notes / Math 55b / chapter 3: Completeness, Separation and Quotients
9 Completeness and Compactifications
· Denis Auroux
Recap: compactness in metric spaces
Definition 9.1.
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.
Theorem 9.2. For metric spaces, the 3 notions are equivalent.
We proved this except for a lemma (Lemma 2) which remained to prove:
Lemma 9.3 (Lebesgue numbers). If is a sequentially compact metric space, then every open cover has a Lebesgue number ( such that any subset of diameter is entirely in one ).
(We’ve seen this hold for compact metric spaces, so it should hold!)
Proof. Suppose there exists an open cover with no Lebesgue number, i.e. with diameter which isn’t contained in any single . Take .
By sequential compactness, there exists a subsequence of that converges to some . Now for some , and so 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.”
Completeness
Theorem 9.4. Every compact metric space is complete, i.e. every Cauchy sequence converges.
Proof. Let be a Cauchy sequence. By sequential compactness there exists a subsequence . Now such that ,, and such that . Hence ,∎
Corollary 9.5., (with the usual distances) are complete.
Proof. Every Cauchy sequence is bounded, hence contained in a compact subset, hence convergent. ∎
Corollary 9.6. with the uniform metric is complete.
Proof. Given a Cauchy sequence (i.e. such that ):, is a Cauchy sequence in (since ), hence converges to some limit (i.e. we have a pointwise limit).
Now: given , take such that . Then ,,
i.e. ,, which implies uniformly. ∎
Note. 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 (Munkres §29)
Definition 9.7. 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 is the subspace topology of ), with open and dense in ().
Example 9.8. as in HW2; this is in fact homeomorphic to (the unit sphere in ).
This is not the only option: e.g. compactifies to or to ; and compactifies e.g. to , to , or to the torus ().
The one-point compactification
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 we define:
Definition 9.9.
Theorem 9.10. is a topology on , and is a compactification of (in particular, is compact), except: fails when is compact! Also, Hausdorff doesn’t necessarily imply Hausdorff.
Proof.
Axioms of a topology: case by case for ’s and ’s. Arbitrary unions and finite intersections of a single type of open set are still of the same type (note: , and a finite union of compact subsets of is compact). Moreover,
where is closed in , hence compact. ✓
is compact: if is an open cover of , then for some , and now the form an open cover of , so such that . Thus is a finite subcover.
∎
However, this is not always Hausdorff! One-point compactifications are only useful if Hausdorff.
Definition 9.11. is locally compact if , compact which contains a neighborhood of .
Example 9.12. is locally compact (), and so is . isn’t (for any of the usual topologies). Neither is with the usual topology (as a subspace of ).
Theorem 9.13. The one-point compactification is Hausdorff iff is locally compact and Hausdorff.
Proof.
Hausdorff we can separate points of by open subsets (in or in ).
locally compact open sets and such that , i.e. we can separate points of from by open subsets in .
∎
Notes not reached in this lecture, covered at the start of Lecture 10
Countability axioms (§30)
Definition 9.14. is first-countable if , a countable basis of neighborhoods at , i.e. open such that every neighborhood contains one of the .
Example 9.15. Metric spaces are first-countable: at , take .
In a first-countable space, a sequence with . (Otherwise only holds.)
Definition 9.16. is second-countable if its topology has a countable basis.
Example 9.17. is second-countable, e.g. basis or .
with the product topology is second-countable (basis = products of a finite number of ,, where all remaining factors are ), while the uniform topology isn’t (because there exist uncountably many disjoint open subsets: the balls of radius centered at the points of ).
Second-countable a countable dense subset (e.g. take one point in each basis open set!). The converse holds for metric spaces (take balls of radius around points of the dense subset) but is false in general: is first-countable and has a countable dense subset, but a countable basis (so isn’t metrizable; HW4).
Regular and normal spaces (§31–32)
Recall: is Hausdorff := we can separate points: ,, disjoint open (aka ).(: is closed .)
Stronger separation axioms:
Definition 9.18. Suppose one-point subsets are closed (). Then say:
is regular if , closed disjoint from , disjoint open sets ,.
is normal if disjoint closed subsets, disjoint open sets ,.
(The first implication: will see.)
Example 9.19. is normal but not metrizable; is regular but not normal. See Munkres §31 for these and more.
Theorem 9.20.
regular + second-countable normal
Hausdorff + compact normal
We won’t prove this, cf. Munkres §32. We did see, when proving that compact subsets of Hausdorff spaces are closed, that compact + Hausdorff regular; normal was on HW2 (Munkres 26.5).
Theorem 9.21. Metric spaces are normal.
Proof. Let be disjoint closed subsets of .
, such that ., such that . Let
(clearly open: unions of balls). We claim . Indeed, if then , such that . So
But this is a contradiction (e.g. if then !).∎
We can now ask which topological spaces are metrizable. We’ve seen: metrizable first-countable and normal (; counterexample: ).
Theorem 9.22 (Urysohn metrization theorem). If is regular and has a countable basis, then it is metrizable.
(The first condition is necessary; the second one is stronger than needed. The Nagata–Smirnov theorem gives a sharper criterion but is more technical to state and prove.)
Urysohn’s lemma
Urysohn’s lemma is the key ingredient in the proof of the metrization theorem.
Theorem 9.23 (Urysohn’s lemma). Let be a normal space and disjoint closed subsets. Then continuous such that and .
Idea:
Construct open sets such that , and moreover . Also set for .
Define , and show is continuous.
Step 1 uses the following reformulation of normality:
Lemma 9.24. is normal closed, open, open such that and . (In fact .)
Proof. and are disjoint closed sets, so since is normal, , open such that . Moreover, is closed and , so . So
∎
Proof of Urysohn’s lemma
Proof.Step 1. Given and disjoint closed, let , and let be open such that .
Next, we construct ,, such that , by induction. Choose a labelling of by an infinite sequence such that and (could e.g. continue: ).
Assuming have already been chosen, we construct using the above lemma: let
so and none of the rationals already considered lie in between. Then by the induction hypothesis, , hence using normality open such that ; let .
By induction, we construct in this way all the ’s, and indeed .
We also set if and if (still true: !).
Step 2. Define , where .
Since and , we have , so . Also, , and and .
So it only remains to show that is continuous! For this, observe:
: indeed, if then , so .
: indeed, if then .
Now given an open interval , we show is open in . Assume , and let such that .
By the above observation, and . So is open, and a neighborhood of . Moreover,
Hence , i.e. contains neighborhoods of its points. ∎
Next we will prove the metrization theorem, namely that if is normal and has a countable basis, then is metrizable.