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I'm trying to show that any subset bounded of $\Bbb{R}^k$ is totally bounded.

Here is what I did:

(1)A subset of a totally bounded Set is bounded:

Proof: Let $X$ be a totally bounded subset and $Y\subset X$ then there exists an $\epsilon /2$-net $\{x_1,x_2,..,x_n\}$ and $X\subset \displaystyle\bigcup_{i=1}^n B(x_i,\epsilon/2)$ . Let $\{x_1,x_2,..,x_m\}$ be the points whose balls contain $Y$ $(m\le n)$. Now $\forall i \in \{1,..,m\} \exists q_i \in A \cap B(x_i,\epsilon/2) $ and $B(x_i,\epsilon/2)\subset B(q_i,\epsilon)$. We have for every $x\in B(x_i,\epsilon/2)$ $$ d(x,q_i)\le d(x,x_i)+d(x_i,q_i) < \frac \epsilon 2 + \frac \epsilon 2 =\epsilon$$

Hence $Y \subset \displaystyle\bigcup_{i=1}^m B(x_i,\epsilon/2) \subset \bigcup_{i=1}^m B(q_i,\epsilon)$ and $q_i \in Y$ for all $i$ hence $Y$ is totally bounded

Back to the problem:

Let $A \subset \Bbb{R}^k$ be a bounded set then $A \subset B(0,R)$ for some $R$ then $A \subset [-R,R] \times [-R,R] \times ...\times [-R,R] $ then $A$ is a subset of a compact set by the Heine-Borel Theorem which is also a totally bounded set, hence by (1) $A$ is totally bounded.

I'm trying to do this problem by only using (1) without invoking the Heine-Borel theorem, can anyone tell me how that can be done? (Is the proof of (1) right for that matter?)

user10444
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    Your proof of (1) is incorrect. There's no reason that the $x_i$ are in $Y$. You have to work a little bit harder. – JSchlather Feb 23 '13 at 13:08
  • @JSchlather Thanks for pointing that out I overlooked it. Is it correct now? I made sure the points are in $Y$ – user10444 Feb 23 '13 at 13:36
  • @user10444 Note that what I wrote is essentially Emanuele's take on the problem. – Pedro Mar 02 '13 at 16:22

2 Answers2

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If $B$ is bounded in $\mathbb R^n$ it is contained in some cube $[-R,R]^n$. Then for any $\epsilon>0$ just consider all balls of radius $\epsilon$ centered in the set $$ (\frac \epsilon 2\mathbb Z)^n \cap [-R,R]^n $$ these balls are a finite number (less then $(4R/\epsilon+1)^n$) and cover the whole cube $[-R,R]^n(edit this)$.

Community
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Emanuele Paolini
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  • Could you please clarify your notations? What is $(\frac \epsilon 2\mathbb Z)^n$ – user10444 Feb 23 '13 at 13:38
  • @user10444 $\newcommand{\eps}{\varepsilon}$ In general for a set of real numbers $A$ and a real number $c$ we can define $cA={cx \mid x \in A}$ in your case $\eps \mathbb Z={\eps z \mid z \in \mathbb Z}$. – JSchlather Feb 23 '13 at 13:47
  • I'm sorry but I can't seem to understand this at all, what is the center of each ball? Why is in contained in B(the center that is)? – user10444 Feb 23 '13 at 13:52
  • @user10444 You don't need to be as careful now that you've proven your lemma, you just need to show each bounded set is contained in a totally bounded set. – JSchlather Feb 23 '13 at 13:59
  • @Emanuele Paolini What is the radius of these balls? – user10444 Feb 23 '13 at 15:28
  • @user10444: the proof is simple: take a grid of squares with side $\epsilon/2$ and notice that the balls of radius $\epsilon$ centered in the grid, cover the whole space (radius $\sqrt 2 \epsilon/2$ would suffice) – Emanuele Paolini Feb 23 '13 at 15:36
  • @Emanuele Paolini I really thank you for your time, its just that I cant understand the proof. Even with the illustration :/ (What do you mean by centered in the grid?) – user10444 Feb 23 '13 at 15:43
  • @Emanuele Paolini How can the balls be less than $2(R/\epsilon)^n$ taking $\epsilon = 1$ and $n=1$ and $R=2$ $\frac 1 2\mathbb Z \cap [-2,2] = {-2, \frac {-3} 2, -1 ,\frac {-1} {2}, 0, \frac 1 2 , 1, \frac 3 2 , 2 }$ we get 9 centers while $2(R/\epsilon)^n$ in this case is 4 – user10444 Feb 23 '13 at 16:03
  • You are right! I have corrected the number. It's a little boring to find the correct number, but you can easily see that it is a finite number... – Emanuele Paolini Feb 23 '13 at 16:12
  • @user10444 I added an answer, see if it helps. – Pedro Feb 23 '13 at 23:02
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I'll be systematic here, I think it can help.

D Let $S$ be any subset of $\Bbb R^n$. Given $\epsilon >0$, we say that $N$ is an $\epsilon$-net for $S$ if the set of open balls

$$B_\epsilon(N)=\{B(x,\epsilon):x\in N\}$$

covers $S$. That is, the set of open balls of radius $\epsilon$ centered at the points of $N$ cover $S$.

D We say a subset $S$ of $\Bbb R^n$ is precompact or totally bounded if for every $\epsilon >0$; there exists a finite $\epsilon$-net for $S$.

T Let $S$ be bounded in $\Bbb R^n$. Then $S$ is precompact.

P Boundedness implies $S$ is contained in some closed ball $B$. But each of these balls contain but a finite number of elements of the form

$${\bf x}_{\ell,{\bf k}}= \left(\frac{k_1}{2^\ell},\dots,\frac{k_n}{2^\ell}\right)$$ for $k_i,\ell \in \Bbb Z\;\;;\ell \geq 0$, a fixed number, while the $k_i$ varies independently through the integers. But then, given $\epsilon >0$, we can take $\ell $ sufficiently large so that $\frac 1 {2^\ell}<\epsilon$, and the set of such points ${\bf x}_{\ell,k}$ contained in $B$ will be a finite $\epsilon$-net for $S$.

NOTE Observe the proof simply relies in producing what we usually think a net is: we show the intersection of our ball with the grid of "mesh" $1/2^\ell$ is finite, and then show that this intersection is an $\epsilon$-net (since we make $2^{-\ell}$ small) of the underlying set $S$ inside $B$. Note that we use the $k$ in the denominator the eventually "get out" of the open ball (since it is bounded, some natural will make $k/2^\ell$ "leave" the ball, no matter how small $2^{-\ell}$ is.)

Pedro
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