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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- January 13th 2010, 09:53 PMpoornaclosed 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 PMDrexel28
- January 14th 2010, 12:37 AMDrexel28
For example. Define with . It is easy to prove that this is monotonically decreasing and bounded below by , and so we know it converges in , 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, .

- January 14th 2010, 03:05 AMputnam120
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 then you can find such that . (note that if then x isn't rational)