# Set that is bounded but not totally bounded: Reading textbook

I've been reading a Real Analysis textbook that my friend loaned to me. I have come across a proposition that says that a totally bounded set is bounded, but a bounded set is not always totally bounded. This makes sense, however I am having trouble thinking of an example of a set that is bounded but not totally bounded. Could anyone shed some light on this? Thanks!

• For finite sets, the concepts of boundedness and total boundedness coincide, since any finite set in a metric space is always totally bounded. For each $\epsilon > 0$, you can always cover a finite set with a finite number of $\epsilon$-balls (in fact, with at most as many balls as there are points in the set). Commented May 16, 2023 at 11:02

I know this is an old post, but any infinite set $$M$$ with a discrete metric is bounded by any $$N>1$$ but it is not totally bounded for open balls with $$\epsilon\leq 1$$ because for it to be totally bounded it must have a finite number of points. But such a ball is only a singleton, and you would require infinitely many to cover $$M$$.

Define a new metric $$d$$ on $$\Bbb R$$ by $$d(x,y)=\min\{|x-y|,1\}$$; you can easily check that $$d$$ generates the usual topology on $$\Bbb R$$. Every subset of $$\Bbb R$$ is bounded with respect to $$d$$, so we need only find a subset that is not totally bounded. $$\Bbb N$$ will do: if $$F\subseteq\Bbb N$$, then

$$\Bbb N\cap\bigcup_{x\in F}B_d\left(x,\frac12\right)=F\;,$$

since $$\Bbb N\cap B_d\left(x,\frac12\right)=\{x\}$$ for each $$x\in F$$. Thus, no finite family of open $$\frac12$$-balls centred at points of $$F$$ can even cover $$\Bbb N$$.

This idea generalizes. Start with any complete metric space $$\langle X,\rho\rangle$$ that is not compact. A metric space is compact if and only if it’s complete and totally bounded, so $$\langle X,\rho\rangle$$ is not totally bounded, and there is some $$\epsilon>0$$ such that no finite family of open $$\epsilon$$-balls covers $$X$$. Without loss of generality we may assume that $$\epsilon<1$$. Now define a metric $$d$$ on $$X$$ by

$$d(x,y)=\min\{\rho(x,y),1\}\;;$$

$$d$$ generates the same topology as $$\rho$$, and every subset of $$X$$ is bounded with respect to $$d$$. Finally, since $$\epsilon<1$$, $$\epsilon$$-balls with respect to $$d$$ are the same as $$\epsilon$$-balls with respect to $$\rho$$, and $$X$$ is bounded but not totally bounded with respect to $$d$$.

• .@Brian M. scott sir How $\Bbb N\cap B_d\left(x,\frac12\right)=\{x\} ?$ im thinking that $\Bbb N\cap B_d\left(x,\frac12\right)= \mathbb{N} \cap ( x- 1/2 , x+ 1/2)= \emptyset$ since $x \in \mathbb{N}$ and $F\subseteq \mathbb{N}$ Commented Sep 1, 2020 at 5:01
• @jasmine: $x\in B_d\left(x,\frac12\right)$, and $x\in\Bbb N$, so $x\in\Bbb N\cap B_d\left(x,\frac12\right)$. (In fact, $x\in B_d(x,r)$ for any $x$ and any $r>0$.) Commented Sep 1, 2020 at 5:04

Take $U=\{e_n|\:n\in\mathbb N\}\subset \ell^\infty (\mathbb R)$.

It is obviously bounded since $\forall x\in U \:\|x\|=1$, but $\forall x,y\in \ell^\infty$ we have $d(x,y)=1$, so obviously for $\epsilon=1$ there is no finite number of open balls with radius $\epsilon$ that cover $U$ - cause each ball would contain at most one member of $U$.

• What is $\ell^\infty ( \mathbb{R})$ in this case? All bounded sequences in $\mathbb{R}$? Commented Sep 17, 2017 at 0:06
• I think what is meant, that by restricting the norm on $\ell ^\infty$ to $U$, we obtain a discrete metric hence picking $\varepsilon < 1$, we need infinitely many $\epsilon$-balls to cover $U$. Commented Mar 14, 2018 at 14:17
• Are you taking the Sup norm on $l^\infty$? If yes, how do we tell $d(x,y)=1$? Commented Dec 22, 2020 at 5:35