notes / Math 55b / chapter 2: Connectedness and Compactness
8 Compactness in Metric Spaces
· Denis Auroux
Last time we saw a number of foundational results about compactness. In metric spaces, compactness implies uniform estimates.
Lemma 8.1 (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 .
This uses the notion of distance to a subset: recall that given ,, we define
Since is continuous, when is compact the inf is achieved, by the extreme value theorem. Moreover, is continuous (the triangle inequality gives ).
Proof of the Lebesgue number lemma. By compactness, we can assume is a finite cover . The function
is continuous (check: distance to a subset is a continuous function of , as above), 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 8.2. is uniformly continuous if for all there is such that for all ,.
(Compare with continuity: the same must work for every !)
Theorem 8.3. 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 8.4.
is compact if every open cover of has a finite subcover.
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 8.5. In , has a limit point (), and the sequence has a convergent 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 8.6. 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. ∎
Theorem 8.7. sequentially compact limit point compact.
Proof. Given an infinite subset , pick a sequence of distinct points of and take a convergent subsequence: this gives a sequence in , for all , converging to some limit . Then every neighborhood of contains for all large , hence infinitely many points of , including some . So is a limit point of .∎
The converse implications don’t hold in general, but in metric spaces the 3 notions agree:
Theorem 8.8. For a metric space : compact limit point compact sequentially compact.
Proof.
Compact limit point compact: already done (for all topological spaces).
Limit point compact sequentially compact: suppose is a metric space and limit point compact, and consider a sequence in . If is finite, then there is such that for infinitely many , which gives a subsequence that converges to .
Otherwise, is infinite, so it has a limit point . So: for all there is such that . First choose such that ; then inductively, given , let
Then take such that . By construction, and . So is a subsequence converging to .
Sequentially compact compact: this is the hardest part. First we show two lemmas.
∎
Lemma 8.9 (Covering by ε-balls). If a metric space is sequentially compact, then for every , can be covered by finitely many open balls of radius .
(As we expect if is to be compact: should have a finite subcover!)
Proof. Assume not, and choose ; then inductively choose . This yields a sequence in , which by sequential compactness must have a convergent subsequence. But this is impossible, since no two terms of the sequence are within of each other! Contradiction. ∎
Notes not reached in this lecture, covered at the start of Lecture 9
Lemma 8.10 (Lebesgue numbers). If a metric space is sequentially compact, then every open cover has a Lebesgue number (there is such that any subset of diameter is entirely in one ).
(We’ve seen this holds for compact metric spaces, so it should hold!)
Proof. Suppose there is an open cover with no Lebesgue number, i.e. for every there is with diameter which isn’t contained in any single . Take .
By sequential compactness, there is a subsequence of that converges to some . Now for some , and so there is such that . Take sufficiently large so that and.
Since has diameter ,, contradiction. ∎
(Remark: this proof illustrates how arguments using sequential compactness are often more intuitive than those involving open covers: “if some property fails to hold uniformly, take a sequence of points where things get worse, extract a convergent subsequence, and see what goes wrong at the limit.”)
Now we can prove sequentially compact compact:
Proof. Given an open cover , by the Lebesgue numbers lemma there is such that every subset of diameter is entirely inside a single . Fix : by the ε-ball lemma, is covered by finitely many -balls. Each of these has diameter , so is contained in some . This gives a finite subcover, replacing each -ball by one containing it (and discarding the rest of the ’s). ∎
Completeness
Theorem 8.11. Every compact metric space is complete, i.e. every Cauchy sequence converges.
Proof. Let be a Cauchy sequence. By sequential compactness there is a subsequence . Now for all there is such that for all ,, and there is such that . Hence for all ,.∎
Corollary 8.12., (with the usual distances) are complete.
Proof. Every Cauchy sequence is bounded, hence contained in a compact subset, hence convergent. ∎
Corollary 8.13. with the uniform metric is complete.
Proof. Given a Cauchy sequence (i.e. for all there is such that ): for all , is a Cauchy sequence in (), hence converges to some limit (i.e. we have a pointwise limit).
Now: given , take such that . Then for all and all ,
i.e. for all ,, which implies uniformly. ∎
(When is a topological space, we’ve seen that uniform limits of continuous functions are continuous, so we also have completeness of , uniform topology. More generally: closed subsets of complete metric spaces are complete!)
Compactification
Definition 8.14. A compactification of a (Hausdorff) topological space is a compact (Hausdorff) space with an inclusion which is an embedding (i.e. a homeomorphism onto its image, i.e. the topology on = the subspace topology of ), with open and dense in ().
Example 8.15. as in HW2; this is in fact homeomorphic to (the unit sphere in ).
This is not the only option: e.g. compactifies to or , and to e.g. ,, or the torus ():
The one-point compactification, if it exists, is unique.
Let (add a new point). The requirements of a compactification imply:
a subset is open in iff it is open in (subspace topology );
a subset containing is open in iff is closed, hence compact (we want compact), and a subset of (since ).
So:
Definition 8.16.
Theorem 8.17. For Hausdorff, is a topology on , and is a compactification of (in particular, is compact), except that fails when is compact!
Proof: next time.
However: even if is Hausdorff, it doesn’t follow that is Hausdorff. We’ll see a criterion next time.