Past exam of the mathematics course of the University of Cambridge / 2016 / ib / Paper 1 / 11G (Analysis II)

A Cauchy sequence in a metric space satisfies: for every $\epsilon >0$ there is $N$ such that $d(x_m,x_n)<\epsilon$ whenever $m,n\ge N$. Suppose a subsequence $x_{n_k}$ converges to $x\in X$. Choose $N$ so that the Cauchy bound is $\epsilon /2$, and then choose $k$ with $n_k\ge N$ and $d(x_{n_k},x)<\epsilon /2$. For every $n\ge N$, the triangle inequality givesThus the whole Cauchy sequence converges to the subsequential limit.

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 $y_0$.

We argue by contradiction using the previous subsequence construction. If $X$ is not complete, choose a Cauchy sequence with no limit in $X$. None of its $d_m$ can be zero: if $d_m=0$, then $d(x_m,x_n)\to 0$, giving a limit in $X$. The construction therefore produces pairwise distinct $y_j$ withLet $Y=\{y_1,y_2,\dots \}$. This subset is closed in $X$. Indeed, if a point $x\notin Y$ lay in its closure, there would be points of $Y$ tending to $x$. Their indices must be unbounded, because the distance from $x$ to any finite subset of $Y$ is positive. Passing to increasing indices gives a convergent subsequence of the original Cauchy sequence. Part (a) would then make the whole sequence converge to $x$, a contradiction.

Now define $F:Y\to Y$ by $F(y_j)=y_{j+1}$. Distinctness makes this well-defined, and the displayed inequality makes it a contraction mapping with constant at most $1/2$. It has no fixed point, contradicting the assumed property of closed subsets. ConsequentlyThis is the converse direction of contraction property of closed subsets characterizes completeness.