Hausdorff metric inherits completeness

Theorem 1.

If $(X,d)$ 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 $(x_{n})_{n\\geq N}$ in $X$ such that $x_{n}\\in A_{n}$ and $d(x_{n},x_{n+1})<2^{-n}$ . Any such sequence is Cauchywith 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 $\\overline{A}$ isnonempty, 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 [1].An exercise for the reader: is the set $A$ constructed above closed?

References

1 J. Munkres, Topology (2nd edition), Prentice Hall, 1999.

单词	HausdorffMetricInheritsCompleteness
释义	Hausdorff metric inherits completeness Theorem 1. If $(X,d)$ 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 $(x_{n})_{n\\geq N}$ in $X$ such that $x_{n}\\in A_{n}$ and $d(x_{n},x_{n+1})<2^{-n}$ . Any such sequence is Cauchywith 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 $\\overline{A}$ isnonempty, 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 [1].An exercise for the reader: is the set $A$ constructed above closed? References 1 J. Munkres, Topology (2nd edition), Prentice Hall, 1999.
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