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 (x_n) 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.
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)
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!
So the notation D(E) means the set of all limit points of sequences in E? But that's the definition of closure of E. So D(E) is equal to the closure of E?
For the second part of the proof, I don't follow why x is not an element of E U D(E). Can you explain a little more, please?