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

Theorem

For every setandis closed if and only if

Ifis closed thenis open.

Ifthena neighbourhoodexists such that

henceand

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

and

henceandis closed.

Sinceeachhas a neighbourhoodsuch thathenceis open andis closed.