Contents

# Contents

## Idea

Urysohn’s lemma (prop. below) states that on a normal topological space disjoint closed subsets may be separated by continuous functions in the sense that a continuous function exists which takes value 0 on one of the two subsets and value 1 on the other (called an “Urysohn function”, def. ) below. In fact the existence of such functions is equivalent to a space being normal (remark below).

Urysohn’s lemma is a key ingredient for instance in the proof of the Tietze extension theorem and in the proof of the existence of partitions of unity on paracompact topological spaces. See the list of implications below.

## Statement

###### Definition

(Urysohn function)

Let $X$ be a topological space, and let $A,B \subset X$ be disjoint closed subsets. Then an Urysohn function for this situation is

to the closed interval equipped with its Euclidean metric topology, such that

• it takes the value 0 on $A$ and the value 1 on $B$:

$f(A) = \{0\} \phantom{AAA} \text{and} \phantom{AAA} f(B) = \{1\} \,.$
###### Proposition

(Urysohn’s lemma)

Assuming excluded middle then:

Let $X$ be a normal (or $T_4$) topological space, and let $A,B \subset X$ be two disjoint closed subsets of $X$. Then there exists an Urysohn function (def. ).

###### Remark

Beware that the function in prop. may take the values 0 or 1 even outside of the two subsets. The condition that the function takes value 0 or 1, respectively, precisely on the two subsets corresponds to perfectly normal spaces.

###### Remark

It is immediate that, conversely, the existence of an Urysohn function (def. ) implies that the topological space is normal. For let $A, B \subset X$ be disjoint closed subsets, and consider a continuous function $f \colon X \to [0,1]$ with $f(A) = \{0\}$ and $f(B) = \{1\}$ then

$U_A \coloneqq f^{-1}([0,1/3) \phantom{AAA} U_B \coloneqq f^{-1}((2/3,1])$

are disjoint open neighbourhoods of these subsets.

Hence Urysohn’s lemma shows that a topological space being normal is equivalent to it admitting Urysohn functions.

## Proof

###### Proof

of Urysohn’s lemma, prop.

Set

$C_0 \coloneqq A \phantom{AAA} U_1 \coloneqq X \backslash B \,.$

Since by assumption

$A \cap B = \emptyset \,.$

we have

$C_0 \subset U_1 \,.$

Notice that (by this lemma) if a space is normal then every open neighbourhood $U \supset C$ of a closed subset $C$ contains a smaller neighbourhood $V$ together with its closure $Cl(V)$

$C \subset V \subset Cl(V) \subset U \,.$

Apply this fact successively to the above situation to obtain the following infinite sequence of nested open subsets $U_r$ and closed subsets $C_r$

$\array{ C_0 && && && &\subset& && && && U_1 \\ C_0 && &\subset& && U_{1/2} &\subset& C_{1/2} && &\subset& && U_1 \\ C_0 &\subset& U_{1/4} &\subset& C_{1/4} &\subset& U_{1/2} &\subset& C_{1/2} &\subset& U_{3/4} &\subset& C_{3/4} &\subset& U_1 }$

and so on, labeled by the dyadic rational numbers $\mathbb{Q}_{dy} \subset \mathbb{Q}$ within $(0,1]$ $\{ U_{r} \subset X \}_{r \in (0,1] \cap \mathbb{Q}_{dy}}$

with the property

$\underset{r_1,r_2 \in (0,1] \cap \mathbb{Q}_{dy}}{\forall} \left( \left( r_1 \lt r_2 \right) \Rightarrow \left( U_{r_1} \subset Cl(U_{r_1}) \subset U_{r_2} \right) \right) \,.$

Define then the function

$f \;\colon\; X \longrightarrow [0,1]$

to assign to a point $x \in X$ the infimum of the labels of those open subsets in this sequence that contain $x$:

$f(x) \coloneqq \underset{U_r \supset \{x\}}{\lim} r$

Here the limit is over the directed set of those $U_r$ that contain $x$, ordered by reverse inclusion.

This function clearly has the property that $f(A) = \{0\}$ and $f(B) = \{1\}$. It only remains to see that it is continuous.

To this end, first observe that

$\array{ (\star) && \left( x \in Cl(U_r) \right) &\Rightarrow& \left( f(x) \leq r \right) \\ (\star\star) && \left( x \in U_r \right) &\Leftarrow& \left( f(x) \lt r \right) } \,.$

Here it is immediate from the definition that $(x \in U_r) \Rightarrow (f(x) \leq r)$ and that $(f(x) \lt r) \Rightarrow (x \in U_r \subset Cl(U_r))$. For the remaining implication, it is sufficient to observe that

$(x \in \partial U_r) \Rightarrow (f(x) = r) \,,$

where $\partial U_r \coloneqq Cl(U_r) \backslash U_r$ is the boundary of $U_r$.

