proof of Urysohn’s lemma

First we construct a family $U_{p}$ of open sets of $X$ indexed by the rationals such that if $p<q$ , then $\bar{U_{p}}\subseteq U_{q}$ . These are the sets we will use to define our continuous function.

Let $P=\mathbb{Q}\cap[0,1]$ . Since $P$ is countable, we can use induction (or recursive definition if you prefer) to define the sets $U_{p}$ . List the elements of $P$ is an infinite sequence in some way; let us assume that $1$ and $0$ are the first two elements of this sequence. Now, define $U_{1}=X\backslash D$ (the complement of $D$ in $X$ ). Since $C$ is a closed set of $X$ contained in $U_{1}$ , by normality of $X$ we can choose an open set $U_{0}$ such that $C\subseteq U_{0}$ and $\bar{U_{0}}\subseteq U_{1}$ .

In general, let $P_{n}$ denote the set consisting of the first $n$ rationals in our sequence. Suppose that $U_{p}$ is defined for all $p\in P_{n}$ and

\textrm{if}\ p<q,\textrm{then}\ \bar{U_{p}}\subseteq U_{q}.

(1)

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

\bar{U_{p}}\subseteq U_{r}\ \textrm{and}\ \bar{U_{r}}\subseteq U_{q}.

We now show that (1) holds for every pair of elements in $P_{n+1}$ . If both elements are in $P_{n}$ , then (1) is true by the inductive hypothesis. If one is $r$ and the other $s\in P_{n}$ , then if $s\leq p$ we have

\bar{U_{s}}\subseteq\bar{U_{p}}\subseteq U_{r}

and if $s\geq q$ we have

\bar{U_{r}}\subseteq U_{q}\subseteq U_{s}.

Thus (1) holds for ever pair of elements in $P_{n+1}$ and therefore by induction, $U_{p}$ is defined for all $p\in P$ .

We have defined $U_{p}$ for all rationals in $[0,1]$ . Extend this definition to every rational $p\in\mathbb{R}$ by defining

\begin{array}[]{ll}U_{p}=\emptyset&\textrm{if}\ p<0\\ U_{p}=X&\textrm{if}\ p>1.\end{array}

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

Now, given $x\in X$ , define $\mathbb{Q}(x)=\{p:x\in U_{p}\}$ . This set contains no number less than $0$ and contains every number greater than $1$ by the definition of $U_{p}$ for $p<0$ and $p>1$ . Thus $\mathbb{Q}(x)$ is bounded below and its infimum is an element in $[0,1]$ . Define

f(x)=\textrm{inf}\ \mathbb{Q}(x).

Finally we show that this function $f$ we have defined satisfies the conditions of lemma. If $x\in C$ , then $x\in U_{p}$ for all $p\geq 0$ so $\mathbb{Q}(x)$ equals the set of all nonnegative rationals and $f(x)=0$ . If $x\in D$ , then $x\notin U_{p}$ for $p\leq 1$ so $\mathbb{Q}(x)$ equals all the rationals greater than 1 and $f(x)=1$ .

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

(a) $x\in\bar{U_{r}}\Rightarrow f(x)\leq r$

Proof. If $x\in\bar{U_{r}}$ , then $x\in U_{s}$ for all $s>r$ so $\mathbb{Q}(x)$ contains all rationals greater than $r$ . Thus $f(x)\leq r$ by definition of $f$ .

(b) $x\notin U_{r}\Rightarrow f(x)\geq r$ .

Proof. If $x\notin U_{r}$ , then $x\notin U_{s}$ for all $s<r$ so $\mathbb{Q}(x)$ contains no rational less than $r$ . Thus $f(x)\geq r$ .

Let $x_{0}\in X$ and let $(c,d)$ be an open interval of $\mathbb{R}$ containing $f(x)$ . We will find a neighborhood $U$ of $x_{0}$ such that $f(U)\subseteq(c,d)$ . Choose $p,q\in\mathbb{Q}$ such that

c<p<f(x_{0})<q<d.

Let $U=U_{q}\backslash\bar{U_{p}}$ . Then since $f(x_{0})<q$ , (b) implies that $x\in U_{q}$ and since $f(x_{0})>p$ , (a) implies that $x_{0}\notin\bar{U_{p}}$ . Hence $x_{0}\in U$ .

Finally, let $x\in U$ . Then $x\in U_{q}\subseteq\bar{U_{q}}$ , so $f(x)\leq q$ by (a). Also, $x\notin\bar{U_{p}}$ so $x\notin U_{p}$ and $f(x)\geq p$ by (b). Thus

f(x)\in[p,q]\subseteq(c,d)

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