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Topic: intersection of embedded closed sets (Read 1984 times) 

MonicaMath
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intersection of embedded closed sets
« on: Sep 17^{th}, 2009, 11:55am » 
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Hi, I need to prove that? if {A_k}, k=1,..., infinity, is a collection of nonempty embedded closed sets of real numbers in decreasing order with A_j is bounded for one j, then : the intersection is nonempty ??


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Eigenray
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Re: intersection of embedded closed sets
« Reply #1 on: Sep 17^{th}, 2009, 6:12pm » 
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Are you familiar with the open cover definition of compactness? If the intersection were empty, we would have A_{j} = _{k>j} U_{k}, where U_{k} = A_{j} \ A_{k} is open in A_{j}. Since A_{j} is compact, this open cover has a finite subcover. But the U_{k} are nested increasing, so we must have A_{j} = U_{k} for some k, meaning A_{k} is empty, a contradiction. There is a more general version here.


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