Hausdorff metric inherits completeness
Theorem 1.
If $\mathrm{(}X\mathrm{,}d\mathrm{)}$ is a complete metric space, then the Hausdorff metric induced by $d$ is also complete^{}.
Proof.
Suppose $({A}_{n})$ is a Cauchy sequence^{} with respect to the Hausdorff metric. By selecting a subsequence if necessary, we may assume that ${A}_{n}$ and ${A}_{n+1}$ are within ${2}^{-n}$ of each other, that is, that ${A}_{n}\subset K({A}_{n+1},{2}^{-n})$ and ${A}_{n+1}\subset K({A}_{n},{2}^{-n})$. Now for any natural number^{} $N$, there is a sequence ${({x}_{n})}_{n\ge N}$ in $X$ such that ${x}_{n}\in {A}_{n}$ and $$. Any such sequence is Cauchy with respect to $d$ and thus converges^{} to some $x\in X$. By applying the triangle inequality^{}, we see that for any $n\ge N$, $$.
Define $A$ to be the set of all $x$ such that $x$ is the limit of a sequence ${({x}_{n})}_{n\ge 0}$ with ${x}_{n}\in {A}_{n}$ and $$. Then $A$ is nonempty. Furthermore, for any $n$, if $x\in A$, then there is some ${x}_{n}\in {A}_{n}$ such that $$, and so $A\subset K({A}_{n},{2}^{-n+1})$. Consequently, the set $\overline{A}$ is nonempty, closed and bounded^{}.
Suppose $\u03f5>0$. Thus $\u03f5>{2}^{-N}>0$ for some $N$. Let $n\ge N+1$. Then by applying the claim in the first paragraph, we have that for any ${x}_{n}\in {A}_{n}$, there is some $x\in X$ with $$. Hence ${A}_{n}\subset K(\overline{A},{2}^{-n+1})$. Hence the sequence $({A}_{n})$ converges to $A$ 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 $A$ 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 |