Set of cluster points being closed

[supertopolo]

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]

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.

 

[supertopolo]

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]

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

 

[supertopolo]

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]

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]

Small point but G must be the set of cluster points, not 'a set containing the cluster points'.

Suppose {y_n} is a sequence of elements in G, and that {y_n} converges to y. We need to show that y is in G.

Remember that each {y_n} is a cluster point of the sequence {x_n}, so each y_i is arbitrarily close to some x_j. Now fill in the details.