= 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 $d(x_m,x_n)\to0$, giving a limit in $X$. The construction therefore produces pairwise distinct $y_j$ with
$$
d(y_{m+1},y_{n+1})\leq\tfrac12d(y_m,y_n).
$$
Let $Y=\{y_1,y_2,\ldots\}$. \b[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. Consequently
$$
\boxed{X\text{ is complete}.}
$$
This is the converse direction of <contraction property of closed subsets characterizes completeness>.
Back to article page
