E Show that the closure of a set S is the smallest closed se

(E) Show that the closure of a set S is the smallest closed set containing S.

Solution

let x belongs to S

then x belongs to A(alpha) for every alpha

so , x belongs to intersection of A = S-bar

thus S subset of S-bar

Now suppose there is some closed set C with S subset of C subset of S-bar

but C is superset of S which implies that

C belongs to A

so S-bar subset of C

=> S-bar =C

therefore S-bar is the smallest closed set containing S