Proof that the Smallest Closed Set Containing a Set is the Closure of the Set
Ifis a set, the smallest closed set containingis the closure of
We can also writewhereis closed andfor each
Letbe a set andsuch that
SupposeandSince eachis closedis also closed andis open.
Hence x has a neighbourhoodofsuch that
Now we must proveSupposethen there is a neighbourhoodof such that
Thereforeis closed and containsfor some