Section 1.20 Things Urysohn
Theorem 1.20.1. Urysohn lemma.
Let \(X\) be a normal space. Given two nonempty disjoint closed sets \(A\) and \(B\) there is a continuous function \(f\colon X\rightarrow [0,1]\) such that \(f(A)=\{0\}\) and \(f(B)=\{1\}\text{.}\)
It follows easily that the result is still true if we replace the interval \([0,1]\) with any closed interval \([a,b]\text{.}\)
Proof.
It would be difficult to improve upon Munkres’ exposition of this. I give essentially a full proof below nonetheless.
Step 1: construction of open sets \(U_p\).
The first crucial step in the proof is to build a collection of open sets
\(\{U_p\}_{p\in \Q}\) satisfying
\(\overline{U_p}\subseteq U_{q}\) if
\(p< q\text{.}\) This is done by first picking an enumeration of
\(\Q\cap [0,1]=\{p_1,p_2,\dots,\}\) where for convenience
\(p_1=1\) and
\(p_2=0\text{.}\) We use the enumeration to define open sets
\(U_p\) for each
\(p\in \Q\) recursively as follows. Set
\(U_1=X-B\text{.}\) Since
\(A\subseteq U_1\text{,}\) by
Theorem 1.19.8 we can find an open
\(U_0\) containing
\(A\) such that
\(\overline{U}_0\subseteq U_1\text{.}\) Now take any
\(n\geq 2\) and assume by induction that we constructed open sets
\(U_{p_1}, U_{p_2},\dots, U_{p_n}\) satisfying the desired inclusion condition. Write
\(p< p_{n+1} < q\) where
\(p\) and
\(q\) are the immediate predecessor and successor, respectively, of
\(p_{n+1}\) among the set
\(\{p_1,p_2,\dots, p_n\}\text{.}\) By induction we have
\(\overline{U_p}\subseteq U_q\text{.}\) Pick
\(U_{p_{n+1}}\) to be any open set satisfying
\begin{equation*}
\overline{U_p}\subseteq U_{p_{n+1}}\subseteq \overline{U_{p_{n+1}}}\subseteq U_q\text{.}
\end{equation*}
Lastly define \(U_p=\emptyset\) for all \(p< 0\) and \(U_p=X\) for all \(p> 0\text{.}\) It is easy to see that thus defined, our collection \(\{U_p\}_{p\in \Q}\) satisfies the desired inclusion condition.
Step 2: definition of \(f\) and continuity.
Define \(f\colon X\rightarrow \R\) as
\begin{equation*}
f(x)=\inf\{p\in \Q\colon x\in U_p\}\text{.}
\end{equation*}
Some easy properties of \(f\text{:}\)
\(0\leq f(x)\leq 1\) for all \(x\in X\text{.}\)
\(f(x)\leq p\) for all \(x\in \overline{U_p}\text{.}\)
\(f(x)\geq p\) for all \(x\notin U_p\text{.}\)
We prove continuity by showing that \(f\) is continuous at each \(x\text{.}\) Given \(x\in X\) and open set \((a,b)\) containing \(y=f(x)\text{,}\) choose rational numbers \(p, q\) satisfying
\begin{equation*}
a< p< y < q < b\text{.}
\end{equation*}
I claim \(U_q-\overline{U_p}\) is an open set containing \(x\) satisfying
\begin{equation*}
f(U_q-\overline{U_p})\subseteq [p,q]\subseteq (a,b)\text{.}
\end{equation*}
First, it is clear that \(U_q-\overline{U_p}\) is open, and it follows easily from the elementary properties of \(f\) quoted above that \(x\in U_q-\overline{U_p}\text{.}\) Next, given any \(x'\in U_q-\overline{U_p}\text{,}\) since \(x'\in \overline{U_q}\text{,}\) we have \(f(x')\leq q\text{,}\) and since \(x'\notin U_p\text{,}\) we have \(f(x')\geq p\text{.}\) This concludes the proof.
Definition 1.20.2. Separated by a continuous function.
Let \(X\) be a topological space. Subsets \(A, B\subseteq X\) are separated by a continuous function if there is a continuous function \(f\colon X\rightarrow [0,1]\) such that \(f(A)=\{0\}\) and \(f(B)=\{1\}\text{.}\)
We can explore this idea of “separated by a continuous function” in relation to some of the other Trennnungsaxiome. For example, taking the regularity axiom (\(T_3\)) and replacing the separated by open sets condition with a separated by a continuous function condition, we obtain the notion of complete regularity.
