# Urysohn's Lemma

## Lemma

Let $T = \struct {S, \tau}$ be a $T_4$ topological space.

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

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

## Proof

Let $T = \struct {S, \tau}$ be a $T_4$ space.

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

Let $P = \Q \cap \closedint 0 1$ where $\closedint 0 1$ is the closed 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:

$\Bbb U = \set {U_i: i \in P}$

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 $\sequence z$.

Let $z_0 = 1, z_1 = 0$.

In general, let $P_n$ denote the set consisting of the first $n$ elements of $\sequence z$.

Let $\map \PP n$ 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:

$\forall A, B \in \map \complement \tau, A \cap B = \O: \exists U_1, V \in \tau: A \subseteq U_1, B \subseteq V$

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 $\map \PP 1$ is shown to hold.

This is our basis for the induction.

#### Induction Hypothesis

This is our induction hypothesis:

Let $\map \PP k$ 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 $\map \PP k$ 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

This is our induction step:

Let $r = z_{k + 1}$ be the next element in $\sequence z$.

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

It is a finite subset of the closed unit interval $\closedint 0 1$.

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

From Finite Non-Empty Subset of Totally Ordered Set has Smallest and Greatest Elements, $P_{k + 1}$ 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 every 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 $\closedint 0 1$.

We now extend this definition to every rational $p$ by defining:

$U_p = \begin{cases} \O & : p < 0 \\ S & : p > 1 \end{cases}$

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

### Definition of Function

Let $x \in S$.

Define $\map \Q x = \set {p: x \in U_p}$.

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 $\map \Q x$ is bounded below and its infimum is an element in $\closedint 0 1$.

We define:

$\map f x = \map \inf {\map \Q x}$

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 $\map \Q x$ equals the set $\Q_+$ of all non-negative rationals.

Hence $\map f x = 0$.

Let $x \in B$.

Then $x \notin U_p$ for $p \le 1$ so $\map \Q x$ equals the set of all the rationals greater than $1$.

Hence $\map f x = 1$.

### Continuity of Function

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

$(a): \quad x \in {U_r}^- \implies \map f x \le r$

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

Thus $\map f x \le r$ by definition of $f$.

$\Box$

$(b): \quad x \notin U_r \implies \map f x \ge r$

We have: $x \notin U_r \implies \forall s < r: x \notin U_s$ so $\map \Q x$ contains no rational less than $r$.

Thus $\map f x \ge r$.

$\Box$

Let $x_0 \in S$ and let $\openint c d$ be an open real interval containing $\map f x$.

We will find a neighborhood $U$ of $x_0$ such that $\map f U \subseteq \openint c d$.

Choose $p, q \in \Q$ such that:

$c < p < \map f {x_0} < q < d$

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

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

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

Hence $x_0 \in U$ and thus $U$ is a neighborhood of $x_0$ because $U$ is open.

Finally, let $x \in U$.

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

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

Thus: $\map f x \in \closedint p q \subseteq \openint c d$

Therefore $f$ is continuous.

$\blacksquare$

## Source of Name

This entry was named for Pavel Samuilovich Urysohn.