[Munkres Topology] 33. The Урысо́н Lemma

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Urysohn Lemma is really important in topology: it opens metrization theorems and manifold theory.

This lemma gives continuous path connecting function on normal space.

Theorem 33.1 (Urysohn lemma). Let $X$ be a normal space; let $A$ and $B$ be disjoint closed subsets of $X$. Let $[a,b]$ be a closed interval in the real line. Then there exists a continuous map $$f: X\rightarrow [a,b]$$ such that $f(x)=a$ for every $x$ in $A$, and $f(x)=b$ for every $x$ in $B$.

Proof. Let assume $[a,b]$ is $[0,1]$. The first step is to construct, using normality, a certain family $U_p$ of open sets of $X$, indexed by the rational numbers. Then one uses these sets to define the continuous function $f$.

Step 1. Let $P$ be the set of all rational numbers in the interval $[0,1]$. We shall define, for each $p$ in $P$, an open set $U_p$ of $X$, in such a way that whenever $p<q$, we have $$\bar{U}_p \subset U_q.$$ Thus, the sets $U_p$ will be simply ordered by inclusion in the same way their subscripts are ordered by the usual ordering in the real line.

Because $P$ is countable, we can use induction to define the sets $U_p$ (or rather, the principle of recursive definition). Arrange the elements of $P$ in an infinite sequence in some way; for convenience, let us suppose that the numbers $1$ and $0$ are the first two elements of the sequence.

Now define the sets $U_p$, as follows: First, define $U_1=X-B$. Second, because $A$ is a closed set contained in the open set $U_1$, we may by normality of $X$ choose an open set $U_0$ such that $$A\subset U_0\quad \mbox{and}\quad \bar{U}_0\subset U_1.$$

In general, let $P_n$ denote the set consisting of the first $n$ rational numbers in the sequence. (just adding any rational number in $[0,1]$ without considering order!) Suppose that $U_p$ is defined for all rational numbers $p$ belonging to the set $P_n$, satisfying the condition $$ (*)\quad p<q \Rightarrow \bar{U}_p\subset U_q.$$ Let $r$ denote the next rational number in the sequence; we wish to denote $U_r$.

Consider the set $P_{n+1}=P_n \cup \{r\}$ ($r$ itself is arbitrary.). It is a finite subset of the interval $[0,1]$, and as such, it has a simple ordering derived from the usual order relation $<$ on the real line. In a finite simple ordered set, every element (other than the smallest and the largest) has an immediate predecessor and an immediate successor. (See Theorem 10.1.) The number $0$ is the smallest element, and $1$ is the largest element, of the simply ordered set $P_{n+1}$, and $r$ is neither $0$ nor $1$. So $r$ has an immediate predecessor $p$ in $P_{n+1}$ and an immediate successor $q$ in $P_{n+1}$. The sets $U_p$ and $U_q$ are already defined, and $\bar{U}_p\subset U_q$ by the induction hypothesis. Using normality of $X$, we can find an open set $U_r$ of $X$ such that $$\bar{U}_p \subset U_r \quad \mbox{and}\quad \bar{U}_r \subset U_q.$$ We assert that $(*)$ now holds for every pair of elements of $P_{n+1}$. If both elements lie in $P_n$, $(*)$ holds by the induction hypothesis. If one of them is $r$ and the other is a point $s$ of $P_n$, then either $s\le p$, in which case $$\bar{U}_s\subset \bar{U}_p\subset U_r,$$ or $s\ge q$, in which case $$\bar{U}_r \subset U_q \subset U_s.$$ Thus, for every pair of elements of $P_{n+1}$, relation $(*)$ holds. (for induction)

By induction, we have $U_p$ defined for all $p\in P$.

