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?

Printable View

- Jan 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?

- Jan 13th 2010, 10:09 PMDrexel28
- Jan 14th 2010, 12:37 AMDrexel28
For example. Define $\displaystyle \gamma_{n+1}=\frac{1}{2}\left(\gamma_n+\frac{2}{\g amma_n}\right)$ with $\displaystyle \gamma_1=\frac{3}{2}$. It is easy to prove that this is monotonically decreasing and bounded below by $\displaystyle 1$, and so we know it converges in $\displaystyle \mathbb{R}$, 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, $\displaystyle \lim\text{ }\gamma_n=\sqrt{2}$.

- Jan 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 $\displaystyle x^2\ge 3$ then you can find $\displaystyle \epsilon$ such that $\displaystyle x^2\pm\epsilon\ge 3$. (note that if $\displaystyle x^2=3$ then x isn't rational)