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Hurkyl

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Note that A is a closed subset of A....

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Hmm, I'm not sure. I never said A was closed. A is in B which is in X. And some other set in A, maybe call it U, is closed. I think the answer would be U is closed in X but I don't quite see why.Note that A is a closed subset of A....

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HallsofIvy

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Ahh i see. Thanks. But is that the only problem here? If we insist that A is compact in B, then that fixes the problem and the statement is true, right?anytopological space is both open and closed as a subset of itself. If A is NOT closed as a subset of topological space B, since it IS closed as a subset of itself, the statement "any closed subset of A is also a closed subset of B" is false.

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matt grime

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HallsofIvy

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Oh, I see. I started to give the proof and then realized I was saying "given points p and q construct neighborhoods about p and q that do not intersect". That's not possible in some topological spaces.

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matt grime

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U is open if and only if U contains the interval (0,1) - the set (0,1) is in this and is certainly compact, but not closed.

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