proof of Urysohn’s lemma

First we construct a family Up of open sets of X indexed by the rationals such that if p<q, then Up¯Uq. These are the sets we will use to define our continuous functionMathworldPlanetmathPlanetmath.

Let P=[0,1]. Since P is countableMathworldPlanetmath, we can use inductionMathworldPlanetmath (or recursive definition if you prefer) to define the sets Up. List the elements of P is an infiniteMathworldPlanetmath sequence in some way; let us assume that 1 and 0 are the first two elements of this sequence. Now, define U1=X\D (the complement of D in X). Since C is a closed setPlanetmathPlanetmath of X contained in U1, by normalityPlanetmathPlanetmath of X we can choose an open set U0 such that CU0 and U0¯U1.

In general, let Pn denote the set consisting of the first n rationals in our sequence. Suppose that Up is defined for all pPn and

ifp<q,thenUp¯Uq. (1)

Let r be the next rational number in the sequence. Consider Pn+1=Pn{r}. It is a finite subset of [0,1] so it inherits the usual ordering < of . In such a set, every element (other than the smallest or largest) has an immediate predecessor and successorMathworldPlanetmathPlanetmathPlanetmath. We know that 0 is the smallest element and 1 the largest of Pn+1 so r cannot be either of these. Thus r has an immediate predecessor p and an immediate successor q in Pn+1. The sets Up and Uq are already defined by the inductive hypothesis so using the normality of X, there exists an open set Ur of X such that


We now show that (1) holds for every pair of elements in Pn+1. If both elements are in Pn, then (1) is true by the inductive hypothesis. If one is r and the other sPn, then if sp we have


and if sq we have


Thus (1) holds for ever pair of elements in Pn+1 and therefore by induction, Up is defined for all pP.

We have defined Up for all rationals in [0,1]. Extend this definition to every rational p by defining


Then it is easy to check that (1) still holds.

Now, given xX, define (x)={p:xUp}. This set contains no number less than 0 and contains every number greater than 1 by the definition of Up for p<0 and p>1. Thus (x) is bounded below and its infimumMathworldPlanetmath is an element in [0,1]. Define


Finally we show that this function f we have defined satisfies the conditions of lemma. If xC, then xUp for all p0 so (x) equals the set of all nonnegative rationals and f(x)=0. If xD, then xUp for p1 so (x) equals all the rationals greater than 1 and f(x)=1.

To show that f is continuous, we first prove two smaller results:

(a) xUr¯f(x)r

Proof. If xUr¯, then xUs for all s>r so (x) contains all rationals greater than r. Thus f(x)r by definition of f.

(b) xUrf(x)r.

Proof. If xUr, then xUs for all s<r so (x) contains no rational less than r. Thus f(x)r.

Let x0X and let (c,d) be an open interval of containing f(x). We will find a neighborhoodMathworldPlanetmathPlanetmath U of x0 such that f(U)(c,d). Choose p,q such that


Let U=Uq\Up¯. Then since f(x0)<q, (b) implies that xUq and since f(x0)>p, (a) implies that x0Up¯. Hence x0U.

Finally, let xU. Then xUqUq¯, so f(x)q by (a). Also, xUp¯ so xUp and f(x)p by (b). Thus


as desired. Therefore f is continuous and we are done.

Title proof of Urysohn’s lemma
Canonical name ProofOfUrysohnsLemma
Date of creation 2013-03-22 13:09:23
Last modified on 2013-03-22 13:09:23
Owner scanez (1021)
Last modified by scanez (1021)
Numerical id 4
Author scanez (1021)
Entry type Proof
Classification msc 54D15