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THEOREM
Let us define $$F(X) := \{ A \subseteq X | A \neq \emptyset, A - \text{closed and bounded} \} $$ Then if $(X, \rho ) $ is complete, then $(F(X), d_h )$ is complete.

Are there any interesting proofs of this theorem other than the ones with long computations with $\epsilon$? I have been searching the internet for a while, but could not find anything.

Alex Ravsky
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Kuba
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  • Do you have a reference for the long computations with $\epsilon$? – Thibaut Dumont Jun 02 '15 at 21:09
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    You might look at Henrikson, Jeff (1999). "Completeness and total boundedness of the Hausdorff metric" . MIT Undergraduate Journal of Mathematics: 69–80. https://web.archive.org/web/20020623095720/http://www-math.mit.edu/phase2/UJM/vol1/HAUSF.PDF (that does have $\epsilon$'s) – Robert Israel Jun 03 '15 at 00:32
  • I'm wondering if there are some more general proofs, maybe something with a metric $$ d (A,B) = \underset{x \in X}{\sup} { |dist(x,A)-dist(x,B)|e^{-d(p,x)} } $$ on closed nonempty sets (assuming that X is complete, $p \in X$). This metric is equivalent to Hausdorff on bounded X. – Kuba Jun 03 '15 at 07:37
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    @Thibaut Dumont https://web.archive.org/web/20020623095720/http://www-math.mit.edu/phase2/UJM/vol1/HAUSF.PDF Theorem 4 – Kuba Jun 03 '15 at 07:41
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    I knew this theorem but with "closed and bounded" replaced by "compact". Are you sure that the version that you use does hold? For instance, does it hold in infinite-dimensional normed spaces where closed balls are bounded but not compact? – Alex M. Jun 08 '15 at 14:48
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    @Alex M: if you consider the metric space of closed non-void subsets of a compact metric space you get again a $\it{compact}$ metric space. Moreover, the topology depends only on the topology of $X$. – orangeskid Jun 12 '15 at 08:13
  • I think that only assuming "closed and bounded" is not enough, as this example shows. https://math.stackexchange.com/questions/3106567/in-what-metric-spaces-is-the-intersection-of-nested-closed-sets-non-empty You really need compactness. – No-one Oct 06 '23 at 20:35

1 Answers1

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HINT:

Step 1. Reduction to a simpler case. Assume that we have $A_n$ bounded and closed such that $d(A_m, A_n) \to 0$. Consider $A'_n \colon = \overline{\cup_{m \ge n} A_m}$. Check that $d(A_n, A'_n) \to 0$ ( simple). Now we have $(A'_n)$ decreasing and Cauchy.

Step2: Proof in the simpler case : if $A_n$ is a decreasing sequence of nonvoid, closed and bounded subsets of $X$ such that $d(A_n, A_m) \underset{m, n \to \infty}{\longrightarrow} 0$, then $A\colon =\cap A_n$ is nonvoid and $A_n \to A$.

A bit of $\epsilon$ work : take $\epsilon > 0$ and consider $(n_k)$ a subsequence of the naturals so that $d(A_m, A_n) \le \epsilon/2^k$ for $m,n \ge n_k$. Every point in $A_{n_1}$ is at distance at most $\epsilon$ from a point in $A$. Indeed, take $x_1$ in $A_{n_1}$ arbitrary, then $x_2 \in A_{n_2}$ , $d(x_2, x_1) \le \epsilon/2$, then $x_3 \in A_{n_3}$, $d(x_3, x_2) \le \epsilon/2^2$, and so on. The sequence $x_n$ converges to a point in $x$ in $A$ and we have $d(x, x_1) \le \epsilon/2 + \epsilon/2^2 + \cdots = \epsilon$.

Basically we show that $A_n$ converges to $\lim \sup A_n$ ( limsup in the lattice of closed subsets).

orangeskid
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  • I fail to see $d(A_n, A'_n) \to 0$. Indeed, $d(A_n, A'_n)=\max{\underset{a\in A_n} \sup d(a, A'_n), \underset{a' \in A'_n} \sup d(a', A_n)}$. $(A_n)$ is Cauchy, hence for any $\epsilon>0$ there is a $N \in \mathbb{N}$ such that $d(A_m, A_n)=\max{\underset{a \in A_n} \sup d(a, A_m), \underset{a' \in A_m} \sup d(a', A_n)}<\epsilon$ for all $m, n \geq N$. I managed to show that $\underset{a \in A_n} \sup d(a, A'_n) \leq \underset{a \in A_n} \sup d(a, A_m)$ when $m, n \geq N$; but could not progress further. Please help-many thanks in advance! – Dick Grayson Feb 26 '22 at 23:32
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    @Dick Grayson: Check this inequality and things get clearer $d(A, \cup B_i) \le \max d (A, B_i)$ – orangeskid Feb 27 '22 at 05:14
  • I think my main struggle is with $A'n=\overline{\cup{m \ge n} A_m}\neq \cup_{m \ge n} A_m$ at this point-I'll keep thinking about it. Thank you very much! – Dick Grayson Feb 27 '22 at 16:02
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    @DickGrayson: the distance to the closure is still the same, due to continuity of distance. – orangeskid Feb 27 '22 at 22:15