Closure proof

mario22

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Dec 11, 2010
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Assume that (X,d) is a metric space and that A,B an open subsets of X such that closure A = closure B = X.
Show that the closure of (A intersect B) is X.

who can help with that?

Thanks,
Mario
 
Just a sketch: The closure of (A intersect B) is a subset of both the closure of A and the closure of B which are both equal to X. Conversely, any point x in X is a limit of some sequence in A, so let {an} be a sequence in A tending to x. We may construct a subsequence of the an's which belong to B. This is because B is open. Take any delta>0, then for any an we can find an element of B, call it bn which is closer than delta to an. Chosing delta appropriately, we have that bn must belong to A. This produces a sequence in (A intersect B) which tends to x.
 
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