# closed, bounded, compact...

• Jun 18th 2009, 02:39 AM
Aileys.
closed, bounded, compact...
I know closed & bounded doesn't imply compact for any old space. My question is where does this argument go wrong: ?

If it's bounded then any sequence has a convergent subsequence

And if it's closed then it contains all it's limit points

So any sequence has a convergent subsequence which converges to a point in the set

But that's the definition of sequentially compact... Uh oh.
• Jun 18th 2009, 03:14 AM
putnam120
Quote:

Originally Posted by Aileys.

If it's bounded then any sequence has a convergent subsequence

I think that is where you mistake is.
Consider $\displaystyle f_n(x)=e^{inx}$.

With $\displaystyle d(f_n,f_m)=\left\{\int_U|f_n-f_m|^2d\mu\right\}^{\frac{1}{2}}$, where $\displaystyle U$ is the interval $\displaystyle [0,2\pi]$. Then $\displaystyle d(f_n,f_m)=K$ if $\displaystyle n\neq m$, where $\displaystyle K$ is some nonzero constant (I'll leave u to figure out the exact value). So clearly you won't have a convergent subsequence since all the points are the same distance (with respect to this metric) apart.
• Jun 18th 2009, 06:33 AM
HallsofIvy
Quote:

Originally Posted by Aileys.
I know closed & bounded doesn't imply compact for any old space. My question is where does this argument go wrong: ?

If it's bounded then any sequence has a convergent subsequence

This statement is not necessarily true in an arbitrary Topological space. For example, in the set of real numbers with the "discrete" topology, that is d(x,y)= 0 if x= y, d(x,y)= 1 if $\displaystyle x\ne y$, the sequence of positive integers is bounded (the largest possible distance between points is 1) but has no convergent subsequence. In the discrete topology, every set is both closed and bounded but only the finite sets are compact.

Quote:

And if it's closed then it contains all it's limit points

So any sequence has a convergent subsequence which converges to a point in the set

But that's the definition of sequentially compact... Uh oh.
• Jun 18th 2009, 07:59 AM
Sampras
Quote:

Originally Posted by Aileys.
I know closed & bounded doesn't imply compact for any old space. My question is where does this argument go wrong: ?

If it's bounded then any sequence has a convergent subsequence

And if it's closed then it contains all it's limit points

So any sequence has a convergent subsequence which converges to a point in the set

But that's the definition of sequentially compact... Uh oh.

Consider the $\displaystyle \ell^{\infty}$ metric space. The unit ball, $\displaystyle \overline{B}(\underline{0}, 1)$ is closed and bounded, but not compact. We can produce an infinite set with no limit point. An example would be: $\displaystyle \underline{x}_{n} = (1,1, \dots, 1, 0, 0, \dots, 0, \dots)$.
• Jun 18th 2009, 03:54 PM
Aileys.
Ahh yes I see. Bolzano Weierstrass is only for R^n. Thanks guys.
• Jun 18th 2009, 10:23 PM
Chandru1
Compact
the exact definition of compact: its complete and totally bounded. A set X is said to be totally bounded if $\displaystyle X \subset \cup_{i=1}^{n} I_{n}$ where the $\displaystyle I_{n}$ are sets of diameter each $\displaystyle <\epsilon.$

In $\displaystyle \mathbb{R}$ totally bounded and bounded mean the same. AND Bolzano Wierestrass is not only for $\displaystyle \mathbb{R}^{n}$ but for $\displaystyle \mathbb{R}$ also. We have any bounded sequence having a convergent subsequence dont we?
• Jun 19th 2009, 01:36 AM
Aileys.
Oh... is that definition equivalent to "any open cover has a finite subcover" for any space? Or is it an even bigger generalisation?

Also, isn't R a subset of R^n?
• Jun 19th 2009, 03:34 AM
Chandru1
Heine Borel
Well, yes if any metric spaces satisfies the Heine Borel property then its compact.

The definition which i gave is used for proving the above result.

I apologize for the mistake of R and R^{n}. If u are in need of a proof using the above property gimme ur email. I shall send u directly.

A good reference would be Methods of Real analysis by R.R.Goldberg
• Jun 21st 2009, 07:41 AM
HallsofIvy
Quote:

Originally Posted by Chandru1
Well, yes if any metric spaces satisfies the Heine Borel property then its compact.

I'm not clear on what you are saying here. "Hein-Borel" says "if a set is both closed and bounded then it is compact", but what do you mean by "satisfies" it? Is both closed and bounded? Then it is certainly not true. The example I gave above, any infinite set with the discrete metric, is a metric space that is both closed and bounded but not compact. If you meant "any metric space in which the Heine-Borel property is true", then what does "its" refer to? Grammatically, it should refer to "any metric space" but again that is not true. The set of real numbers with the usual metric has the Heine-Borel property but the set of all real numbers is certainly not compact.

Quote:

The definition which i gave is used for proving the above result.

I apologize for the mistake of R and R^{n}. If u are in need of a proof using the above property gimme ur email. I shall send u directly.

A good reference would be Methods of Real analysis by R.R.Goldberg