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  • What does it mean to be compact in the $w^{*}$-topology?
  • How is it related to polytopes unit balls?
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Do you know what the "w^$*$-topology" means? (I might guess that it's an abbreviation for the "weak-$*$ topology", but I'm not sure.) –  Pete L. Clark Feb 16 '13 at 20:08
    
@PeteL.Clark if you can explian I'll be greatful –  0x90 Feb 16 '13 at 20:14
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Weak* topology is normally used in infinite-dimensional spaces (since all reasonable topologies coincide in finite-dimensional spaces). If by "polytope" you mean something in a finite-dimensional space, then I imagine weak* is of no help. –  GEdgar Feb 16 '13 at 20:17

1 Answer 1

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A set $\rm K$ is compact in the $\ast$-weak topology if and only if given any covering of $\rm K$ by $\ast$-weak open sets $(\rm U_i)_{i \in \rm I}$, there exists a finite subset $\rm J \subset \rm I$ such that the $(\rm U_j)_{j \in \rm J}$ is a covering of $\rm K$.

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I suspect the problem OP is having is not the definition of compact, but the definition of $w^*$, and whether the definition of "compact" has any other characterization in this sort of topology –  Thomas Andrews Feb 15 '13 at 13:28
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Seriously? $ $ $ $ –  Martin Feb 15 '13 at 13:42

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