is compact if every open cover of admits a finite subcover, i.e. there are such that .
is compact.
Theorem 7.1. If is compact and is continuous, then (with the subspace topology) is compact.
(Remark: an open cover of with the subspace topology open, .)
Proof. Let be an open cover of . Then is an open cover of , hence there are such that . So for all ,, i.e. , a finite subcover. ∎
Taking , once we know subsets of () are compact iff closed and bounded, this gives the extreme value theorem.
Theorem 7.2. compact, closed is compact.
Proof. Given an open cover of , i.e. open with , let , which is open: then is an open cover of , hence there is a finite subcover. Discarding , this gives a finite subcover for .∎
The converse is true in Hausdorff spaces!
Theorem 7.3. Hausdorff, compact is closed in .
Proof. We show that is open. Let . Since is Hausdorff, for every there are disjoint open subsets ,.
Now is an open cover, so by compactness there are such that .
Let , which is open. Then , so .
Hence: for all , there is an open such that .
∎
(If we tried this for an arbitrary subset of , we’d find that isn’t a neighborhood of anymore. Compactness lets us reduce an infinite process to a finite one.)
Remark. We’ve actually shown more: Hausdorff, compact, there exist disjoint open subsets ,,. I.e.: we can separate points from compact subsets!
Remark. When isn’t Hausdorff, compact closed in : e.g. with the finite complement topology: any subset is compact. Indeed, a nonempty open subset contains all but finitely many points, so given an open cover it is easy to find a finite subcover: take one nonempty , with finite complement , then take containing for .
Another instance of compactness allowing us to intersect infinitely many opens (or rather, reduce to a finite intersection) is the tube lemma:
Proposition 7.4 (Tube lemma). Let be a topological space, a compact topological space, . If is open and , then there exists a neighborhood of in such that .
Proof. For all , open, so there is a basis open set , a neighborhood of in , a neighborhood of in , such that .
Now is an open cover. (Remark: , but is not open!)
Since is compact, there are such that . Let . Then is a neighborhood of in , and .∎
As a consequence:
Theorem 7.5. compact is compact.
Proof. Let be an open cover of . For any given , is compact, so there is a finite subcollection which suffices to cover .
is open, so by the tube lemma there is a neighborhood in such that . Now is compact, and form an open cover, so there are such that .
Now ,,, is a finite subcover for .∎
Compact subsets of
Theorem 7.6. is compact iff is closed and bounded.
Proof.
If is compact then it is closed (by the theorem above: is Hausdorff) and bounded: is an open cover, so there is a finite subcover, so there is such that .
If is closed and bounded, then it’s a closed subset of for some . is a finite product of compact sets (), hence compact; and a closed subset of a compact space is compact.
∎
Remark.
Closed and bounded are necessary conditions for compactness of a subspace of any metric space (HW!), but in “most” metric spaces, closed + bounded compact.
There are easy counterexamples (find one for HW!). More interesting: let be any infinite-dimensional vector space with a norm, . E.g. , continuous functions with the sup norm (uniform topology). Then is closed and bounded but never compact. (The proof uses sequential compactness.) (Don’t use this for Munkres 26.4.)
Applications of compactness
We now look at applications of compactness. We’ve seen:
Theorem 7.7. If is compact and is continuous, then (with the subspace topology) is compact.
Corollary 7.8 (Extreme value theorem). compact (nonempty), continuous attains its max and min.
Example 7.9. Let be a metric space, nonempty, . Define
(the distance of to the subset ). If is compact then the inf is always achieved! See HW3 Problem 1 = Munkres 27.2.
Similarly, the diameter of a bounded subset is . The sup is attained for compact ( is continuous, so achieves its max).
Another corollary:
Corollary 7.10. If is compact and is Hausdorff, then any continuous bijection is a homeomorphism.
Proof. We need to check is continuous as well (so open open).
open closed, hence compact compact. Since is Hausdorff, this implies is closed, i.e. is open in .∎
(We’ve seen that without such assumptions a continuous bijection need not be a homeomorphism, e.g. ,.)
The Lebesgue number lemma and uniform continuity
In metric spaces, compactness implies uniform estimates.
Notes not reached in this lecture, covered at the start of Lecture 8
Lemma 7.11 (Lebesgue number lemma). Let be a compact metric space and an open cover of . Then there is such that any subset of diameter is entirely contained in a single open .
Proof. By compactness, we can assume is a finite cover . The function
is continuous (check: distance to a subset is a continuous function), so it achieves its min, which is therefore (for every there is such that , and then , as is a closed set, hence compact). Hence there is such that for all . Thus for every there is such that , i.e. . Since a subset of diameter is contained in a ball of radius , the result follows. ∎
This is the magic of compactness!
Noncompact counterexamples (where there is no Lebesgue number):
intervals with overlaps of lengths , e.g. with .
, with containing the hyperbola , containing the -axis, and the distance between the two curves :
Uniform continuity
This only makes sense for metric spaces! There is no notion of uniform size of neighborhood without a metric.
Definition 7.12. is uniformly continuous if for all there is such that for all ,.
(Compare with continuity: the same must work for every !)
Theorem 7.13. If and are metric spaces, is continuous, and is compact, then is uniformly continuous.
Proof. Take , and consider the open cover of by balls of radius (so if land in the same ball, they’re less than apart). Then is an open cover, so by the Lebesgue number lemma there is such that if then and lie in the same element of the cover, hence .∎
Alternative notions of compactness
Definition 7.14.
is limit point compact if every infinite subset of has a limit point.
is sequentially compact if every sequence in has a convergent subsequence.
Example 7.15. In , has a limit point (), and the sequence has a convergent subsequence (); so does (e.g. the subsequence ).
But has no limit point and the sequence doesn’t have a convergent subsequence, so is neither limit point compact nor sequentially compact.
Theorem 7.16. is compact is limit point compact.
Proof. Assume is not limit point compact, i.e. there is an infinite with no limit point. Since contains all of its limit points (there are none), is closed in , hence compact. However, for all , isn’t a limit point, so there is a neighborhood such that . is now an infinite open cover of , without any finite subcover, since each only belongs to and not to any other element of the cover. Contradiction. ∎
Next time we’ll see:
Theorem. sequentially compact limit point compact.
Theorem. For a metric space : compact limit point compact sequentially compact.