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How do I show that the real numbers are not compact?

  1. Nov 19, 2006 #1
    A trivial, yet difficult question. How would one prove that the real numbers are not compact, only using the definition of being compact? In other words, what happens if we reduce an open cover of R to a finite cover of R?

    I let V be a collection of open subset that cover R
    Then I make the assumption that that this open cover can be reduced to a finite subcover.
    (Clearly this is not possible) I am struggling to see/show what happens to R when I make this assumption. Should I simply find a counterexample, if so, what could it look like?

    I have proven several, similar problems, but this one is so general that its tricky.
     
    Last edited: Nov 19, 2006
  2. jcsd
  3. Nov 19, 2006 #2

    0rthodontist

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    You have to check two properties:
    Closed?
    Bounded?
    What do you think?
     
  4. Nov 19, 2006 #3
    But the question states that I should do this proof directly from the definition of being compact. Ie, if an open cover can be reduced to a finite subcover.
     
  5. Nov 19, 2006 #4

    matt grime

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    Compact: every open cover has a finite subcover.
    So what is the definition of non-compact? Now show the reals satisfy this definition. If you can negate logical statements then this problem is not at all tricky.
     
  6. Nov 19, 2006 #5
    So, find one open cover that cannot be reduced to a finite subcover?
    But wouldn't that imply that I have to state what this open cover looks like? Should I maybe let R be an open cover of itself?
     
  7. Nov 19, 2006 #6

    matt grime

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    That open cover has exactly one set in it, R. One is a finite number.

    Find an open cover of infinitely many sets that does not have a finite subcover. Post what you're thinking, i.e. how you might make such a cover.

    Please, no one post a solution right away.
     
  8. Nov 19, 2006 #7
    I think I sorted some stuff out, and {n-1,n+1 n in Z} I believe must work!? Thanks for the support
     
  9. Nov 19, 2006 #8

    HallsofIvy

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    Well, I would write it (n-1,n+1) for n contained in Z. {n-1, n+1} means the set containing exactly two members, n-1 and n+1. That's not open. (I don't know what "n in Z" inside the braces could mean!)
     
  10. Nov 19, 2006 #9

    arildno

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    You might try and have a closer look at the family of sets [itex]S=\{(-n,n),n\in\mathbb{N}\}[/itex]
     
  11. Nov 19, 2006 #10

    Hurkyl

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    Fred's works too (and was essentially my first thought), assuming he meant { (n-1, n+1) | n in Z }.
     
  12. Nov 19, 2006 #11

    arildno

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    Agreed. I was unsure of what he meant with the type of parentheses he used.
     
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