Theorem

A spaceis regular if and only if, givenand a neighbourhoodofwiththere is a neighbourhoodofsuch that

Proof

Supposeis regular. Letrepresent an open neighbourhood ofthenis closed and

Hence open setsandexist such thatand

Sincewe haveand sincewe have

Hence

Now supposeandis an open neighbourhood ofThen an open neighbourhoodofexists such that

Letand letbe a closed subset ofwithis a neighbourhood of Then an open setexists such thatand

is open andandis an open subset ofcontaining

Henceandandare the required sets andis regular.