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Set of cluster points being closed

  • #1

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!
 

Answers and Replies

  • #2
HallsofIvy
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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.
 
  • #3
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.
 
  • #4
Dick
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You can show C is closed by showing it's complement is open. Isn't that what you are trying to do?
 
  • #5
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.
 
  • #6
Dick
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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?
 
  • #7
matt grime
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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.
 

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