Extreme Value Theorem states that continuous functions on a closed interval 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 is a set of sets such that . An **open cover** is a cover all of whose elements are open sets. Occasionally we’ll instead deal with a subspace and say that an **open cover** of is a set of open sets in such that includes . Since is then an open cover of by the first definition, and we can likewise go backwards from open sets in the subspace topology of to open sets in , the distinction is immaterial.

A **subcover** of a given cover is a subset that also covers . We can now formulate our first definition:

**Compactness**. A space 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 is compact as a subspace of because any open set containing 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. is not compact because is an open cover with no finite subcover. Likewise, is not compact because the set is an open cover with no finite subcover.

Subspaces of compact spaces aren’t necessarily compact, but *closed* subspaces are. As a proof, let , with compact, and let be an open cover of (with open in ). Then is an open cover of , with open since is closed. This has a finite subcover; throwing out gives a finite subset of the covering .

Likewise, compactness is preserved by a continuous function. This is probably my favorite theorem in topology. Let be continuous and let be an open cover of . Then is an open cover of . If is compact, then we can find a finite subcover, and applying again gives a finite cover of . This is simple, but so insanely useful. For instance, the Extreme Value Theorem is a corollary: any continuous function maps the compact set onto a compact subset of , which has a maximum and minimum by what we’re about to prove.

**Theorem** (Heine-Borel). *A subspace of is compact iff it is closed and bounded.*

**Proof**. The set of cubes for 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 is compact but not closed. Let be a limit point of that is not in . For every , let be a neighborhood of disjoint from some neighborhood of . Then any finite cover is disjoint from , which is also a neighborhood of , because it is a finite intersection of open sets. But then we have a neighborhood of disjoint from , contradicting ‘s limit point status. Thus, compact subspaces of are closed and bounded.

Finally, let be closed and bounded. Then for some . If is compact, then , a closed subspace, is as well. So we only have to prove that cubes are compact!

The final step is a bit trickier. Let be covered by , an open cover without a finite subcover. By bisecting the cube in all the cardinal directions, we get 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 . Inductively, let be a subcube of of side length , such that no finite subcover of covers it. Then , in order of decreasing size.

We want to show that the intersection of all is just one point. If we can do this, we’re done, and here’s why. Choose an element of containing this point. Since all elements of are open, contains some open cube (as these cubes form a basis for the Euclidean topology). This cube has some finite side length , which is greater than for all but finitely many . So this cube, and hence , contains all but finitely many of the . This is a contradiction, since we assumed that each needed infinitely many elements of to cover it. This proves that closed and bounded subspaces of are compact.

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 . Don’t let them! One insidious example is the **finite intersection property**, which states that if 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, is then a collection of open sets such that no finite subset covers , and the conclusion is then that the collection itself does not cover , which is the contrapositive of the compactness property.

Fortunately, we can use properties of to prove that this property holds, at least in the very specific case where and is a closed interval such that . I prefer to use for uncountable index sets, so I’m changing to .

**Lemma**. *Let be a nested sequence of closed intervals. Then is nonempty. If moreover , then the intersection is just one point.*

**Proof**. For the first part, observe that we have two sequences: which is increasing, and , which is decreasing. For any , , and then for any , if , then , and if , then . So for any latex a_i\le b_j$. In particular, for all (recall that the infimum of a set is its greatest lower bound, and the supremum of a set is its least upper bound). Then . If we let , then is nonempty. But for all , so . This proves the first part of the lemma.

For the second part, notice also that if , then for some by the definition of supremum (otherwise would be a smaller upper bound). Similarly, if , then for some . In either case, . So is actually equal to .

Now we have to invoke the definition of sup again. The limit point of the sequence , which we’ll call , has some $a_i$ in for each $\epsilon$. We must have for all , for otherwise, for all $i$, then is a neighborhood of not intersecting the sequence. Thus, . Similarly, . But , so , meaning . Then .

In proving this, we used a property of metric spaces that’s very special to called **completeness**. This is going to get its own post, but for , 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 : given with an open cover , let be the least upper bound of the set . But then for some , so the same open cover actually covers for some small . This is a contradiction unless .

The problem is that this doesn’t extend to for , 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 , projecting onto each coordinate gives sequences of nested intervals , where is the coordinate and indexes the sequence. By the lemma, each of these sequences has a single point in its intersection. Then is the intersection of the cubes. The rest of the proof, listed above, follows.

This monumental theorem about 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 . 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!

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[…] 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 […]

Pingback by Limit Point and Sequential Compactness « Gracious LivingDecember 30, 2010 @ 02:51[…] 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 LivingDecember 31, 2010 @ 01:55Thank 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.

Comment by beroalJuly 16, 2012 @ 14:30