Completeness of the Hausdorff hyperspace 2026-10-25
For any nonempty complete metric space , the family of its nonempty closed bounded subsets is a complete metric space under the Hausdorff distance. Boundedness of the entire ambient space is unnecessary in this general form.
For a Cauchy sequence of these sets, define and . A Cauchy sequence in any metric space is bounded, so its tail sets are contained in a common finite-radius neighbourhood of one bounded ; hence every is bounded. The distance to a set is one-Lipschitz, so estimates pass to the closure, giving and for . Choose with . The Summably nested Hausdorff intersection theorem gives . The triangle inequality now gives , proving completeness.
The intersection is a closed set because arbitrary intersections of closed sets are closed. Its nonemptiness needs a construction: bounded closed sets need not be compact sets in this setting.
Fix , , and . Recursively choose , for , using the approximation property of an infimum so thatThis is possible because and . The two bounds form convergent geometric series, so the triangle inequality shows is a Cauchy sequence. By completeness it converges to some .
For each fixed , all sufficiently late lie in , by nesting. Closedness of gives , and hence . In particular, starting with any proves . Moreover,For each starting point , let in this bound on its distance to a set . This does not assert that a nearest point exists. It proves . Since , the other directed Hausdorff distance is zero, andThis is the Summably nested Hausdorff intersection theorem; the summable movement replaces any need for compactness.