Topology, Urysohn's Lemma

Urysohn's Lemma

Urysohn (biography) developed an equivalent criterion for a normal space. We present the criterion first, then prove it is equivalent to the original definition.

A Hausdorff space S is normal iff, for every nonempty closed set E and every open set O containing E, there is some intermediate open set U such that O contains closure(U) contains U contains E.

Assume S is normal. If O is the entire space S set U = O. Otherwise let F be the complement of O and separate E and F in open sets U and V respectively. Now the closure of U is the intersection of all closed sets containing U, including the complement of V. Thuse the closure of U is contained in the complement of V, which is contained in O.

Assume S has the intermediate open set property and Let E and F be nonempty disjoint closed sets. Let O be the complement of F and let U be the intermediate open set whose closure is contained in O. Let C be the closure of U, a closed set that is disjoint from F. Now find an intermediate open set V containing F, inside the complement of C. The open sets U and V separate E and F, and since S is Hausdorff, it is normal.

A Continuous Map from S into [0,1]

We can use Urysohn's lemma to build a continuous function on S that is 0 on E and 1 on F. (I know we haven't defined continuous yet, but I'll assume you know what it means.) Assume S is normal and let E and F be any nonempty closed disjoint sets in S. Let O be the complement of F. Choose any intermediate open set and call it U½. Then choose intermediate open sets U¼ between E and U½, and U¾ between the closure of U½ and O. Repeat this process until we have Uq for every rational q with a terminating binary expansion in (0,1).

Let U0 = E, and let U1 = S.

Build a map from the space S into the closed interval [0,1] as follows. For any point x in S, let g(x) be the greatest lower bound of all the rationals q such that x is in Uq. Note that g(E) = 0 and g(F) = 1.

We wish to show g() is a continuous function. That is, the preimage of every open interval in [0,1] is an open set in S. Given an open interval I in [0,1], choose any z in I, and rational numbers 0 ≤ b ≤ z ≤ c ≤ 1 where b and c lie in I, and have terminating binary expansions. Every point in Uc has image at most c. Every point outside the closure of Ub has image at least b. The open set Uc-closure(Ub), rather like an annulus, has image inside I. Furthermore, every x with g(x) = z lives in this annulus. This applies to every point z in I. Therefore the preimage of I is covered by open sets in S, and is open.

A Hausdorff space S is normal iff, for every pair of disjoint nonempty closed sets E and F, there is a continuous map g(S) into [0,1] such that g(E) = 0 and g(F) = 1. We already proved normal implies g(x).

Given g(x), let U be the preimage of the points below ½ and V the preimage of the points above ½. This gives separation, and combined with Hausdorff, S is normal.

Completely Regular

A hausdorff space S is completely regular if, for every closed set E and every point x not in E, there is a continuous real valued function that is 0 on E and 1 on x.

Each normal space is completely regular by Urysohn's lemma, as shown above. And any completely regular space is regular. Take the preimages of [0,½) and (½,1] to separate E and x. Thus completely regular is sometimes referred to as T3.5.