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 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 [1]. An exercise for the reader: is the set constructed above closed?
References
-
1
J. Munkres, Topology
(2nd edition), Prentice Hall, 1999.
Title | Hausdorff metric inherits completeness |
---|---|
Canonical name | HausdorffMetricInheritsCompleteness |
Date of creation | 2013-03-22 14:08:51 |
Last modified on | 2013-03-22 14:08:51 |
Owner | mps (409) |
Last modified by | mps (409) |
Numerical id | 8 |
Author | mps (409) |
Entry type | Theorem![]() |
Classification | msc 54E35 |
Related topic | Complete |