Urysohn's Lemma

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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.

$\blacksquare$


Also see


Source of Name

This entry was named for Pavel Samuilovich Urysohn.


Sources

This article incorporates material from Urysohn's Lemma on PlanetMath, which is licensed under the Creative Commons Attribution/Share-Alike License.