Question: I need to show that $A\subset \mathbb{R}$ is totally bounded iff A is bounded. This is what I've tried thus far:
Proof $(\Leftarrow)$: $A$ is bounded so there exists an $x\in\mathbb{R}$ and a $r>0$ such that for every $a \in A$, we have that $d(x,a) <r$, and thus that $A\subset B_r(x)$. Since $A\subset B_r(x)$, we have that $A$ is covered by finitely many $r$-balls and thus that A is totally bounded.
Proof $(\Rightarrow)$ $A$ is totally bounded, so $A$ is covered by finitely many $\epsilon$-balls. We know that for any point $x$ in $\bigcup_{i = 1}^{n} B_e(x_i)$ and for $r = diameter(z,y)$ with $z,y\in B_\epsilon(x_i)$, we have that $\bigcup_{i = 1}^{n} B_e(x_i) \subset B_r(x)$. So $A$ is bounded.
Am I going in the right direction?
Best Answer
Suppose that $A$ is totally bounded. Then there are finitely many real numbers $x_1,\ldots,x_n$ such that $A\subset\bigcup_{k=1}^nB_1(x_k)$. But this expresses $A$ as a finite union of bounded sets. Therefore, $A$ is bounded.
Suppose now that $A$ is bounded. Fix $\varepsilon>0$ and consider that intervals $\left(n\varepsilon-\frac\varepsilon2,n\varepsilon+\frac\varepsilon2\right)$ ($n\in\mathbb Z$). They are all open balls with radius $\varepsilon$. Since $A$ is bounded, you can find $m,n\in\mathbb Z$ such that $m<n$ and that $A\subset\left(m\varepsilon-\frac\varepsilon2,n\varepsilon+\frac\varepsilon2\right)$. But then $A\subset\bigcup_{k=m}^n(k\varepsilon-\frac\varepsilon2,k\varepsilon+\frac\varepsilon2)$