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Is C^n or R^n compact?

  1. Feb 11, 2008 #1
    1. The problem statement, all variables and given/known data
    Is C^n or R^n compact?



    3. The attempt at a solution
    They are not bounded so can't be compact.
     
  2. jcsd
  3. Feb 11, 2008 #2

    morphism

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    Right. Of course you can also exhibit an infinite open cover with no finite subcover, namely {B_n : n>=1}, where B_n={x:|x|<n} is the open ball centered at 0 of radius n.
     
  4. Feb 11, 2008 #3
    What does your example mean? That C^n or R^n could be compact?
     
  5. Feb 11, 2008 #4

    morphism

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  6. Feb 11, 2008 #5
    What does it mean then?
     
  7. Feb 11, 2008 #6

    morphism

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    Hmm. I take it by "compact" you mean "closed and bounded"? This isn't the most general definition of compactness, which usually goes as follows: a topological space X is compact if every collection of open sets whose union is X has a finite subcollection whose union is still X (more succinctly put: if every open cover of X admits a finite subcover). In R^n and C^n with their standard topologies, it's a theorem (Heine-Borel) that this definition is equivalent to "closed and bounded"; in a general metric space this can be adapted to "complete and totally bounded".

    So what I did in my post was exhibit an open cover of R^n (and C^n) that doesn't have a finite subcover, which means (by the general definition) that they're not compact.
     
  8. Feb 11, 2008 #7
    Is C^n/{0} compact? How about R^n/{0}?
     
  9. Feb 11, 2008 #8

    quasar987

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    Why do you have doubts?
     
  10. Feb 12, 2008 #9
    Those spaces are unbounded so they must not be compact even after deleting a fintie number of zeroes.
     
  11. Apr 13, 2009 #10
    Pivoxa15, your definition of compactness is not the most general one. It relies on defining what a bounded set is , which in turn relies on some concept of "size". This is generally done by using norms but sometimes you don't have a norm to work with.
    The hierarchy of spaces is as follows: Inner product spaces < Normed Spaces < Metric Spaces < Topological Spaces, where "A<B" means that the type of Spaces "A" induces a type of space B but there are spaces of type B that are not of type A. For example, there is a method to determine if a norm could be generated from an inner product. It is called the "polarization identity". Look it up.

    With this in mind, I would say that the problem you posted is not as trivial as saying "it's not bounded" if the definition you are supposed to use is the general one (but still, it's not that much of a difficult problem).
     
  12. Apr 14, 2009 #11
    Or you could use the extra structure you know there is on C^n or R^n and the Heine-Borel theorem. As long as you mention that, not bounded is quite enough, even if you're working with the open cover definition.
     
  13. Apr 21, 2009 #12
    yes , but I'm sure he was using books where you're not supposed to assume the extra structure :)
     
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