Proof. To prove this, we need to show that for any \(x,y\in E\), \(d(x,y)\) is finite. Choose \(\varepsilon =1\), then there is a finite number of non-empty open sets \(E_1,E_2,\dots ,E_n\) of \(X\) such that each \(E_i\) has diameter less than \(1\) and that \[E\subset \bigcup ^n_{i=1}E_i.\] Choose a point \(a_k\) arbitrary in \(E_i\), \(i=1,2,\dots ,n\).
Let \(M=d(a_1,a_2)+d(a_2,a_3)+\cdots +d(a_{n-1},a_n)\) since \(E_i's\) cover \(E\), hence \(x,y\in E\) implies \(x\in E_i\) and \(y\in E_j\) for some \(i,j\). Without loss of generality, we assume \(i<j\). Now, \[d(x,y)\leq d(x,a_i)+d(a_i,a_{i+1})+\cdots +d(a_{j-1},a_j)+d(a_j,y)\] Now, \(\diam (E_j)<1\), \(d(x,a_i)<1\), \(d(a_j,y)<1\). Therefore, \[d(x,y)\leq 1+M+1=M+2\hspace {0.5cm} \forall x,y\in E\] This proves that \(E\) is bounded.
□

Proof. Assume \(E\) is totally bounded, thus \(\displaystyle { E\subset \bigcup ^n_{i=1}E_i}\) and \(\diam (E_i)<\varepsilon \) for \(i=1,2,\dots ,n\) for a given \(\varepsilon >0\). If \(x\in E\), then \(x\in E_i\) for some \(i\).
For some \(a_i\in E_i,\hspace {0.5cm} i=1,2,\dots ,n\). Then, \(B=\{a_1,a_2,\dots ,a_n\}\) is \(\varepsilon \)-dense in \[\bigcup ^n_{i=1}E_i\] Thus, \(d(x,a_i)<\varepsilon \). So, if \(E\) is totally, then it is a finite \(\varepsilon \)-dense subset.
Conversely, Let \(B=\{x_1,x_2,\dots ,x_n\}\) be \(\varepsilon \)/3 dense in \(E\). Then let \(x\in E\), and we have that there exist \(x_i\in B\ni d(x,x_i)<\dfrac {\varepsilon }{3}\). This implies \(x\in B_{\frac {\varepsilon }{3}}(x_i)\). Thus \(\hspace {0.2cm} B_{\frac {\varepsilon }{3}}(x_1), B_{\frac {\varepsilon }{3}}(x_2),\dots ,B_{\frac {\varepsilon }{3}}\hspace {0.2cm}\) forms a covering of \(E\).

If \(y\in E\), then \(y\in B_{\frac {\varepsilon }{3}}(x_j)\) for some \(x_j\in B\). Now, \begin {align*} d(x,y) &\leq d(x,x_i)+d(x_i,x_j)+d(x_j,y)\\ &\leq \frac {\varepsilon }{3}+\frac {\varepsilon }{3}+\frac {\varepsilon }{3}\\ &=\varepsilon \end {align*}

Hence \(\displaystyle { E\subset \bigcup ^n_{i=1}B_{\frac {\varepsilon }{3}}(x_i)}\hspace {0.2cm}\) and each have diameter less than \(\varepsilon \). Thus \(E\) is totally bounded. □

Solution. Here \(e_i\) is the sequence with \(1\) in the \(i\)th place and \(0\) elsewhere.

The distances

For \(i\neq j\) the sequences \(e_i\) and \(e_j\) differ in exactly two places, and in each the difference has modulus \(1\), so \[d(e_i,e_j)=\sup _k|(e_i)_k-(e_j)_k|=1 .\]

Bounded

Every pair of points of \(E\) is at distance \(1\), so \(\diam (E)=1<\infty \) and \(E\) is bounded.

Not totally bounded

Take \(\varepsilon =\frac 12\). Any set of diameter less than \(\frac 12\) can contain at most one \(e_i\), since two of them are a distance \(1\) apart. So covering the infinite set \(E\) would need infinitely many such sets, and no finite cover by sets of diameter less than \(\frac 12\) exists.

The contrast with \(l_2\), where the same set has \(d(e_i,e_j)=\sqrt 2\), is only in the constant: in both spaces the unit sequences stay a fixed distance apart no matter how far out one goes, which is exactly what total boundedness forbids. In \(\mathbb {R}^n\) this cannot happen — there, bounded and totally bounded coincide.

Proof. We prove both directions; the statement is an equivalence, and each half is used later.

Totally bounded \(\implies \) every sequence has a Cauchy subsequence

Let \((x_n)\) be a sequence in \(A\). For \(\varepsilon =1\), cover \(A\) by finitely many sets of diameter less than \(1\); one of them must contain \(x_n\) for infinitely many \(n\), giving a subsequence \((x_n^{(1)})\) all of whose terms lie within \(1\) of each other. Repeat inside that subsequence with \(\varepsilon =\frac 12\), then \(\frac 13\), and so on, obtaining nested subsequences, the \(k\)-th having all terms within \(\frac 1k\).

Take the diagonal sequence whose \(k\)-th term is the \(k\)-th term of the \(k\)-th subsequence. Beyond position \(k\) it is a subsequence of the \(k\)-th one, so its terms are eventually within \(\frac 1k\) of each other for every \(k\); that is precisely the Cauchy condition.

Every sequence has a Cauchy subsequence \(\implies \) totally bounded

Suppose \(A\) is not totally bounded. Then for some \(\varepsilon >0\) no finite collection of \(\varepsilon \)-balls covers \(A\). Pick \(x_1\in A\); since \(B(x_1;\varepsilon )\) does not cover \(A\), pick \(x_2\in A\) outside it; since \(B(x_1;\varepsilon )\cup B(x_2;\varepsilon )\) does not cover \(A\), pick \(x_3\) outside both, and so on. The resulting sequence has \(d(x_m,x_n)\geq \varepsilon \) for all \(m\neq n\), so no subsequence of it is Cauchy — contradicting the hypothesis. □

Proof. Let \(A\) be a bounded subset of \(\mathbb {R}\). Let \((x_n)\) be a sequence in \(A\). Since \(A\) is bounded, \((x_n)\) is bounded. i.e \(|x_n|<M\).

Since \((x_n)\) is bounded, it contains a convergent sub-sequence \((x_{n_k})\). We know that every convergent sequence is a Cauchy sequence. Therefore \((x_{n_k})\) is a Cauchy sequence. So, every sequence \((x_n)\) of \(A\) has a Cauchy sub-sequence. Thus by the previous theorem, \(A\) is totally bounded. □