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Math Help - compact sets

  1. #1
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    compact sets

    Can anyone help me to understand why set S =(0,2) for each n in the Natural Numbers is not compact in the most basic way possible?

    I know so far that S is compact iff every open cover of S contains a finite subcover.

    and i also know that a Set (1/n,3) is a open cover of the set (0,2).

    thank you.
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  2. #2
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    The collection \left\{ {\left( {0,2 - \frac{1}{n}} \right):n \in \mathbb{Z}^ +  } \right\} is an open covering of (0,2).
    Can you give us a finite subcover of that cover?
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  3. #3
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    i don't think you can.. because you have infinite many numbers in that set (0,2-(1/n))

    but a cover of that would be (0,1) ?? but this would be infinite?? if i'm not wrong..
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  4. #4
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    wait i think what i said before is non sense..
    couldn't the set (1/n,2), be a subcover of the set you gave??
    Last edited by pandakrap; January 27th 2009 at 11:56 AM.
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  5. #5
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    Quote Originally Posted by pandakrap View Post
    couldn't the set (1/n,2), be a subcover of the set you gave??
    The set \left(\frac{1}{n},2\right) is not even in the cover I gave you!
    It must be a finite collection of subsets of the given collection.
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  6. #6
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    i'm sorry.. it's a bit difficult to picture this...


    would (0,2-(2/n)) be a subcover of your cover (0,2-(1/n)) ??
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  7. #7
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    Quote Originally Posted by pandakrap View Post
    i'm sorry.. it's a bit difficult to picture this...


    would (0,2-(2/n)) be a subcover of your cover (0,2-(1/n)) ??
    (0, 2-(2/n)) is not an open cover. O = \{(0, 2-(2/n))\}_{n=2}^{\infty} is an open cover for (0,2). Same with O = \{(0, 2-(1/n))\}_{n=2}^{\infty}.

    Even though the above O is an open cover for (0,2), it has no finite subcover for (0,2). That means, n should go to infinity to cover (0,2).

    The above one is a counterexample why (0,2) is not compact, because a subspace of a space X is compact if and only if every open cover of A by open sets in X has a finite subcover
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