Urysohn's lemma

Two subsets ${\displaystyle A}$ and ${\displaystyle B}$ of a topological space ${\displaystyle X}$ are said to be separated by neighbourhoods if there are neighbourhoods ${\displaystyle U}$ of ${\displaystyle A}$ and ${\displaystyle V}$ of ${\displaystyle B}$ that are disjoint. In particular ${\displaystyle A}$ and ${\displaystyle B}$ are necessarily disjoint.

Two plain subsets ${\displaystyle A}$ and ${\displaystyle B}$ are said to be separated by a continuous function if there exists a continuous function ${\displaystyle f:X\to [0,1]}$ from ${\displaystyle X}$ into the unit interval ${\displaystyle [0,1]}$ such that ${\displaystyle f(a)=0}$ for all ${\displaystyle a\in A}$ and ${\displaystyle f(b)=1}$ for all ${\displaystyle b\in B.}$ Any such function is called a Urysohn function for ${\displaystyle A}$ and ${\displaystyle B.}$ In particular ${\displaystyle A}$ and ${\displaystyle B}$ are necessarily disjoint.

It follows that if two subsets ${\displaystyle A}$ and ${\displaystyle B}$ are separated by a function then so are their closures. Also it follows that if two subsets ${\displaystyle A}$ and ${\displaystyle B}$ are separated by a function then ${\displaystyle A}$ and ${\displaystyle B}$ are separated by neighbourhoods.

A normal space is a topological space in which any two disjoint closed sets can be separated by neighbourhoods. Urysohn's lemma states that a topological space is normal if and only if any two disjoint closed sets can be separated by a continuous function.

The sets ${\displaystyle A}$ and ${\displaystyle B}$ need not be precisely separated by ${\displaystyle f}$, i.e., it is not necessary and guaranteed that ${\displaystyle f(x)\neq 0}$ and ${\displaystyle \neq 1}$ for ${\displaystyle x}$ outside ${\displaystyle A}$ and ${\displaystyle B.}$ A topological space ${\displaystyle X}$ in which every two disjoint closed subsets ${\displaystyle A}$ and ${\displaystyle B}$ are precisely separated by a continuous function is perfectly normal.

Urysohn's lemma has led to the formulation of other topological properties such as the 'Tychonoff property' and 'completely Hausdorff spaces'. For example, a corollary of the lemma is that normal T₁ spaces are Tychonoff.

A topological space ${\displaystyle X}$ is normal if and only if, for any two non-empty closed disjoint subsets ${\displaystyle A}$ and ${\displaystyle B}$ of ${\displaystyle X,}$ there exists a continuous map ${\displaystyle f:X\to [0,1]}$ such that ${\displaystyle f(A)=\{0\}}$ and ${\displaystyle f(B)=\{1\}.}$

Illustration of the first few sets built as part of the proof.

The proof proceeds by repeatedly applying the following alternate characterization of normality. If ${\displaystyle X}$ is a normal space, ${\displaystyle Z}$ is an open subset of ${\displaystyle X}$, and ${\displaystyle Y\subseteq Z}$ is closed, then there exists an open ${\displaystyle U}$ and a closed ${\displaystyle V}$ such that ${\displaystyle Y\subseteq U\subseteq V\subseteq Z}$.

Let ${\displaystyle A}$ and ${\displaystyle B}$ be disjoint closed subsets of ${\displaystyle X}$. The main idea of the proof is to repeatedly apply this characterization of normality to ${\displaystyle A}$ and ${\displaystyle B^{\complement }}$, continuing with the new sets built on every step.

The sets we build are indexed by dyadic fractions. For every dyadic fraction ${\displaystyle r\in (0,1)}$, we construct an open subset ${\displaystyle U(r)}$ and a closed subset ${\displaystyle V(r)}$ of ${\displaystyle X}$ such that:

• ${\displaystyle A\subseteq U(r)}$ and ${\displaystyle V(r)\subseteq B^{\complement }}$ for all ${\displaystyle r}$,
• ${\displaystyle U(r)\subseteq V(r)}$ for all ${\displaystyle r}$,
• For ${\displaystyle r<s}$, ${\displaystyle V(r)\subseteq U(s)}$.

Intuitively, the sets ${\displaystyle U(r)}$ and ${\displaystyle V(r)}$ expand outwards in layers from ${\displaystyle A}$:

${\displaystyle {\begin{array}{ccccccccccccccc}A&&&&&&&\subseteq &&&&&&&B^{\complement }\\A&&&\subseteq &&&\ U(1/2)&\subseteq &V(1/2)&&&\subseteq &&&B^{\complement }\\A&\subseteq &U(1/4)&\subseteq &V(1/4)&\subseteq &U(1/2)&\subseteq &V(1/2)&\subseteq &U(3/4)&\subseteq &V(3/4)&\subseteq &B^{\complement }\end{array}}}$

This construction proceeds by mathematical induction. For the base step, we define two extra sets ${\displaystyle U(1)=B^{\complement }}$ and ${\displaystyle V(0)=A}$.

Now assume that ${\displaystyle n\geq 0}$ and that the sets ${\displaystyle U\left(k/2^{n}\right)}$ and ${\displaystyle V\left(k/2^{n}\right)}$ have already been constructed for ${\displaystyle k\in \{1,\ldots ,2^{n}-1\}}$. Note that this is vacuously satisfied for ${\displaystyle n=0}$. Since ${\displaystyle X}$ is normal, for any ${\displaystyle a\in \left\{0,1,\ldots ,2^{n}-1\right\}}$, we can find an open set and a closed set such that

${\displaystyle V\left({\frac {a}{2^{n}}}\right)\subseteq U\left({\frac {2a+1}{2^{n+1}}}\right)\subseteq V\left({\frac {2a+1}{2^{n+1}}}\right)\subseteq U\left({\frac {a+1}{2^{n}}}\right)}$

The above three conditions are then verified.

Once we have these sets, we define ${\displaystyle f(x)=1}$ if ${\displaystyle x\not \in U(r)}$ for any ${\displaystyle r}$; otherwise ${\displaystyle f(x)=\inf\{r:x\in U(r)\}}$ for every ${\displaystyle x\in X}$, where ${\displaystyle \inf }$ denotes the infimum. Using the fact that the dyadic rationals are dense, it is then not too hard to show that ${\displaystyle f}$ is continuous and has the property ${\displaystyle f(A)\subseteq \{0\}}$ and ${\displaystyle f(B)\subseteq \{1\}.}$ This step requires the ${\displaystyle V(r)}$ sets in order to work.

The Mizar project has completely formalised and automatically checked a proof of Urysohn's lemma in the URYSOHN3 file.

• Mollifier

• Willard, Stephen (2004) [1970]. General Topology. Mineola, N.Y.: Dover Publications. ISBN 978-0-486-43479-7. OCLC 115240.
• Willard, Stephen (1970). General Topology. Dover Publications. ISBN 0-486-43479-6.

• "Urysohn lemma", Encyclopedia of Mathematics, EMS Press, 2001 [1994]
• Mizar system proof: http://mizar.org/version/current/html/urysohn3.html#T20