# Closed, Bounded but not Compact.

• Nov 6th 2010, 03:58 PM
matt.qmar
Closed, Bounded but not Compact.
The set of rationals Q forms a metric spce by $\displaystyle d(p,q) = | p - q |$

Then a subset E of Q is defined by $\displaystyle E = \{ p \in Q : 2 < p^2 < 3 \}$
So I am trying to show that E is closed and bounded, but not compact.

To me, it is clear than E is bounded (by 2 and 3?!).

I am having trouble showing E is closed. I believe 2 and 3 are limit points of E but E does not contain them, so E is not closed? also, E does not contain e = 2.718281828.... but that is also a limit point of E, is it not? Maybe I am confused with my definition of limit point.... are any of these (2, 3, e) actually limit points of E?

I think I can show E is not compact, since the open cover of E,
$\displaystyle \bigcup\ (2 + \frac{1}{n}, 3 - \frac{1}{n})$ for n = 2, 3, 4, ....
has no finite subcover?

I am worried I am perhaps misunderstanding some definitions... Any help or direction would be greatly appriciated.

Thank you in advanced for you time!
• Nov 6th 2010, 04:27 PM
Plato
Do you understand that $\displaystyle E = \left( {\sqrt 2 ,\sqrt 3 } \right) \cap \mathbb{Q}~?$

Consider this cover $\displaystyle O_n = \left( {\sqrt 2 + \frac{{\sqrt 3 - \sqrt 2 }}{{3n}},\sqrt 3 - \frac{{\sqrt 3 - \sqrt 2 }}{{3n}}} \right)$
• Nov 6th 2010, 04:35 PM
matt.qmar
Wow, thank you for pointing that out. I somehow goofed on what the set E was!

However, even in this new light, E does not appear to be closed... $\displaystyle \sqrt 2$ is not in E, but it is a limit point of E?

Thanks!
• Nov 6th 2010, 05:09 PM
Plato
You are still missing the point $\displaystyle \sqrt2\notin \mathbb{Q}$.
Does the set contain all of its limit points in $\displaystyle \mathbb{Q}~?$
• Nov 6th 2010, 05:44 PM
Drexel28
Quote:

Originally Posted by matt.qmar
The set of rationals Q forms a metric spce by $\displaystyle d(p,q) = | p - q |$

Then a subset E of Q is defined by $\displaystyle E = \{ p \in Q : 2 < p^2 < 3 \}$
So I am trying to show that E is closed and bounded, but not compact.

To me, it is clear than E is bounded (by 2 and 3?!).

I am having trouble showing E is closed. I believe 2 and 3 are limit points of E but E does not contain them, so E is not closed? also, E does not contain e = 2.718281828.... but that is also a limit point of E, is it not? Maybe I am confused with my definition of limit point.... are any of these (2, 3, e) actually limit points of E?

I think I can show E is not compact, since the open cover of E,
$\displaystyle \bigcup\ (2 + \frac{1}{n}, 3 - \frac{1}{n})$ for n = 2, 3, 4, ....
has no finite subcover?

I am worried I am perhaps misunderstanding some definitions... Any help or direction would be greatly appriciated.

Thank you in advanced for you time!

Another simplistic theorem tells you that if $\displaystyle X,Y,Z$ are metric spaces then $\displaystyle Z$ is a compact subspaces of $\displaystyle Y$ if and only if $\displaystyle Z$ is a compact subspaces of $\displaystyle Y$. Note though that with this in hand $\displaystyle E$ will be a compact subspaces of $\displaystyle \mathbb{Q}$ if and only if it's a compact subspace of $\displaystyle \mathbb{R}$. But, considering that it isn't closed ($\displaystyle \sqrt{2}$ is a limit point not in the set) it follows that $\displaystyle E$ is not compact.