Let (X,d) be a metric space and E is a subset of X. Prove that
(c means complement, E bar means the closure of E)
The Attempt at a Solution
Let (X,d) be a metric space and B(r,x) is the open ball of radius r about x.
Definition: Let F be a subset of X. The interior of F, int(F), is the set of all x E F such that there is an r > 0 such that B(r,x) is contained in F.
Definition: Let D be a subset of X. By definition, D is open iff for all a E D, there exists r>0 such that B(r,a) is contained in D.
Definition: Let F be a subset of X. F is called closed iff whenever (xn) is a sequence in F which converges to a E X, then a E F. (i.e. F contains all limit points of sequences in F) The closure of F is the set of all limit points of sequences in F.
Theorem: x is in the closure of F iff for all r>0, B(r,x) intersect F is nonempty.
I wrote all the relevant definitions in front of me, but I still can't figure out this one. I know that for proving equality of sets, I have to prove that A is contianed in B(x E A => x E B) and B is contianed in A (x E B => x E A). However, for this problem, I can write down the basic definitions for each direction, but I just can't figure out the connection between the two sets...
Any help is greatly appreciated!