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Attouch-Wets topology on Function spaces. (English) Zbl 0842.54028
If $$(X,d)$$ is a metric space then the Hausdorff metric on the space of closed, bounded subsets of $$X$$ must be modified to accommodate unbounded sets. One possibility is the Attouch-Wets topology [see, for example, G. Beer and A. Di Concilio, Proc. Am. Math. Soc. 112, 235-244 (1991; Zbl 0677.54007)] which is metrizable and which can be constructed via the metric: $m_d (A,B) : = \sum^\infty_{n = 1} 2^{-n} d_n (A,B)/ \bigl( 1 + d_n (A,B) \bigr)$ where $$d_n (A,B) : = \sup \{(x,A) - d(x,B) |: x \in B (x^*, n)\}$$ and $$x^*$$ is an arbitrary (but fixed) point of $$X$$. This topology has advantages for handling closed, convex sets in an infinite-dimensional normed space $$X$$. If $$(X,d)$$ and $$(Y,e)$$ are metric spaces then a continuous function from $$X$$ to $$Y$$ may be identified with its graph and the space $$C(X,Y)$$ of continuous functions given a uniformity via the relative Attouch-Wets topology on the collection of graphs (viewed as closed subsets of $$X \times Y$$ with the product metric $$d \times e)$$.
One aim of the paper is to establish conditions that imply that a collection of functions is complete in the above uniformity. The authors prove that if $${\mathcal F}$$ is a pointwise equicontinuous, pointwise bounded family of functions that is closed in $$(C(X,Y), m_{d \times e})$$ then $${\mathcal F}$$ is complete. They give an example to show that neither the equicontinuity nor the boundedness is necessary for the conclusion. Moreover, if $$(Y,e)$$ is complete and has a nontrivial path then $$(X,d)$$ is complete if and only if every family $${\mathcal F}$$ as above is complete. The ideas extend naturally, as is shown in the final section, to families of set-valued mappings (multifunctions, correspondences) from $$X$$ to $$Y$$ whose graphs are closed.
Reviewer: A.C.Thompson

##### MSC:
 54C35 Function spaces in general topology 54B20 Hyperspaces in general topology 54C60 Set-valued maps in general topology