# What is the definition of continuity of set-valued functions?

According to the wiki of Kakutani's fixed-point theorem, A set-valued mapping $\varphi$ from a topological space $X$ into a powerset $\wp(Y)$ called upper semi-continuous if for every open set $W \subseteq Y$, $\lbrace x| \varphi(x) \subseteq W \rbrace$ is an open set in $X$.

My question:

1. What is the definition of continuity of a multi valued map $\varphi$?
2. What's the definition of open sets in $\wp(Y)$, in other words, what topology does $\wp(Y)$ have?
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Note that $Y$ is also a space in the statement of the theorem, albeit $Y=X$. What you've written is garbled, as you're talking about $W \subseteq Y$ and also $W \subseteq X$ –  David Roberts May 20 '13 at 11:50

$\phi$ is upper semicontinuous if, for every open $W\subset Y$, the set $\lbrace x | \phi(x)\subset W\rbrace$ is open in $X$.

$\phi$ is lower semicontinuous if, for every open $W\subset Y$, the set $\lbrace x | \phi(x)\cap W\neq \emptyset\rbrace$ is open in $X$.

$\phi$ is continuous if it is both upper semincontinuous and lower semicontinuous.

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Which begs the question (which I think is what the OP might have been getting at): what is the topology on $\mathcal{P}(Y)$ such that $\phi : X\to \mathcal{P}(Y)$ is continuous if and only if its upper- and lower- semicontinuous? –  Mark Grant May 20 '13 at 12:00
@MarkGrant: If $X$ and $Y$ are Hausdorff, a closed-valued map $\phi\colon X \to \operatorname{Cl}(Y)$ is upper- and lower semicontinuous iff $\phi$ is continuous with respect to the Vietoris topology. A sub-basis for the Vietoris topology is given by $U^- = \lbrace F \in \operatorname{Cl}(Y) \mid F \cap U \neq \emptyset\rbrace$ and $U^+ = \lbrace F \in \operatorname{Cl}(Y) \mid F \subseteq U\rbrace$ where $U$ runs through the open sets of $Y$. [Specialize to $Y$ discrete to get $\mathcal{P}(Y)$]. Gerald points out that upper and lower continuity are more related to order than to a topology. –  Martin May 20 '13 at 13:06
Mark Grant: It's immediate from the definitions that the relevant topology is the one Martin described. I hadn't known, though, that this is called the Vietoris topology. –  Steven Landsburg May 20 '13 at 15:03
@Martin Thank you very much. Btw, I have found a paper Topologies on Spaces of Subsets by Ernest Michael introduced topology on subsets in its 5th section. –  Heng Gu May 20 '13 at 16:16
Also, this Vietoris topology is the one induced by the Hausdorff metric, in the special case where $X$ is a compact metric space and we take only the closed non-empty subsets (instead of the whole power set). The closed non-empty subsets of a space $X$ in this topology are called the hyperspace of $X$, denoted $H(X)$ or sometimes $2^X$. –  Henno Brandsma May 22 '13 at 4:04

The definition quoted is an "order" notion of upper semicontinuous, not a "topology" notion. For real-valued functions, the two coincide. But in other settings you can have one but not the other.

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What about in this case? Is there no topology to define continuous compatibly? –  Heng Gu May 20 '13 at 12:56

One sensible way of generalizing continuity to set-valued functions (from $X$ to subsets of $Y$) is to require the graph of the function to be closed in the product $X\times Y$. This would be equivalent to the continuity of the function if $Y$ is compact. Thus, the Heaviside function is not continuous because one of the points 0 or 1 on the $y$-axis is not in the graph, but if one redefines it to take both values at 0, the graph becomes closed subset of the plane. See http://en.wikipedia.org/wiki/Closed_graph_theorem for a related (but different) notion.

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This is upper semi-continuity in multi-valued lingo. –  alvarezpaiva Feb 24 '14 at 12:54
The heaviside function can be chosen to be uppersemicontinuous but you can't make its graph closed. –  katz Feb 24 '14 at 13:46
To make the graph closed, you make the function set-vauled. All values but one are singletons, but the value at $0$ is the set $[0,1]$. –  Gerald Edgar Feb 24 '14 at 14:26
Fine, but this is rather different from the notion of semicontinuity for ordinary functions. By the way, is there a reason to chose the interval [0,1] rather than the two-point set? –  katz Feb 24 '14 at 15:44
Yes, it is defferent from the definition for ordinary functions. Some people call it upper hemicontinuity to avoid confusion. –  alvarezpaiva Feb 24 '14 at 18:08