Definition 1.20.4. Completely regular.
A \(T_1\)-space \(X\) is completely regular (or \(T_{3\frac{1}{2}}\)) if for any closed set \(A\) and element \(x_0\notin A\text{,}\) there is a continuous function \(f\colon X\rightarrow [0,1]\text{,}\) \(a,b\in \R\) such that \(f(x_0)=1\) and \(f(A)=\{0\}\text{.}\)
The notion of complete regularity strikes a nice balance between regularity and normality. In particular, it is stronger than regularity, but still behaves well with respect to subspaces and products.
Theorem 1.20.6. Complete regularity: subspace and product properties.
Subspaces of completely regular spaces are completely regular.
Products of completely regular spaces are completely regular.
Proof.
The proof of (1) is elementary. (See text.) The proof of (2) is more interesting. Assume \(\{X_i\}_{i\in I}\) is a collection of completely regular spaces, and let \(X=\prod_{i\in I}X_i\text{.}\) Let \(A\subseteq X\) be a closed set, and let \(b=(b_i)_{i\in I}\) be any element not contained in \(A\text{.}\) Since \(X-A\) is open, we can find a base open neighborhood of \(b\) of the form \(U=\prod_{k=1}^n U_{i_k}\times \prod_{i\ne i_k}X_i\text{.}\) For each \(1\leq k\leq n\) let \(f_k\colon X_{i_k}\rightarrow [0,1]\) be a continuous function satisfying \(f_k(b_{i_k})=1, f_k(U_{i_k}^c)=\{0\}\text{.}\) The function
\begin{align*}
f\colon X \amp \rightarrow [0,1]\\
x=(x_i)_{i\in I} \amp \mapsto \prod_{k=1}^nf_{i_k}(x_{i_k})
\end{align*}
is continuous (product of continuous functions) and satisfies \(f(b)=1\) and \(f(A)=\{0\}\text{.}\)
Theorem 1.20.7. Urysohn metrization theorem.
If \(X\) is a regular second countable space, then \(X\) is metrizable.
Proof.
The main step in the proof, one that uses
Theorem 1.20.1, is the construction of a countable family of continuous functions
\(\{f_n\colon X\rightarrow [0,1]\}_{n\in \Z_+}\) satisfying the following property: for all
\(x\in X\) and for all open sets
\(U\) containing
\(x\text{,}\) there is an
\(n\in \Z_+\) such that
\(f_n(x)>0\) and
\(f_n(X-U)=\{0\}\text{.}\) It then follows from
Theorem 1.20.8 that the map
\(f\colon X\rightarrow \R^\omega\) defined as
\(f(x)=(f_n(x))_{n\in \Z_+}\) is an embedding into the metric space
\(\R^\omega\text{.}\) Since subspaces of metric spaces are metric spaces, we conclude that
\(X\) is metrizable.
We now proceed with the construction. Let
\(\mathcal{B}=\{B_1, B_2,\dots, \}\) be a countable basis of
\(X\text{.}\) For all
\(m,n\in \Z_+\) satisfying
\(\overline{B_m}\subseteq B_n\text{,}\) let
\(g_{m,n}\colon X\rightarrow [0,1]\) be a continuous function satisfying
\(g_{m,n}(\overline{B_m})=\{1\}\) and
\(g_{m,n}(X-B_n)=\{0\}\text{:}\) this guaranteed to exist by
Theorem 1.20.1 since
\(X\text{,}\) being regular and second countable, is normal. Since the collection
\(\{g_{m,n}\}\) is countable we can enumerate it as
\(\{g_{m,n}\}=\{f_1, f_2,\dots\}\text{.}\) Now, for any
\(x\) and any open set
\(U\ni x\text{,}\) we have
\(x\in B_n\subseteq U\) for some
\(n\in \Z_+\text{.}\) Since
\(X\) is regular, we can find a basis element
\(B_m\) such that
\(x\in \overline{B_m}\subseteq B_n\text{.}\) We have
\(g_{m,n}=f_k\) for some
\(k\in \Z_+\text{,}\) \(f_k(x)=1\text{,}\) and
\(f_k(X-U)\subseteq f_k(X-B_m)=\{0\}\text{,}\) as desired.
Theorem 1.20.8. Embedding theorem.
