1. ## Totally bounded subset

I need to prove the following:

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

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

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

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.

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

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

3. Originally Posted by Plato
You cannot conclude that $\displaystyle \text{diam}(A)\le \epsilon.$ Because that is not true.

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

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

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

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

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

Right?

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