Urysohn’s Lemma

This lemma is one crucial result in general point-set topology, essential to help us establish conditions for topological spaces to be metrizable. I’m most certainly not the one who came up with this idea, but this is the lemma in its full glory.

Theorem 1 (Urysohn’s Lemma). Let $K$ be a topological space. Then $K$ is normal if and only if for disjoint closed subsets $L, M \subseteq K$ , there exists a continuous function $f : K \to [0, 1]$ such that $f(L) = \{0\}$ and $f(M) = \{1\}$ .

Proof. For the easy direction $(\Leftarrow)$ , the disjoint closed subsets $L, M$ are separated by the neighbourhoods $f^{-1}([0, 1/2))$ and $f^{-1}((1/2, 1])$ .

For the difficult direction $(\Rightarrow)$ , fix disjoint closed subsets $L, M \subseteq K$ . Define $\bar V_0 := L$ and $V_1 := K \backslash M$ . We note that $\bar V_0 \subseteq V_1$ . By normality, there exists a neighbourhood $V_{1/2}$ of $V_0$ such that

$\bar V_0 \subseteq V_{1/2} \subseteq \bar{V}_{1/2} \subseteq V_1.$

In fact, for any two rational numbers $r < s$ , there exists a neighbourhood $V_{(r+s)/2}$ of $\bar V_r$ such that

$\bar V_r \subseteq V_{(r+s)/2} \subseteq \bar{V}_{(r+s)/2} \subseteq V_s.$

Define the dyadic rationals $D$ as rational numbers of the form $m/2^n$ , where $m,n$ are positive integers and $0 \leq m \leq 2^n$ . Then inductively, for each $r \in D$ , a $V_r$ exists as per our construction above.

The rough idea is to define $f(x)$ to be the “effectively smallest” possible $r$ such that $x \in V_r$ . If $x \notin V_r$ for any $r \in D$ , then $x \notin V_1$ , which means that $x \in M$ . Define $f(x) = 1$ whenever $x \notin V_1$ , so that $f(M) = \{1\}$ . Otherwise, $f(x) \in V_r$ for some $r \in D$ . Define the we;;-defined mapping $f : K \to [0, 1]$ by

$f(x) = \inf \{r : x \in V_r \}.$

We leave it as an exercise in real analysis to verify the following implications:

If $x \in \bar V_r$ , then $f(x) \leq r$ .
If $x \notin V_r$ , then $f(x) \geq r$ .

In particular, if $x \in L = \bar V_0$ , then $f(x) = 0$ , which means $f(L) = \{0\}$ . All that remains is to prove that $f$ is continuous on $K$ .

Fix $x \in K$ and $\epsilon > 0$ . We need to prove that $f^{-1}(f(x) + (-\epsilon, \epsilon)) =: U$ is open. Since $\bar D = [0, 1]$ , find dyadic rationals $r, s$ such that

$f(x) - \epsilon < r < f(x) < s < f(x) + \epsilon.$

By the construction of $f$ , $x \in V_s \backslash \bar V_r$ . Furthermore, for any $y \in V_s \backslash \bar V_r$ , $r \leq f(y) \leq s$ . This implies

$f(x) \in f(V_s \backslash \bar V_r) \subseteq [r, s] \subseteq f(x) + (-\epsilon, \epsilon).$

Hence, $x \in V_s \backslash \bar V_r \subseteq U$ , so that $U$ is open, as required.

With Urysohn’s lemma, we can state the big result that we aim to prove…eventually.

Theorem 2 (Urysohn Metrization Theorem). If $K$ is regular and second-countable, then it is metrizable.

The goal is to find a continuous injection $f : K \to L$ that is a homemorphism with $f(K)$ , where $L$ is metrizable. Since subspaces of metrizable spaces are metrizable, $f(K)$ is metrizable, and thus $K$ will be metrizable too. The challenge is constructing an $L$ that is metrizable, and then constructing the required injection $f$ . As such, we need to take a detour into metrizable spaces in order to establish this result. This we will do, next time.

—Joel Kindiak, 25 Apr 25, 1531H

KindiakMath

Urysohn’s Lemma

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Urysohn’s Lemma

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