If a subset of a top. space is closed, then it contains all its limit points. So I need to show that if is closed then . Since A is closed, is open. Also, A' is the set of all limit points of A. I am not sure if this helps, but
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Originally Posted by dwsmith If a subset of a top. space is closed, then it contains all its limit points. So I need to show that if is closed then . Since A is closed, is open. Also, A' is the set of all limit points of A. I am not sure if this helps, but I'm confused, you're trying to prove that if is closed then where is DEFINED to be ? Well, isn't so that trivially ?
Originally Posted by Drexel28 I'm confused, you're trying to prove that if is closed then where is DEFINED to be ? Well, isn't so that trivially ? No. The limit points don't have to to be in A.
Originally Posted by dwsmith The limit points don't have to to be in A. They must belong to A if A is closed. In the OP you wrote: if is closed then . Note that one way of defining as the smallest closed set that contains .
I think there is an issue. The also part is clarifying notation not part of the statement to prove.
Originally Posted by dwsmith I think there is an issue. The also part is clarifying notation not part of the statement to prove. This is your OP. Originally Posted by dwsmith;682765 So I need to show that if [TEX A\subset X[/TEX] is closed then . Since A is closed, There it is. You said it twice. Because is open then by definition of limit point, all .
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