Past exam of the analytic topology and axiomatic set theory course of the University of Oxford / 2008 / A / 2 / b

Urysohn's lemma states that, in a normal space, disjoint closed subsets $A,B$ can be separated by a continuous function $u:X\to [0,1]$ with $u=0$ on $A$ and $u=1$ on $B$.

First use the shrinking consequence of the separation property of a normal space: if a closed set $F$ lies in an open set $G$, there is an open $W$ with $F\subseteq W\subseteq \overline{W}\subseteq G$. To prove it, separate the disjoint closed sets $F$ and $X\setminus G$ by disjoint open neighborhoods $W,V$. Then $\overline{W}\subseteq X\setminus V\subseteq G$.

Let $D$ be the dyadic rationals in $[0,1]$. Set $U_1=X\setminus B$, and shrink around $A$ to choose $U_0$ with $A\subseteq U_0\subseteq \overline{U_0}\subseteq U_1$. At each subsequent dyadic stage, insert between each adjacent pair $s<t$ an open set $U_{(s+t)/2}$ such thatThe shrinking lemma applies because the first set is closed and contained in the last open set. Induction gives $\overline{U_s}\subseteq U_t$ for all dyadic $s<t$. Part (a) now constructs a continuous $u$. Each point of $A$ lies in $U_0$, so has value zero. Each point of $B$ lies in none of the $U_t$, so has value one. This proves the lemma.

For the embedding, let $I=C(X,[0,1])$, a set of continuous functions, and define the evaluation mapEvery coordinate is continuous, so $e$ is continuous in the product topology. The Hausdorff assumption makes singletons closed. For distinct points, Urysohn's lemma applied to their singletons gives a coordinate with different values, so $e$ is injective.

To check the inverse topology, take $x\in O$ with $O$ open. Apply the lemma to $\{x\}$ and $X\setminus O$, choosing a coordinate $u$ with $u(x)=1$ and $u=0$ outside $O$. ThenThis is the preimage of an open coordinate cylinder under $e$. Thus every original neighborhood is detected by coordinate neighborhoods, and $e^{-1}:e[X]\to X$ is continuous. The evaluation map is a topological embedding.