Hausdorff metric inherits completeness
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 Cauchy with 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 is nonempty, 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 . An exercise for the reader: is the set constructed above closed?
- 1 J. Munkres, Topology (2nd edition), Prentice Hall, 1999.
|Title||Hausdorff metric inherits completeness|
|Date of creation||2013-03-22 14:08:51|
|Last modified on||2013-03-22 14:08:51|
|Last modified by||mps (409)|