# Urysohn's lemma

In topology, **Urysohn's lemma** is a lemma that states that a topological space is normal if and only if any two disjoint closed subsets can be separated by a continuous function.^{[1]}

Urysohn's lemma is commonly used to construct continuous functions with various properties on normal spaces. It is widely applicable since all metric spaces and all compact Hausdorff spaces are normal. The lemma is generalised by (and usually used in the proof of) the Tietze extension theorem.

The lemma is named after the mathematician Pavel Samuilovich Urysohn.

## Discussion

[edit]Two subsets and of a topological space are said to be separated by neighbourhoods if there are neighbourhoods of and of that are disjoint. In particular and are necessarily disjoint.

Two plain subsets and are said to be separated by a continuous function if there exists a continuous function from into the unit interval such that for all and for all Any such function is called a **Urysohn function** for and In particular and are necessarily disjoint.

It follows that if two subsets and are separated by a function then so are their closures. Also it follows that if two subsets and are separated by a function then and 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 and need not be precisely separated by , i.e., it is not necessary and guaranteed that and for outside and A topological space in which every two disjoint closed subsets and 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 T_{1} spaces are Tychonoff.

## Formal statement

[edit]A topological space is normal if and only if, for any two non-empty closed disjoint subsets and of there exists a continuous map such that and

## Proof sketch

[edit]The proof proceeds by repeatedly applying the following alternate characterization of normality. If is a normal space, is an open subset of , and is closed, then there exists an open and a closed such that .

Let and be disjoint closed subsets of . The main idea of the proof is to repeatedly apply this characterization of normality to and , continuing with the new sets built on every step.

The sets we build are indexed by dyadic fractions. For every dyadic fraction , we construct an open subset and a closed subset of such that:

- and for all ,
- for all ,
- For , .

Intuitively, the sets and expand outwards in layers from :

This construction proceeds by mathematical induction. For the base step, we define two extra sets and .

Now assume that and that the sets and have already been constructed for . Note that this is vacuously satisfied for . Since is normal, for any , we can find an open set and a closed set such that

The above three conditions are then verified.

Once we have these sets, we define if for any ; otherwise for every , where denotes the infimum. Using the fact that the dyadic rationals are dense, it is then not too hard to show that is continuous and has the property and This step requires the sets in order to work.

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

## See also

[edit]## Notes

[edit]**^**Willard 1970 Section 15.

## References

[edit]- 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.