Urysohn Lemma

The Urysohn lemma is one of the deepest and most useful results in point-set topology. It states that in a normal space, any two disjoint closed sets can be separated by a continuous real-valued function. This connects the topological separation axioms with the algebraic/analytic structure of continuous functions.

Statement

Theorem6.8Urysohn Lemma

Let $X$ be a normal topological space. If $A$ and $B$ are disjoint closed subsets of $X$ , then there exists a continuous function $f: X \to [0, 1]$ such that $f|_A \equiv 0$ and $f|_B \equiv 1$ .

RemarkConverse

The converse is also true: if for every pair of disjoint closed sets there exists a continuous function to $[0,1]$ separating them, then $X$ is normal. (Take $U = f^{-1}([0, 1/2))$ and $V = f^{-1}((1/2, 1])$ .)

Thus normality is equivalent to the functional separation of closed sets.

Proof Sketch

Proof

The proof constructs the function $f$ by building a family of open sets indexed by the dyadic rationals $D = \{p/2^n : p, n \in \mathbb{Z}, 0 \leq p \leq 2^n\} \cap [0, 1]$ .

Step 1: Constructing the open sets.

We construct open sets $\{U_r\}_{r \in D}$ such that:

$A \subseteq U_r$ for all $r \in D$ .
$U_0 \supseteq A$ and $U_1 = X \setminus B$ .
If $r < s$ (both in $D$ ), then $\overline{U_r} \subseteq U_s$ .

Start with $U_1 = X \setminus B$ , which is open and contains $A$ . By the normality characterization (Theorem 6.4), there exists an open $U_0$ with $A \subseteq U_0 \subseteq \overline{U_0} \subseteq U_1$ .

Now inductively define $U_r$ for all dyadic rationals. At stage $n$ , we have defined $U_r$ for all $r = k/2^{n-1}$ . For each new dyadic rational $r = (2m+1)/2^n$ , the sets $U_{m/2^{n-1}}$ and $U_{(m+1)/2^{n-1}}$ satisfy $\overline{U_{m/2^{n-1}}} \subseteq U_{(m+1)/2^{n-1}}$ . Apply normality to find $U_r$ with: $\overline{U_{m/2^{n-1}}} \subseteq U_r \subseteq \overline{U_r} \subseteq U_{(m+1)/2^{n-1}}.$

Step 2: Defining the function.

For $x \in X$ , define: $f(x) = \inf\{r \in D : x \in U_r\}.$

If $x \in A$ , then $x \in U_r$ for all $r \in D$ , so $f(x) = 0$ . If $x \in B$ , then $x \notin U_1 = X \setminus B$ and $x \notin U_r$ for any $r \leq 1$ , so $f(x) = 1$ .

Step 3: Continuity.

We verify $f^{-1}([0, a))$ and $f^{-1}((a, 1])$ are open for all $a \in [0, 1]$ :

$f^{-1}([0, a)) = \bigcup_{r < a, r \in D} U_r$ , which is a union of open sets, hence open.
$f^{-1}((a, 1]) = \bigcup_{r > a, r \in D} (X \setminus \overline{U_r})$ , since $f(x) > a$ iff $x \notin \overline{U_r}$ for some $r > a$ . This is a union of open sets, hence open.

Since sets of the form $[0, a)$ and $(a, 1]$ form a subbasis for $[0, 1]$ , $f$ is continuous.

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Applications

ExampleSeparating Closed Sets in $\mathbb{R}^n$

Since $\mathbb{R}^n$ is normal (it is metrizable), the Urysohn lemma applies. For disjoint closed sets $A, B \subseteq \mathbb{R}^n$ , there exists a smooth "bump function" separating them. The Urysohn lemma provides a continuous (not necessarily smooth) separator. With additional techniques (partitions of unity), one obtains smooth separators.

ExampleUrysohn Lemma for Metric Spaces

In a metric space $(X, d)$ , the Urysohn function can be written explicitly: $f(x) = \frac{d(x, A)}{d(x, A) + d(x, B)}.$ This is continuous (as $d(\cdot, A)$ and $d(\cdot, B)$ are continuous and their sum is positive on $X$ since $A \cap B = \emptyset$ and both are closed), equals $0$ on $A$ , and equals $1$ on $B$ .

This explicit formula is much simpler than the general construction, but it relies on having a metric.

RemarkUrysohn's Legacy

The Urysohn lemma (proved by Pavel Urysohn in 1925) is fundamental for:

The Tietze extension theorem (extending continuous functions from closed subsets).
The Urysohn metrization theorem (second-countable normal spaces are metrizable).
The Stone--Cech compactification (constructed using continuous functions to $[0,1]$ ).
Establishing that normal spaces are completely regular.