Totally bounded subset

**tarheelborn** · November 4th 2010, 08:36 AM

I need to prove the following:

Suppose $M$ is a metric space and $A \subseteq M$ . Then $A$ is totally bounded if and only if, for every $\epsilon >0$ , there is a finite $\epsilon$ -dense subset of $A$ .

$(\Rightarrow )$ : Let $\epsilon > 0$ . Then $\epsilon > 0$ . So there exist sets $A_1, A_2, \cdots, A_k$ , with $diam(A_i)<\epsilon$ and $A=\displaystyle \cup_{i=1}^{k} A_i$ . Now $\forall i$ let $a_i \in A_i$ . Let $X=\{ a_i \} _{i=1}^{k}$ . So $X$ is a finite subset of $A$ . I claim that $X$ is $\epsilon$ -dense in $A$ . To see this, let $a \in A$ . I must show that $\exists$ some element of $X$ so that $\rho (a,x)< \epsilon$ . If $a \in A$ , then $a \in A_i$ for some $i$ . Since $a_i \in A_i$ , $diam(A_i) < \epsilon$ , $\rho (a, a_i) < \epsilon$ .

$(\Leftarrow )$ : Now suppose that for $\epsilon > 0$ , $A$ has a finite $\epsilon$ -dense subset. I must prove that $A$ is totally bounded. Since $A$ has a finite $\epsilon$ -dense subset, $\exists \ a \in A$ so that $\rho (x,a)< \epsilon$ . So $diam(A)=$ l.u.b. $\{ \rho (a_1, a_2)|a_1, a_2 \in A \}< \epsilon$ so $diam(A)< \epsilon \ \forall a$ .

Now for the problem: I am not sure if I have included all the definition for totally bounded; that is, I think I still have another step, but am not sure. Thanks for your help.

**Plato** · November 4th 2010, 09:08 AM

You cannot conclude that $\text{diam}(A)\le \epsilon.$ Because that is not true.

Can you show that $A\subseteq \bigcup\limits_{k = 1}^n {\mathcal{B}\left( {a_k ;\epsilon } \right)}~?$

**tarheelborn** · November 4th 2010, 09:46 AM

Originally Posted by Plato

You cannot conclude that $\text{diam}(A)\le \epsilon.$ Because that is not true.

Can you show that $A\subseteq \bigcup\limits_{k = 1}^n {\mathcal{B}\left( {a_k ;\epsilon } \right)}~?$

Since $A$ is $\epsilon$ -dense in $M, \ \forall a \in A \ \exists \ y \in B$ so that $\rho (a,y)< \epsilon$ , which means $A\subseteq \bigcup\limits_{k = 1}^n {\mathcal{B}\left( {a_k ;\epsilon } \right)}$ .

**Plato** · November 4th 2010, 10:11 AM

The statement that $X$ is totally bounded means that there exist an $\epsilon \text{-net}$ for $X$ .

If you have a finite $\epsilon \text{-dense}$ set on $X$ then is that an $\epsilon\text{-net}?$

**tarheelborn** · November 4th 2010, 10:26 AM

We have not used the term $\epsilon$ -net in my class. I am in an advanced calculus class. I know that if $X$ is totally bounded, then $X$ can be covered by a finite number of subsets of M whose diameters are all less than $\epsilon$ . Is that what you mean by $\epsilon$ -net?

**tarheelborn** · November 4th 2010, 11:03 AM

Since there is an $\epsilon$ -dense set in $A$ , say $\{ x_1, x_2, \cdots, x_n \}$ is $\epsilon$ -dense in $A$ , then $B[x_i; \epsilon], \cdots, B[x_n; \epsilon]$ form a covering of $A$ by sets of diameter $< \epsilon$ . Hence $A$ is totally bounded. QED.

Right?

**Plato** · November 4th 2010, 12:11 PM

Well that is what I would say is correct.
I have never seen the term $\epsilon \text{-dense}$ .
I was guessing that it meant $\epsilon\text{-net}$ which is finite.

Thread: Totally bounded subset

LinkBack

Thread Tools

Display

Totally bounded subset

Similar Math Help Forum Discussions

supremum if bounded subset

Closure of a totaly bounded set is totally bounded

Totally bounded set

Bounded subset of R

Need to prove that bounded <=> totally bounded in Rn

Search Tags