semi continuity

Question

I've read about semi continuity on set valued-maps from topological space and real functions. Given $X$ and $Y$ are two topological spaces and set valued function $$F:X\to \mathcal{P}(Y).$$ Function $F$ is said to be upper semi continuous to $x\in X$ if for every open set containing $F(x)$, there exists $U\in\mathcal{U}(x)$ such that $$F(U)=\displaystyle \cup\{F(u):u\in U\}\subseteq V.$$ Meanwhile, in real-valued functions, $f$ is upper semi continuous to $x\in X$ if for every $k>f(x)$, there is a neighbourhood $V$ of $x$ such that $k>f(y)$ for every $y\in V$. If we consider set of real numbers as topological space, why two definitions above doesn't look similar?

Answer 1

They are actually similar at a higher level. $\Bbb R$ has a subbase consisting of two types of sets: upper sets $U(a) =\{x \in \Bbb R: x > a\}$, and lower sets $L(a)= \{x \in \Bbb R: x < a\}$, where $a \in \Bbb R$. This holds in fact in all ordered topological spaces by the definition of the order topology. Now $\{L(a),R(a): a \in \Bbb R\}$ generates the topology on $\Bbb R$ and in fact only the upper sets (plus $\{\emptyset, \Bbb R\}$) already form a topology on $\Bbb R$ (not a very nice one, only $T_0$ but not $T_1$ but it's sometimes used) and likewise for only the lower sets, and the usual topology is the supremum of these two weak topologies in the poset of all topologies on $\Bbb R$.

Now $f:X \to \Bbb R$ is upper semicontinuous (also as you state the definition) iff $f^{-1}[U(a)]$ is open in $X$ for all $a \in \Bbb R$. So it's continuity for the upper sets only. So quite literally semicontinuous: only the upper sets behave nicely under $f^{-1}$, hence upper semicontinuous. You could also say $f$ is continuous for the upper-topology on $\Bbb R$.

You can also check that lower semicontinuous functions are those that have $f^{-1}[L(a)]$ always open, and this explains their name. Also clear right away is that a function that is both upper and lower semicontinuous is continuous. Then openness of all inverse images of subbasic elements are open etc.

The same can be said for $\mathcal{P}(Y)$. This too has a standard topology that has a subbase that consists of two types of sets:

$[U]= \{F \in \mathcal{P}(Y): F \cap U \neq \emptyset\}$, where $U$ varies over all non-empty open subsets of $Y$ ("hit"-sets, because they have to "hit" $U$) and

$\langle U \rangle = \{F \in \mathcal{P}(Y): F \subseteq U\}$, where $U$ runs over the same sets ("miss"-sets as these are sets that "miss" (are disjoint from) $Y\setminus U$). All these types of sets together define a subbase for the Vietoris topology, which is normally restricted to the subspace of all closed (or compact even) non-empty sets of $Y$, to make it more nicely behaved, and is the major example of hit-and-miss topologies on power sets (here we have to hit open sets and miss closed sets, but other variants have been considered).

A set-values $F: X \to \mathcal{P}(Y)$ is upper semicontinuous if $F^{-1}[\langle V \rangle]=\{x \in X: F(x) \subseteq V\}$ is open in $X$ for all $V \subseteq Y$ open and non-empty. Again, continuity for half of the topology's subbase. It makes sense to call this the "upper" half as $\langle V \rangle$ are all sets upper bounded by $V$, as it were.

It's straightforward to check that it coincides with your definition (you take $x \in F^{-1}[\langle V \rangle]$ and the condition verifies it's an interior point of that set).

You can hopefully guess what a lower semi-continuous set-valued function must be, and check it against other statements of the definition.

You can also check that if $f:X \to \Bbb R$ is seen as a set valued function, e.g. by $F(x) = \{(-\infty,f(x)]\} \in \mathcal{P}(\Bbb R)$ then usc-ness of $F$ comes down to usc-ness of $f$ and vice versa (please check that, but it seems OK).

So actually, the definitions are similar: it's continuity wrt a half of a standard subbase of the codomain.