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Thread: Closed, Bounded but not Compact.

  1. #1
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    Closed, Bounded but not Compact.

    The set of rationals Q forms a metric spce by $\displaystyle d(p,q) = | p - q |$

    Then a subset E of Q is defined by $\displaystyle E = \{ p \in Q : 2 < p^2 < 3 \}$
    So I am trying to show that E is closed and bounded, but not compact.

    To me, it is clear than E is bounded (by 2 and 3?!).

    I am having trouble showing E is closed. I believe 2 and 3 are limit points of E but E does not contain them, so E is not closed? also, E does not contain e = 2.718281828.... but that is also a limit point of E, is it not? Maybe I am confused with my definition of limit point.... are any of these (2, 3, e) actually limit points of E?

    I think I can show E is not compact, since the open cover of E,
    $\displaystyle \bigcup\ (2 + \frac{1}{n}, 3 - \frac{1}{n}) $ for n = 2, 3, 4, ....
    has no finite subcover?


    I am worried I am perhaps misunderstanding some definitions... Any help or direction would be greatly appriciated.

    Thank you in advanced for you time!
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  2. #2
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    Do you understand that $\displaystyle E = \left( {\sqrt 2 ,\sqrt 3 } \right) \cap \mathbb{Q}~?$

    Consider this cover $\displaystyle O_n = \left( {\sqrt 2 + \frac{{\sqrt 3 - \sqrt 2 }}{{3n}},\sqrt 3 - \frac{{\sqrt 3 - \sqrt 2 }}{{3n}}} \right)$
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  3. #3
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    Wow, thank you for pointing that out. I somehow goofed on what the set E was!

    However, even in this new light, E does not appear to be closed... $\displaystyle \sqrt 2 $ is not in E, but it is a limit point of E?

    Thanks!
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  4. #4
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    You are still missing the point $\displaystyle \sqrt2\notin \mathbb{Q}$.
    Does the set contain all of its limit points in $\displaystyle \mathbb{Q}~?$
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by matt.qmar View Post
    The set of rationals Q forms a metric spce by $\displaystyle d(p,q) = | p - q |$

    Then a subset E of Q is defined by $\displaystyle E = \{ p \in Q : 2 < p^2 < 3 \}$
    So I am trying to show that E is closed and bounded, but not compact.

    To me, it is clear than E is bounded (by 2 and 3?!).

    I am having trouble showing E is closed. I believe 2 and 3 are limit points of E but E does not contain them, so E is not closed? also, E does not contain e = 2.718281828.... but that is also a limit point of E, is it not? Maybe I am confused with my definition of limit point.... are any of these (2, 3, e) actually limit points of E?

    I think I can show E is not compact, since the open cover of E,
    $\displaystyle \bigcup\ (2 + \frac{1}{n}, 3 - \frac{1}{n}) $ for n = 2, 3, 4, ....
    has no finite subcover?


    I am worried I am perhaps misunderstanding some definitions... Any help or direction would be greatly appriciated.

    Thank you in advanced for you time!
    Another simplistic theorem tells you that if $\displaystyle X,Y,Z$ are metric spaces then $\displaystyle Z$ is a compact subspaces of $\displaystyle Y$ if and only if $\displaystyle Z$ is a compact subspaces of $\displaystyle Y$. Note though that with this in hand $\displaystyle E$ will be a compact subspaces of $\displaystyle \mathbb{Q}$ if and only if it's a compact subspace of $\displaystyle \mathbb{R}$. But, considering that it isn't closed ($\displaystyle \sqrt{2}$ is a limit point not in the set) it follows that $\displaystyle E$ is not compact.
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