Let \(X\) be a \(T_1\)-space. Given any collection of continuous functions \(\{f_i\colon X\rightarrow \R\}_{i\in I}\) satisfying the property that for all \(x\in X\) and for all open sets \(U\) containing \(x\) there is an \(i\in I\) such that \(f_i(x)> 0\) and \(f_i(X-U)=\{0\}\text{,}\) the map \(f\colon X\rightarrow \R^I\) defined as \(f(x)=(f_i(x))_{i\in I}\) is an embedding into \(\R^I\text{.}\)
Proof.
\(f\) is injective.
Since \(X\) is \(T_1\text{,}\) given any \(x\ne y\in X\) we can find an open set \(U\) such that \(x\in U\) and \(y\notin U\text{.}\) By assumption, there is some \(i_0\in I\) such that \(f_{i_{0}}(x)> 0\) and \(f_{i_0}(X-U)=\{0\}\text{.}\) It follows that \(f_{i_{0}}(x)\ne f_{i_0}(y)\) and hence that \(f(x)\ne f(y)\text{.}\)
\(f\) is continuous.
This follows immediately from
Theorem 1.10.11 since each
\(f_i\) is continuous.
\(f\) is embedding.
Lastly, we prove that \(f\) is an embedding mapping \(X\) homeomorphically onto \(f(X)\subseteq \R^I\text{.}\) Let \(U\) be an open set containing an element \(x\in X\text{,}\) and let \(y=f(x)\in f(U)\text{.}\) Pick any \(i_0\in I\) such that \(f_{i_0}(x)> 0\) and \(f_{i_0}(X-U)=\{0\}\text{.}\) Let \(\pi_{i_0}\colon \R^I\rightarrow \R\) be the projection map onto the \(i_{0}\)-th coordinate, and define \(V=f(X)\cap \pi_{i_0}^{-1}((0,\infty))\text{.}\) Since \(\pi_{i_0}\) is continuous and \((0,\infty)\) is open, the set \(V\) is open in the subspace topology of \(f(X)\text{.}\) We claim that \(y\in V\subseteq f(U)\text{,}\) proving that \(f\) is an embedding. Since \(\pi_{i_0}(y)=f_{i_0}(x)> 0\text{,}\) we have \(y\in \text{.}\) Similarly, if \(y'=f(x')\in V\text{,}\) then \(\pi_{i_0}(y')=\pi_{i_0}(x')> 0\text{,}\) from whence it follows that \(x'\in U\text{:}\) otherwise we’d have \(f_{i_0}(x')=0\text{.}\)
Corollary 1.20.10.
A space \(X\) is completely regular if and only if it is homeomorphic to a subspace of \([0,1]^I\) for some set \(I\text{.}\)
Proof.
Implication:
\((\Rightarrow)\text{.}\) Assume
\(X\) is completely regular. Recall that in particular this means
\(X\) is
\(T_1\text{.}\) Define
\(I\) to be the set of all pairs
\((x, U)\) where
\(x\in X\) and
\(U\) is an open neighborhood of
\(x\text{.}\) For each
\(i=(x,U)\text{,}\) let
\(f_i\colon X\rightarrow [0,1]\) be a function satisfying
\(f(x)=1\) and
\(f(X-U)=\{0\}\text{.}\) (Such a function exists since
\(X\) is completely regular.) Now, for all
\(x\in X\) and for all open sets
\(U\) containing
\(x\text{,}\) corresponding to the index
\(i=(x,U)\) we have the function
\(f_i\) which satisfies
\(f_i(x)=1> 0\) and
\(f_i(X-U)=\{0\}\text{.}\) We conclude by
Theorem 1.20.8 that the map
\(f(x)=(f_i(x))_{i\in I}\) is an embedding of
\(X\) into
\(\R^I\text{,}\) and thus that
\(X\) is homeomorphic to a subspace of
\(\R^I\text{.}\)
Implication: \((\impliedby)\text{.}\) Since \(\R\) is completely regular, and since complete regularity is preserved under arbitrary products, the space \(\R^I\) is completely regular for any set \(I\text{.}\) Furthermore, since complete regularity is inherited by subspaces, any subspace of \(\R^I\) is completely regular. Thus, if \(X\) embeds into \(\R^I\text{,}\) then it is homeomorphic to a subspace of \(\R^I\text{,}\) and hence is completely regular.
