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!

  • $\begingroup$ 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). $\endgroup$
    – Rainb
    May 16 at 11:02

3 Answers 3


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.


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 $U$.

  • $\begingroup$ What is $\ell^\infty ( \mathbb{R})$ in this case? All bounded sequences in $\mathbb{R}$? $\endgroup$ Sep 17, 2017 at 0:06
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    $\begingroup$ 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 $U$. $\endgroup$
    – AlvinL
    Mar 14, 2018 at 14:17
  • $\begingroup$ Are you taking the Sup norm on $l^\infty$? If yes, how do we tell $d(x,y)=1$? $\endgroup$
    – Saikat
    Dec 22, 2020 at 5:35

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


$d$ 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 $d$.

  • $\begingroup$ .@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 $F\subseteq \mathbb{N}$ $\endgroup$
    – jasmine
    Sep 1, 2020 at 5:01
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    $\begingroup$ @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 $r>0$.) $\endgroup$ Sep 1, 2020 at 5:04

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