Suppose that

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- May 28th 2009, 11:19 AMflower3clousre
Suppose that

- May 28th 2009, 01:40 PMJose27
If is an open subset in then is open in . So, if is closed in then is closed in . Now you define the closure of in as the intersection of all closed subsets of such that they contain , but since closed sets in are closed in then the closure in has more elements to intersect and thus is contained in the closure in .

Well, that is an overview of the proof. I leave the details to you. - May 28th 2009, 02:13 PMPlato
Here is a different approach.

if and only if each for each open set , .

But .