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Math Help - 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 d(p,q) = | p - q |

    Then a subset E of Q is defined by 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,
    \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 E = \left( {\sqrt 2 ,\sqrt 3 } \right) \cap \mathbb{Q}~?

    Consider this cover 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...  \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 \sqrt2\notin \mathbb{Q}.
    Does the set contain all of its limit points in \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 d(p,q) = | p - q |

    Then a subset E of Q is defined by 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,
    \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 X,Y,Z are metric spaces then Z is a compact subspaces of Y if and only if Z is a compact subspaces of Y. Note though that with this in hand E will be a compact subspaces of \mathbb{Q} if and only if it's a compact subspace of \mathbb{R}. But, considering that it isn't closed ( \sqrt{2} is a limit point not in the set) it follows that E is not compact.
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