# Thread: Totally bounded subset

1. ## Totally bounded subset

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.

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

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

3. 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 $A\subseteq \bigcup\limits_{k = 1}^n {\mathcal{B}\left( {a_k ;\epsilon } \right)}$.

4. 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 $\epsilon\text{-net}?$

5. 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?

6. 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?

7. 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.