# Bounded implies totally bounded

I don't understand a proof my textbook provides. Let me write it out first:

Theorem: Let $X \subset \mathbb{R}^n$ and $X$ bounded, then $X$ is totally bounded

Proof: There exists $r >0$ such that $X \subseteq [-r,r]^n$.

Let $\epsilon > 0$ be given. Then we subdivide the interval $[-r,r]$ into $m$ sub-intervals of length $\frac{2r}{m}$ such that $\frac{2nr}{\epsilon} < m$. I.e., we subdivide the interval into the sub-intervals $$\left[-r,-r+\frac{2r}{m}\right], \dots ,\left[r-\frac{2r}{m},r\right]$$

Let's denote those with $I_1, \dots I_m$ respectively. Hence, we obtain a subdivision of the cube $[-r,r]^n$ into sub-cubes of the form $I_{k_1} \times \dots \times I_{k_n}$ with $k_i \in \{1,2, \dots m\}$. Every one of these smaller n-dimensional cubes is contained in a ball with center, the center of the cube and radius $\epsilon$. Hence, $X$ is contained in $m^n$ balls with radius $\epsilon$ and this ends the proof.

Questions: Since the norms $\Vert.\Vert_M, \Vert.\Vert_S, \Vert.\Vert_E$ are equivalent, it doesn't matter which norm we use to do this proof. Do we use the $M$-norm here? ($\Vert x \Vert_M := \max_{i=1}^n|x_i|)$ Are the balls they are speaking off cubes then?

How does it follow that "Every one of these smaller n-dimensional cubes is contained in a ball with center, the center of the cube and radius $\epsilon$"

According to me, this is because such a sub-cube is an open ball for the $M$ norm with radius $r/m$, hence, if we call the center $x$, we have:

$$\text{subcube } = B_M\left(x,\frac{r}{m}\right) \subset B_M\left(x,\frac{2nr}{m}\right) \subset B_M(x, \epsilon)$$

hence it is contained in an open ball (actually an open cube, this wording is confusing!)

Is this correct?

• Anyone care to explain the downvote? – user370967 Dec 28 '17 at 21:32
• I thought it was just, like, Valleyspeak for bounded. – copper.hat Dec 28 '17 at 22:44

Actually, the norm could just as well be the euclidean norm $\|\cdot\|_2$.

An $n$-dimensional cube $I_{k_1} \times \cdots \times I_{k_n}$ of side length $\frac{2r}{m}$ is contained in a closed ball of radius $\frac{2r}{m} \cdot\frac{\sqrt{n}}2 = \frac{r\sqrt{n}}{m}$ centered at the center of the cube.

This is because the length of the spatial diagonal of an $n-$cube of side length $a$ is $a\sqrt{n}$, analogously as the diagonal of the square is $a\sqrt{2}$, and of the $3-$cube is $a\sqrt{3}$.

We have $$\frac{r\sqrt{n}}{m} \le \frac{rn}{m} <\frac\varepsilon2 < \varepsilon$$

so in particular the cube $I_{k_1} \times \cdots \times I_{k_n}$ is contained in an open $\varepsilon$-ball around its center.

Your work with the $\|\cdot\|_\infty$ norm is also correct, of course. Total boundedness can also be defined like this:

$A \subseteq X$ is totally bounded if for every $\varepsilon > 0$ there exists sets $S_1, \ldots, S_n \subseteq X$ such that $A \subseteq \bigcup_{i=1}^n S_i$ and $\operatorname{diam} S_i < \varepsilon$.

So the proof could in fact be finished after you conclude that the cubes $I_{k_1} \times \cdots \times I_{k_n}$ have diameter $\frac{r\sqrt{n}}{m} < \varepsilon$.

• So you are sure my work with the maximumnorm is correct? – user370967 Dec 28 '17 at 21:51
• @Math_QED Sure, it's correct. Probably the author meant some other norm though, because the radius of "balls" in the max-norm does not depend on $n$, and you had $\frac{2nr}{\epsilon} < m$. But it doesn't matter. – mechanodroid Dec 28 '17 at 21:56
• I think the approach with the maximumnorm is easier as you don't have to prove what the length of the diagonal is. I agree that the author probably worked with the usual euclidean norm. Thanks a lot for the help! – user370967 Dec 28 '17 at 21:58