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PROVING INTERSECTION OF Any number of COMPACT SETS is COMPACT?

  1. Oct 24, 2008 #1
    I'm stuck ...
    Ive proved the intersection of any number of closed sets is closed ...
    and Let S = { A_a : a Element of I } be an collection of compact sets...then
    by heine Borel Theorem ...Each A_a in S is closed...so this part is done now I just
    have to show the intersection is bounded...
    so I said since each A_a in S is bounded by Heine Borel ..then A_a is a subset of [a,b]
    and then this is when Im really stuck:
    Let B={[a,b] : There exists an A_a in S such that A_a subset of [a,b]}

    I took the intersection of both sides of the subset ...and said Intersection S is a subset of
    Intersection of [a,b] so Intersection of S is bounded and therefore compact

    but somehow I think this is an error ..I dont know how to do this formally

    Please help
     
  2. jcsd
  3. Oct 24, 2008 #2

    morphism

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    The intersection of a bunch of sets is a subset of each of the sets. And a subset of a bounded set is <fill blank>.
     
  4. Oct 25, 2008 #3

    HallsofIvy

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    Are you talking about sets or real numbers? Or, more generally, in Rn?

    It is true that in any metric topology, a compact set is both closed and bounded.

    It is NOT true that in any metric topology, closed and bounded sets are compact.
    For example, the Heine-Borel theorem is not true of the rational numbers with d(x,y)= |x-y|.

    If you are working in the real numbers, then morphism is giving you a good hint: if A is a bounded set the A intersect ANY other sets is bounded.

    If you are working in any topological space, the theorem is still true. You would need to look at the basic definition of "compact" because without a metric, "bounded" is not even defined: let {U} be an open cover for A and show it has a finite subcover.
     
  5. Oct 26, 2008 #4

    mathwonk

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    this seems to be false, unless by compact you mean the bourbaki definition of compact which includes hausdorff as part of the definition. if that is what you mean by compact then a compact set is also closed. so since a closed subset of a compact set is compact you get it easily. but the disjoint union of two closed unit discs, then identified except at the origin seems to be a space in which both discs are compact but their intersection is not, being a punctured disc.
     
  6. Oct 26, 2008 #5

    morphism

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    I think the OP is working in R^n, because he/she mentioned Heine-Borel.
     
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