proof of Urysohn’s lemma

First we construct a family $U_p$ of open sets of $X$ indexed by therationals such that if , then $\overline{U_p}\subseteq U_q$. Theseare the sets we will use to define our continuous function.

Let $P=\mathbb{Q}\cap [0,1]$. Since $P$ is countable, we can useinduction (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 thissequence. Now, define $U_1=X\backslash \backslash D$ (the complement of $D$ in$X$). Since $C$ is a closed set of $X$ contained in $U_1$, bynormality of $X$ we can choose an open set $U_0$ such that $C\subseteq U_0$ and $\overline{U_0}\subseteq U_1$.

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

(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 inheritsthe usual ordering of $ℝ$. In such a set, every element(other than the smallest or largest) has an immediate predecessor andsuccessor. We know that $0$ is the smallest element and $1$ thelargest of $P_{n+1}$ so $r$ cannot be either of these. Thus $r$ hasan immediate predecessor $p$ and an immediate successor $q$ in$P_{n+1}$. The sets $U_p$ and $U_q$ are already defined by theinductive hypothesis so using the normality of $X$, there exists anopen set $U_r$ of $X$ such that

$$\overline{U_p}\subseteq U_r\text{and}\overline{U_r}\subseteq U_q.$$

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

$$\overline{U_s}\subseteq \overline{U_p}\subseteq U_r$$

and if $s\ge q$ we have

$$\overline{U_r}\subseteq U_q\subseteq U_s.$$

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

We have defined $U_p$ for all rationals in $[0,1]$. Extend thisdefinition to every rational $p\in ℝ$ by defining

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

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

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

Finally we show that this function $f$ we have defined satisfies theconditions of lemma. If $x\in C$, then $x\in U_p$ for all $p\ge 0$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 $p\le 1$ 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) $x\in \overline{U_r}\Rightarrow f(x)\le r$

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

(b) $x\notin U_r\Rightarrow f(x)\ge r$.

Proof. If $x\notin U_r$, then $x\notin U_s$ for all so $\mathbb{Q}(x)$ contains no rational less than $r$. Thus $f(x)\ge r$.

Let $x_0\in X$ and let $(c,d)$ be an open interval of $ℝ$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

Let $U=U_q\backslash \backslash \overline{U_p}$. Then since , (b)implies that $x\in U_q$ and since $f(x_0)>p$, (a) implies that $x_0\notin \overline{U_p}$. Hence $x_0\in U$.

Finally, let $x\in U$. Then $x\in U_q\subseteq \overline{U_q}$, so $f(x)\le q$ by (a). Also, $x\notin \overline{U_p}$ so $x\notin U_p$ and $f(x)\ge p$ by (b). Thus

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

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