MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?
share|cite|improve this question
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
up vote 12 down vote accepted

$\phi$ 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.

share|cite|improve this answer
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.

share|cite|improve this answer
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 for a related (but different) notion.

share|cite|improve this answer
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. – Mikhail 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? – Mikhail 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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.