Hello

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- Jan 27th 2008, 10:15 AMpatricia-donnellymetric space
Hello

- Jan 27th 2008, 02:16 PMThePerfectHacker
Suppose is an accumulation point of a set . Then there exists

**distinct**such that . Let then by definition of convergence it means for . Thus, are**distinct**elements which lie in and hence there are infinitely many.

Quote:

Using this, prove that any finite subset of X is closed.

- Jan 28th 2008, 09:34 AMpatricia-donnelly
You've been a great help. Thank you very much