Proof that Every Subset of a Set is Contained in its Closure and is Closed iff it is Contained in its Closure


For every setandis closed if and only if

Ifis closed thenis open.

Ifthena neighbourhoodexists such that


For the second part, if A is closed thenis open. Eachlies in a neighbourhoodsuch that


henceandis closed.

Sinceeachhas a neighbourhoodsuch thathenceis open andis closed.