Urysohn's Lemma

Lemma
Let $T = \left({S, \tau}\right)$ be a $T_4$ topological space.

Let $A, B \subseteq S$ be closed sets of $T$ such that $A \cap B = \varnothing$.

Then there exists an Urysohn function for $A$ and $B$.

Proof
Let $T = \left({S, \tau}\right)$ be a $T_4$ space.

Let $A, B \subseteq S$ be closed sets of $T$ such that $A \cap B = \varnothing$.

Let $P = \Q \cap \left[{0 \,.\,.\, 1}\right]$ where $\left[{0 \,.\,.\, 1}\right]$ is the unit interval.

$\Q$ is countable, therefore so is $P$.

Creation of Domain
We are going to construct a set $\Bbb U \subseteq \tau$ of open sets with $P$ as an indexing set:
 * $\Bbb U = \left\{{U_i: i \in P}\right\}$

such that:
 * $\forall p, q \in P: p < q \implies U_p^- \subseteq U_q$

where $U_p^-$ denotes the set closure of $U_p$.

We define $U_p$ by induction, as follows.

List the elements of $P$ in the form of an infinite sequence $\left \langle{z}\right \rangle$.

Let $z_0 = 1, z_1 = 0$.

In general, let $P_n$ denote the set consisting of the first $n$ elements of $\left \langle{z}\right \rangle$.

Let $\mathcal P \left({n}\right)$ be the proposition:
 * $U_p$ is defined for all $p \in P_n$, and:
 * $(1): \quad \forall p, q \in P_n: p < q \implies U_p^- \subseteq U_q$

Basis for the Induction
As $S$ is a $T_4$ space we have that:


 * $\forall A, B \in \complement \left({\tau}\right), A \cap B = \varnothing: \exists U_1, V \in \tau: A \subseteq U_1, B \subseteq V$

We have that $A \subseteq U_1$ is a closed set of $S$.

We define $U_1 = S \setminus B$ where $S \setminus B$ denotes the complement of $B$ in $S$.

As $S$ is $T_4$, we can choose an open set $U_0 \in \tau$ such that $A \subseteq U_0$ and $U_0^- \subseteq U_1$.

Thus $\mathcal P \left({1}\right)$ is shown to hold.

Induction Hypothesis
Let $\mathcal P \left({k}\right)$ be the proposition:
 * $U_p$ is defined for all $p \in P_k$, and:
 * $(1): \quad \forall p, q \in P_k: p < q \implies U_p^- \subseteq U_q$

We want to show that if $\mathcal P \left({k}\right)$ holds, then:


 * $U_p$ is defined for all $p \in P_{k+1}$, and:
 * $(1): \quad \forall p, q \in P_{k+1}: p < q \implies U_p^- \subseteq U_q$

Induction Step
Let $r = z_{k+1}$ be the next element in $\left \langle{z}\right \rangle$.

Consider $P_{k+1} = P_k \cup \left\{{r}\right\}$.

It is a finite subset of the unit interval $\left[{0 \,.\,.\, 1}\right]$.

We consider the usual $<$ ordering on $P_{k+1}$, which is a subset of $\left[{0 \,.\,.\, 1}\right]$ which in turn is a subset of $\R$.

From Finite Subset of Totally Ordered Set, $P_{k+1}$ has both a minimal element $m$ and a maximal element $M$.

From Predecessor and Successor of Finite Toset, every element other than $m$ and $M$ has an immediate predecessor and immediate successor.

We already know that $z_1 = 0$ is the minimal element and $z_0 = 1$ is the maximal element of $P_{k+1}$.

So $r$ must be neither of these.

Thus:
 * $r$ has an immediate predecessor $p$
 * $r$ has an immediate successor $q$

in $P_{k+1}$.

The sets $U_p$ and $U_q$ are already defined by the inductive hypothesis.

