Hausdorff metric inherits completeness
Theorem 1.
If is a complete metric space, then the Hausdorff metric induced by is also complete.
Proof.
Suppose is a Cauchy sequence with respect to the Hausdorff metric. By selecting a subsequence if necessary, we may assume that and are within of each other, that is, that and .Now for any natural number
, there is a sequence in such that and . Any such sequence is Cauchywith respect to and thus converges
to some . By applying the triangle inequality
, we see that for any , .
Define to be the set of all such that is the limit of a sequence with and .Then is nonempty.Furthermore, for any , if ,then there is some such that , and so. Consequently, the set isnonempty, closed and bounded.
Suppose . Thus for some . Let .Then by applying the claim in the first paragraph, we have that for any, there is some with . Hence.Hence the sequence converges to in the Hausdorff metric.∎
This proof is based on a sketch given in an exercise in [1].An exercise for the reader: is the set constructed above closed?
References
- 1 J. Munkres, Topology
(2nd edition), Prentice Hall, 1999.