Urysohn’s theorem: a criterion for a topological space to be metrizable.
Countability axioms
Metrizability requires first-countability.
Definition N17.1. is first-countable if for all there is a countable basis of neighborhoods, i.e. there are open sets such that every neighborhood of contains one of the .
Counterexample: . Metric spaces are first-countable:
In a first-countable space, there is a sequence with .
A second, stricter condition: second-countable.
Definition N17.2. is second-countable if its topology has a countable basis.
Example N17.3.
has the countable basis .
with the product topology has a countable basis: products of finitely many with , times copies of .
under the uniform topology doesn’t have a countable basis. (A countable basis gives a countable dense subset: e.g. take one point in each basis open set.) But the balls with all are uncountably many disjoint open sets (their centers are all at distance from each other).
If has a countable dense subset and is metrizable, then there exists a countable basis. This is false in general: is a countable dense subset of , which has no countable basis.
Normal spaces
Recall:
Hausdorff(): any two points have disjoint open neighborhoods.
:regular: for all with closed, there are disjoint open ,.
:normal.
where regular Hausdorff uses (points are closed). The converses fail: is regular but not normal, and is normal but not metrizable.
Theorem N17.4. Hausdorff and compact normal. (HW 2)
Theorem N17.5. Metric spaces are normal.
Proof. Let be disjoint closed sets in . For each choose with , and for each choose with (possible since and are closed). Now
are open, contain resp. , and are disjoint: , because if were in both, then
so or . Contradiction.
∎
Metrizable (first-)countable + normal. (I forgor: necessary & sufficient.)
Theorem N17.6 (Urysohn). If has a countable basis and is normal, then is metrizable.
Lemma N17.7 (Urysohn’s lemma). Let be normal and disjoint closed subsets. Then there is a continuous such that for all and for all .
“building by specifying more and more…”
Proposition N17.8 (Reformulation of normality). is normal for every closed and open with , there is an open such that .
Proof.() is closed and disjoint from . Since are closed and disjoint, there are disjoint open sets and ,. Then
with open and closed, so .∎
Proof of Urysohn’s lemma. Given disjoint closed , let , which is open and contains . Pick open with . Then pick an open with , and so on, inductively for all of . This builds open sets for all such that
Define if and if . Define
Then . Claim: .∎
normal + countable basis construct an embedding (product topology): is injective, and (homeomorphism, with the subspace topology on ). The product topology on is given by the metric