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We know that if $(X,d)$ is a complete metric space, then $(CB(X),H)$ is complete too, where $CB(X)$ is the collection of non-empty closed bounded subset of $X$ and $H$ is the Hausdorff metric induced by $d$. Let $T:X\to CB(X)$ be a mappings. Does all Cauchy sequence $\{x_n\}$ in $(X,d)$ implies that $\{Tx_n\}$ is a Cauchy sequence in $(CB(X),H)$? and the converse is it true, that is, does all Cauchy sequence $\{Tx_n\}$ in $(CB(X),H)$ implies that $\{x_n\}$ is a Cauchy sequence in $(X,d)$?

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    $\begingroup$ I can see how an element $x$ of $X$ defines one of $CB(X)$, namely $\{x\}$. So I can make sense of the first question and the answer is trivially yes: what is $H(\{x\},\{y\})$? I can't make sense of the second question. $\endgroup$ – Julien Nov 26 '13 at 15:52
  • $\begingroup$ Correction in the assumption has been made, thanks! $\endgroup$ – John Nov 26 '13 at 16:49
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If $T:(X,d)\to (CB(X),H)$ is a natural mapping $x\mapsto \{x\}$ then $T$ is an isometry.

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