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

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.