Urysohn's lemma

Statement

Let $X$ be a normal space and let $A, B \subseteq X$ be disjoint closed sets. Then there is a continuous map $f \colon X \to [0, 1]$ with $f(a) = 0$ for all $a \in A$ and $f(b) = 1$ for all $b \in B$.

Remarks

The converse also holds: if such a function exists for all disjoint closed $A$ and $B$, then $X$ is normal. Take $U = f^{-1}([0, 1/2))$ and $V = f^{-1}((1/2, 1])$. The proof builds open sets $U_r$ for the dyadic rationals $r \in [0, 1]$, nested by the order of $r$. Then $f(x)$ is the infimum of the $r$ with $x \in U_r$. In a metric space, the function $f(x) = d(x, A) / (d(x, A) + d(x, B))$ works. The Tietze extension theorem extends this result to all continuous functions on closed sets.

Explanations

L1 human

In a normal space, a continuous function can separate two disjoint closed sets: it is $0$ on one set and $1$ on the other. So normal spaces have many continuous real functions.