Step 2. Now we have defined $U_p$ for all rational numbers $p$ in the interval $[0,1]$. We simply extend this definition to all rational numbers $p$ in $\mathbb{R}$ by defining $$U_p=\varnothing \quad \mbox{if }p<0,$$ $$U_p=X\quad \mbox{if }p>1.$$ It is still true (as you can check) that for any pair of rational numbers $p$ and $q$, $p<q \Rightarrow \bar{U}_p\subset U_q.$$

Step 3. Given a point $x$ of $X$, let us define $\mathbb{Q}(x)$ to be the set of those rational numbers $p$ such that the corresponding open sets $U_p$ contain $x$: $$\mathbb{Q}(x)=\{p|x\in U_p\}.$$ This set contains no number less than $0$, since no $x$ is in $U_p$ for $p<0$. And it contains every number greater than $1$, sine every $x$ is in $U_p$ for $p>1$. Therefore, $\mathbb{Q}(x)$ is bounded below, and its greatest lower bound is a point of the interval $[0,1]$. Define $$f(x)=\inf \mathbb{Q}(x)=\inf \{p| x\in U_p\}.$$

Step 4. We show that $f$ is the desired function. If $x\in A$, then $x\in U_p$ for every $p\ge 0$, so that $\mathbb{Q}(x)$ equals the set of all nonnegative rationals, and $f(x)=\inf \mathbb{Q}(x)=0$. Similarly, if $x\in B$, then $x\in U_p$ for no $p\le 1$, so that $\mathbb{Q}(x)$ consists of all rational numbers greater than $1$, and $f(x)=1$.

Now show $f$ is continuous. For this purpose, we first prove that following elementary facts:

(1) $x\in \bar{U}_r\Rightarrow f(x)\le r.$

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

To prove (1), note that if $x\in \bar{U}_r$, then $x\in U_s$ for every $s>r$. Therefore, $\mathbb{Q}(x)$ contains all rational numbers greater than $r$, so that by definition we have $$f(x)=\inf \mathbb{Q}(x)\le r.$$ To prove (2), note that if $x\notin U_r$, then $x$ is not in $U_s$ for any $s<r$. Therefore, $\mathbb{Q}(x)$ contains no rational numbers less than $r$, so that $$f(x)=\inf \mathbb{Q}(x)\ge r.$$

Now we prove continuity of $f$. Given a point $x_0$ of $X$ and an open interval $(c,d)$ in $\mathbb{R}$ containing the point $f(x_0)$, we wish to find a neighborhood $U$ of $x_0$ such that $f(U)\subset (c,d)$ (this is true for all $x_0$, so $f^{-1}((c,d))$ be open.). Choose rational numbers $p$ and $q$ such that $$c<p<f(x_0)<q<d.$$ We assert that the open set $$U=U_q-\bar{U}_p$$ is the desired neighborhood of $x_0$. See Figure 33.2.

First, we note that $x_0\in U$. For the fact that $f(x_0)<q$ implies by condition (2) that $x_0\in U_q$, while the fact that $f(x_0)>p$ implies by (1) that $x_0\notin \bar{U}_p$.

Second, we show that $f(U)\subset (c,d)$. Let $x\in U$. Then $x\in U_q\subset \bar{U}_q$, so that $f(x)\le q$, by (1). And $x\notin \bar{U}_p$, so that $x\notin U_p$ and $f(x)\ge p$, by (2). Thus, $f(x)\in [p,q]\subset (c,d)$, as desired.

Definition. If $A$ and $B$ are two subsets of the topological space $X$, and if there is a continuous function $f:X\rightarrow [0,1]$ such that $f(A)=\{0\}$ and $f(B)=\{1\}$, we say that $A$ and $B$ can be separated by a continuous function.

Definition. A space $X$ is completely regular if one-point sets are closed in $X$ and if for each point $x_0$ and each closed set $A$ not containing $x_0$, there is a continuous function $f:X\rightarrow [0,1]$ such that $f(x_0)=1$ and $f(A)=\{0\}$.

Theorem 33.2. A subspace of a completely regular space is completely regular. A product of completely regular space if completely regular.

Proof.