yes. but from what HallsofIvy said "Now show that there is no member of G within distance r/2 of p." I don't understand why this had to hold, and if it holds, i don't understand how that will make X\C be open.
I had a problem with wording of the question. C is a subset of X containing only the cluster points of the sequence {xn}. C doesn't contain any other elements.
Homework Statement
(X,d) metric space, we have a sequence xn from n=1 to infinity
G is a subset of X containing all cluster points of sequence xn.
need to show that G is closed.
The Attempt at a Solution
I tried to show that X\G is open. so take any point c in X\G,
there exists an...