Completeness of a metric space with the Hausdorff metric

Asked Oct 24 '13 at 04:24

Active Oct 24 '13 at 04:36

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Let $(Y,d)$ be a metric space and let $K(Y)$ denote the set of all non-empty compact subsets of $Y$. This collection is a metric space when equipped with the Hausdorff distance $h$.

I want to prove that $(Y,d)$ being complete implies that $(K(Y),h)$ is complete.

Any help would be greatly appreciated!

edited Oct 24 '13 at 04:36

dfeuer

9,069

asked Oct 24 '13 at 04:24

RV702

According to answers to this question, in suffices to show that $K(Y)$ is closed in $F(Y)$. Since a subset of a complete metric space is compact iff it is closed and totally bounded, for this purpose it suffices to show that a limit $A$ (in $F(Y)$) of a sequence ${A_n}$ of non-empty compact subsets of $Y$ is totally bounded. – Alex Ravsky Oct 27 '17 at 06:33
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Possible duplicate of Prove that if $(E,d)$ is complete then, $(\mathcal{K}(E), \mathcal{H})$ is also complete – Oct 28 '17 at 15:50
https://math.stackexchange.com/questions/1309827/completeness-of-the-hausdorff-distance – Guy Fsone Jul 14 '22 at 15:16
@user357151 No. It's not a time traveler. The link you gives is dated by 2017, the question here was asked in 2013. – vesszabo Mar 25 '24 at 11:19

Completeness of a metric space with the Hausdorff metric

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