# 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, 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 (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 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

Let $U=U_{q}\backslash\bar{U_{p}}$. Then since $f(x_{0}), (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 ProofOfUrysohnsLemma 2013-03-22 13:09:23 2013-03-22 13:09:23 scanez (1021) scanez (1021) 4 scanez (1021) Proof msc 54D15