This holds because the dyadic numbers are dense in $\mathbb{R}$. (And this would fail if we stopped the above decomposition into $U_{a/2^n}$-s at some finite $n$.) Namely, in one direction, if $x \in \partial U_r$ then for every small positive real number $\epsilon$ there exists a dyadic rational number $r'$ with $r \lt r' \lt r + \epsilon$, and by construction $U_{r'} \supset Cl(U_r)$ hence $x \in U_{r'}$. This implies that $\underset{U_r \supset \{x\}}{\lim} = r$.

Now we claim that for all $\alpha \in [0,1]$ then

1. $f^{-1}(\,(\alpha, 1]\,) = \underset{r \gt \alpha}{\cup} \left( X \backslash Cl(U_r) \right)$

2. $f^{-1}(\,[0,\alpha)\,) = \underset{r \lt \alpha}{\cup} U_r$

Thereby $f^{-1}(\,(\alpha, 1]\,)$ and $f^{-1}(\,[0,\alpha)\,)$ are exhibited as unions of open subsets, and hence they are open.

Regarding the first point:

\begin{aligned} & x \in f^{-1}( \,(\alpha,1]\, ) \\ \Leftrightarrow\, & f(x) \gt \alpha \\ \Leftrightarrow\, & \underset{r \gt \alpha}{\exists} (f(x) \gt r) \\ \overset{(\star)}{\Rightarrow}\, & \underset{r \gt \alpha}{\exists} \left( x \notin Cl(U_r) \right) \\ \Leftrightarrow\, & x \in \underset{r \gt \alpha}{\cup} \left(X \backslash Cl(U_r)\right) \end{aligned}

and

\begin{aligned} & x \in \underset{r \gt \alpha}{\cup} \left(X \backslash Cl(U_r)\right) \\ \Leftrightarrow\, & \underset{r \gt \alpha}{\exists} \left( x \notin Cl(U_r) \right) \\ \Rightarrow\, & \underset{r \gt \alpha}{\exists} \left( x \notin U_r \right) \\ \overset{(\star \star)}{\Rightarrow}\, & \underset{r \gt \alpha}{\exists} \left( f(x) \geq r \right) \\ \Leftrightarrow\, & f(x) \gt \alpha \\ \Leftrightarrow\, & x \in f^{-1}(\, (\alpha,1] \,) \end{aligned} \,.

Regarding the second point:

\begin{aligned} & x \in f^{-1}(\, [0,\alpha) \,) \\ \Leftrightarrow\, & f(x) \lt \alpha \\ \Leftrightarrow\, & \underset{r \lt \alpha}{\exists}( f(x) \lt r ) \\ \overset{(\star \star)}{\Rightarrow}\, & \underset{r \lt \alpha}{\exists }( x \in U_r ) \\ \Leftrightarrow\, & x \in \underset{r \lt \alpha}{\cup} U_r \end{aligned}

and

\begin{aligned} & x \in \underset{r \lt \alpha}{\cup} U_r \\ \Leftrightarrow\, & \underset{r \lt \alpha}{\exists }( x \in U_r ) \\ \overset{}{\Rightarrow}\, & \underset{r \lt \alpha}{\exists }( x \in Cl(U_r) ) \\ \overset{(\star)}{\Rightarrow}\, & \underset{r \lt \alpha}{\exists }( f(x) \leq r ) \\ \Leftrightarrow\, & f(x) \lt \alpha \\ \Leftrightarrow\, & x \in f^{-1}(\, [0,\alpha) \,) \end{aligned} \,.

(In these derivations we repeatedly use that $(0,1] \cap \mathbb{Q}_{dy}$ is dense in $[0,1]$, and we use the contrapositions of $(\star)$ and $(\star \star)$, by excluded middle.)

Now since the subsets $\{ [0,\alpha), (\alpha,1]\}_{\alpha \in [0,1]}$ form a sub-base for the Euclidean metric topology on $[0,1]$, it follows that all pre-images of $f$ are open, hence that $f$ is continuous.

## Implications

Urysohn’s lemma is key in the proof of many other theorems, for instance

Due to Pavel Urysohn.

Lectures notes include

• Tarun Chitra, section 2.1 of The Stone-Cech Compactification 2009 pdf