Past exam of the mathematics course of the University of Cambridge / 2016 / ib / Paper 1 / 11G / b

For fixed $m$, the reverse triangle inequality givesThe right side tends uniformly to zero as $n,k\to \infty$ because $(x_n)$ is a Cauchy sequence. Hence $(d(x_m,x_n){)}_n$ is a real Cauchy sequence. By completeness of $\mathbb{R}$, it has a finite, nonnegative limit:This conclusion uses completeness only of the real line, without assuming that $X$ is complete.

Given $\epsilon >0$, the Cauchy sequence property gives $N$ with $d(x_m,x_n)<\epsilon /2$ for all $m,n\ge N$. Fixing $m\ge N$ and letting $n\to \infty$ yields $0\le d_m\le \epsilon /2<\epsilon$. Therefore

Write the subsequence as $y_n=x_{k_n}$ with strictly increasing indices $k_n$. If $y_{\ell }=x_m$, then $d(y_{\ell },y_n)=d(x_m,x_{k_n})$. This is a subsequence of the real sequence converging to $d_m$, soThe conclusion depends on the point $x_m$ itself, even if that point appears at several indices.

Apply the triangle inequality with a third point $x_k$:Letting $k\to \infty$ gives the desired bounds. Interchanging $m,n$ also gives the sharper symmetric lower bound:

First handle possible zero values. If $d_m=d_n=0$, the previous upper bound forces $x_m=x_n$. Thus, if infinitely many $d_m$ vanish, choosing those indices gives a constant subsequence, which satisfies both inequalities.

Otherwise, discard the finitely many zero values and use $d_m\to 0$ to choose increasing indices $k_j$ withSet $y_j=x_{k_j}$. The preceding bounds give $|{\delta }_j-{\delta }_{\ell }|\le d(y_j,y_{\ell })\le {\delta }_j+{\delta }_{\ell }$. For $m\ge 2$,For $m<n$, use ${\delta }_n\le {\delta }_m/8$ and ${\delta }_{n+1}\le {\delta }_m/64$ to obtainThe same holds when $n<m$ by symmetry; when $m=n$, both distances are zero. This proves both required inequalities. The adjacent-distance inequality is understood for $m\ge 2$, since this subsequence starts at index one and has no $y_0$.