closed bounded and compact

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January 13th 2010, 09:53 PM
poorna

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?
January 13th 2010, 10:09 PM
Drexel28

Quote:

Originally Posted by poorna

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 $q_n\to \sqrt{2}$
January 14th 2010, 12:37 AM
Drexel28

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, $\lim\text{ }\gamma_n=\sqrt{2}$ .
January 14th 2010, 03:05 AM
putnam120

Well not compact is the easy one to prove so I won't walk about it.(Lipssealed)

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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