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Math Help - compactness

  1. #1
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    compactness

    Regard Q, the set of all rational numbers, as a metric space, with
    d(p,q)=|p-q|. Let E be the set of all P in Q such that 2<P^2<3. Show
    that E is closed and bounded in Q, but that E is not compact.
    Is E open in Q?

    I could prove that E is closed, open and bounded. But will the following solution suffice to show that E is not compact?

    I can find a sequence of points {Kn} = 1.7, 1.73, 1.732,.... in E. This is a sequence converging to 3^(1/2).
    So {Kn} is an infinite subset of E which does not have a limit point in E.
    So E is not compact.
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  2. #2
    Senior Member roninpro's Avatar
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    I can find a sequence of points {Kn} = 1.7, 1.73, 1.732,.... in E. This is a sequence converging to 3^(1/2).
    So {Kn} is an infinite subset of E which does not have a limit point in E.
    So E is not compact.
    I think that you need to be careful about your claim that \{K_n\} does not have a limit point in E. How do you know?
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  3. #3
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    @ roninpro
    Let x be a limit point of {Kn}. Then I can find a sequence of points in Kn converging to x. But {Kn} = 1.7, 1.73, 1.732, ..... which is a seq convering to 3^(1/2), so any subsequence also converges to 3^(1/2). So x has to be 3^(1/2).

    Am I right?
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  4. #4
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    Yes, but the point is a little more subtle than that. Because your sequence, as a sequence of real numbers, converges to \sqrt{2} and limits are unique, it cannot converge to any other real number and so, since the rationals are a subset of the reals, it cannot converge to any rational number. Now, that is sufficient to say that your sequence, as a seuquence in the rational numbers, does not converge.
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by poorna View Post
    Regard Q, the set of all rational numbers, as a metric space, with
    d(p,q)=|p-q|. Let E be the set of all P in Q such that 2<P^2<3. Show
    that E is closed and bounded in Q, but that E is not compact.
    Is E open in Q?

    I could prove that E is closed, open and bounded. But will the following solution suffice to show that E is not compact?

    I can find a sequence of points {Kn} = 1.7, 1.73, 1.732,.... in E. This is a sequence converging to 3^(1/2).
    So {Kn} is an infinite subset of E which does not have a limit point in E.
    So E is not compact.
    As you have it we have that \left(\mathbb{Q},d\right) is a subspace of \mathbb{R} with the usual topology. Now, it is true that if X is a topological (specifically metric space) and Y a subspace then Z is a compact subspace of Y iff it is a compact subspace of X. So, now as you noted E would have to be compact in \mathbb{R} to be compact in \mathbb{Q} but it is definitely not for the simple reason that x_n=\frac{1}{2}\left(x_n+\frac{2}{x_n}\right),\tex  t{ }x_0 is a sequence of points in E such that x_n\to \sqrt{2} and so \sqrt{2} is a limit point of E but not a point of E...in other words E isn't even a closed subspace of \mathbb{R}! What does the Heine-Borel theorem say?
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  6. #6
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    @ Drexel28
    Yeah, E has to be a closed and bounded subset of R in order to be compact in R. And ince compactness i independent of the set in which is contained, we have that E is not compact.
    Thanks!
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