Solution. Let \(E=\{x_1,x_2,\dots ,x_n\}\) be finite and non-empty. The distances \(d(x_i,x_j)\) form a finite collection of real numbers, so it has a largest member; call it \(M\). Then \(d(x,y)\leq M\) for all \(x,y\in E\), which is exactly the definition of bounded, and \(\diam (E)=M<\infty \).

Finiteness is doing all the work: an infinite collection of distances need not have a largest member, which is why an infinite set may be unbounded. The empty set is bounded vacuously.

Proof. Let \((x_n)\) be Cauchy in \((X,d)\), this means \(d(x_n,x_m)<\varepsilon \) for all \(n,m\geq n_0\). Now, take \(\varepsilon =1\). Take \(\hspace {0.2cm}\displaystyle { M=\max _{1\leq n\leq n_0}\{d(x_n,x_{n_0})\}<\infty }.\hspace {0.2cm}\) Now \(\hspace {0.2cm} \displaystyle {d(x_n,x_m)\leq d(x_n,x_{n_0})+d(x_{n_0},x_m)}.\hspace {0.2cm}\) Now, \[d(x_n,x_{n_0})<M+1\hspace {0.5cm} \forall n\] \[d(x_{n_0},x_m)<M+1\hspace {0.5cm} \forall n\] Therefore, \(\hspace {0.2cm} \displaystyle {d(x_n,x_m)\leq d(x_n,x_{n_0})+d(x_{n_0},x_m)<2(M+1)\hspace {0.3cm} \forall m,n\geq n_0}.\hspace {0.2cm}\) Let \(K=2(M+1)\), we have that \(\hspace {0.2cm} \displaystyle {d(x_n,x_m)\leq K,\hspace {0.5cm} \forall n,m}.\hspace {0.2cm}\) Thus every Cauchy sequence is bounded in \((X,d)\).

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