# Hausdorff metric inherits completeness

###### 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 $(x_{n})_{n\geq N}$ in $X$ such that $x_{n}\in A_{n}$ and $d(x_{n},x_{n+1})<2^{-n}$. 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 $n\geq N$, $d(x_{n},x)<2^{-n+1}$.

Define $A$ to be the set of all $x$ such that $x$ is the limit of a sequence $(x_{n})_{n\geq 0}$ with $x_{n}\in A_{n}$ and $d(x_{n},x_{n+1})<2^{-n}$. Then $A$ is nonempty. Furthermore, for any $n$, if $x\in A$, then there is some $x_{n}\in A_{n}$ such that $d(x_{n},x)<2^{-n+1}$, and so $A\subset K(A_{n},2^{-n+1})$. Consequently, the set $\overline{A}$ is nonempty, closed and bounded   .

Suppose $\epsilon>0$. Thus $\epsilon>2^{-N}>0$ for some $N$. Let $n\geq N+1$. 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 $d(x_{n},x)<2^{-n+1}$. 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 . An exercise for the reader: is the set $A$ constructed above closed?

## References

Title Hausdorff metric inherits completeness HausdorffMetricInheritsCompleteness 2013-03-22 14:08:51 2013-03-22 14:08:51 mps (409) mps (409) 8 mps (409) Theorem  msc 54E35 Complete