As $T$ is a $T_4$ space, there exists an open set $U_r \subseteq \tau$ such that:
 * $U_p^- \subseteq U_r$
 * $U_r^- \subseteq U_q$

We now show that $(1)$ holds for every pair of elements in $P_{k+1}$.

If both elements are in $P_n$, then $(1)$ is true by the inductive hypothesis.

If one is $r$ and the other is $s \in P_k$, then:
 * $s < p \implies U_s^- \subseteq U_p^- \subseteq U_r$

and:
 * $s \ge q \implies U_r \subseteq U_q^- \subseteq U_s^-$

Thus $(1)$ holds for ever pair of elements in $P_{k+1}$.

Therefore by induction, $U_p$ is defined for all $p \in P$.

We have defined $U_p$ for all rational numbers in $\left[{0 \,.\,.\, 1}\right]$.

We now extend this definition to every rational $p$ by defining:
 * $U_p = \begin{cases}

\varnothing & : p < 0 \\ S & : p > 1 \end{cases}$

It is easily checked that $(1)$ still holds.

Definition of Function
Let $x \in S$.

Define $\Q \left({x}\right) = \left\{{p: x \in U_p}\right\}$.

This set contains no rational number less than $0$ and contains every rational number greater than $1$ by the definition of $U_p$ for $p < 0$ and $p > 1$.

Thus $\Q \left({x}\right)$ is bounded below and its infimum is an element in $\left[{0 \,.\,.\, 1}\right]$.

We define:
 * $f \left({x}\right) = \inf \left({\Q \left({x}\right)}\right)$

Now we need to show that $f$ satisfies the conditions of Urysohn's Lemma.

Let $x \in A$.

Then $x \in U_p$ for all $p \ge 0$, so $\Q \left({x}\right)$ equals the set $\Q_+$ of all nonnegative rationals.

Hence $f \left({x}\right) = 0$.

Let $x \in B$.

Then $x \notin U_p$ for $p \le 1$ so $\Q \left({x}\right)$ equals the set of all the rationals greater than $1$.

Hence $f \left({x}\right) = 1$.

Continuity of Function
To show that $f$ is continuous, we first prove two smaller results:


 * $(a): \qquad x \in U_r \implies f \left({x}\right) \le r$

We have: $x \in U_r^- \implies \forall x > r: x \in U_s$ so $\Q \left({x}\right)$ contains all rationals greater than $r$.

Thus $f \left({x}\right) \le r$ by definition of $f$.


 * $(b): \quad x \notin U_r \implies f \left({x}\right) \ge r$

We have: $x \in U_r \implies \forall s < r: x \notin U_s$ so $\Q \left({x}\right)$ contains no rational less than $r$.

Thus $f \left({x}\right) \ge r$.

Let $x_0 \in S$ and let $\left({c \,.\,.\, d}\right)$ be an open real interval containing $f \left({x}\right)$.

We will find a neighborhood $U$ of $x_0$ such that $f \left({U}\right) \subseteq \left({c \,.\,.\, d}\right)$.

Choose $p, q \in \Q$ such that:
 * $c < p < f \left({x_0}\right) < q < d$

Let $U = U_q \setminus U_p^-$.

Then since $f \left({x_0}\right) < q$, we have that $(b)$ implies vacuously that $x \in U_q$.

Since $f \left({x_0}\right) > p$, $(a)$ implies that $x_0 \notin U_p$.

Hence $x_0 \in U$.

Finally, let $x \in U$.

Then $x \in U_q \subseteq U_q^-$, so $f \left({x}\right) \le q$ by $(a)$.

Also, $x \notin U_p^-$ so $x \notin U_p$ and $f \left({x}\right) \ge p$ by $(b)$.

Thus: $f \left({x}\right) \in \left[{p \,.\,.\, q}\right] \subseteq \left({c \,.\,.\, d}\right)$

Therefore $f$ is continuous.

Also see

 * Urysohn's Lemma Converse