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 ${\displaystyle f(B)=\{1\}.}$ 

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