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Math Help - closed bounded and compact

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    closed bounded and compact

    consider the set of rationals to be a metric space with metric, d(x,y)=|x-y|. Then let E be the set of all rationals q with 2<q^2<3. Prove that E is closed and bounded. But not compact. I can see that it is bounded. How do i prove the other two?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by poorna View Post
    consider the set of rationals to be a metric space with metric, d(x,y)=|x-y|. Then let E be the set of all rationals q with 2<q^2<3. Prove that E is closed and bounded. But not compact. I can see that it is bounded. How do i prove the other two?
    If it were compact then it would be complete. Clearly there exists a rational sequence \left\{q_n\right\} that is Cauchy in this interval and q_n\to \sqrt{2}
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    MHF Contributor Drexel28's Avatar
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    For example. Define \gamma_{n+1}=\frac{1}{2}\left(\gamma_n+\frac{2}{\g  amma_n}\right) with \gamma_1=\frac{3}{2}. It is easy to prove that this is monotonically decreasing and bounded below by 1, and so we know it converges in \mathbb{R}, so it must be Cauchy there. And if it's Cauchy there it must be Cauchy here (since a Cauchy sequence is independent of the space it's embedded in). But, \lim\text{ }\gamma_n=\sqrt{2}.
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    Well not compact is the easy one to prove so I won't walk about it.

    As for closed it is easier to show that the complement is open. WLOG x>0 and x^2\ge 3 then you can find \epsilon such that x^2\pm\epsilon\ge 3. (note that if x^2=3 then x isn't rational)
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