# Set of cluster points being closed

## 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 r>0 s.t. B(c;r) is contained in X\G.
But I can't show how to do this.

There might be another way to do this is that,
if yn from n=1 to infinity is a convergen sequnce in G,
then its limit is contained in G. But how could we construct
such sequence, and show that its limit lies within G?

need help pls. thanks!

HallsofIvy
Homework Helper
If p is a cluster point of {xn}, then, given any r, there exist an xn within distance r of p.

If p is in X\G, then p is NOT a cluster point and so that is not true for some r. Now show that there is no member of G within distance r/2 of p.

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.

Dick
Homework Helper
You can show C is closed by showing it's complement is open. Isn't that what you are trying to do?

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.

Dick
Homework Helper
The C's and G's are getting a little confused here. They are the same thing, yes? Halls is saying to take p to be an element of X/G. If you can show there is an open ball around any such p which is in X/G, then X/G is open. Isn't it?

matt grime