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.