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
Sinceeachhas a neighbourhoodsuch thathenceis open andis closed.