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I've been reading about functional analysis and topology and I came across this sentence:

subset of a topological vector space is compact with respect to the weak topology

I have been trying to search the web for an explanation to understand what this means, and I also tried the chat rooms without finding a satisfactory answer. English is not my native language so that might be one of the reasons I'm not getting the sentence. I do understand the concepts of topological space, compactness and weak topology.

My question is: What does this sentence mean, explicitly?

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  • $\begingroup$ Shouldn't be your "topological" be "linear"? $\endgroup$ – Przemysław Scherwentke Apr 30 '18 at 9:03
  • $\begingroup$ I'm sorry @PrzemysławScherwentke but I'm not 100% sure what you're asking. I added the "vector" term to my question. $\endgroup$ – jjepsuomi Apr 30 '18 at 9:08
  • $\begingroup$ And this "vector" makes the difference. :-) $\endgroup$ – Przemysław Scherwentke Apr 30 '18 at 9:13
  • $\begingroup$ Excellent, thank you :-) $\endgroup$ – jjepsuomi Apr 30 '18 at 9:15
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    $\begingroup$ If you know weak topology the statement simply means any open cover of the subset by open sets in the weak topology has a finite subcover. For example, the closed unit ball of an infinite dimensional Hilbert space is not compact in the norm topology but it is compact with respect to the weak topology. $\endgroup$ – Kavi Rama Murthy Apr 30 '18 at 9:29
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If you know weak topology the statement simply means any open cover of the subset by open sets in the weak topology has a finite subcover. For example, the closed unit ball in an infinite dimensional Hilbert space is not compact in the norm topology but it is compact in the weak topology.

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  • $\begingroup$ Hi @KaviRamaMurthy okay, thank you very much! So every subset $t\in\tau_w$ is compact (where $\tau_w$ is the weak topology). Got it! :-) $\endgroup$ – jjepsuomi May 1 '18 at 17:01

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