Prove that, if S is open, each of its points is a point of accumulation of S.

How can I prove this??

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- January 20th 2011, 06:40 PMmathsohardPoint sets
Prove that, if S is open, each of its points is a point of accumulation of S.

How can I prove this?? - January 20th 2011, 06:54 PMDrexel28
- January 20th 2011, 08:57 PMmathsohard
This is not true??? hmm... I am using Advanced calculus by taylor and mann and on page 517 #5 they asked me to prove :(

- January 20th 2011, 09:05 PMDrexel28
It's clearly untrue in arbitrary spaces, but since this is an advanced calculus book I'm going to take a wild guess and assume you meant and open subset of . If so then this is true if is non-empty. Let and any neighborhood of it, then since is open we know there exists a neighborhood of such that . Note then that is a neighborhood of contained in . Choose some open ball and use the uncountability of open balls in euclidean space to find a point different from in .

- January 21st 2011, 04:36 AMPlato