Upper semicontinuity of set-valued maps with open values

Let $X$ and $Y$ be metric spaces. The $(\varepsilon,\delta)$-definition of continuity of single-valued maps can be rephrased as:

Let $f$ be a single-valued map from $X$ to $Y$. $f$ is continuous at $x_0 \in X$ if for every neighborhood $N_Y$ of $f(x_0)\in Y$, there exists a neighborhood $N_X$ of $x_0$ such that $f(N_X) \subset N_Y$.

In this book (Jean-Pierre Aubin and Hélène Frankowska, MR 2458436 Set-valued analysis, ISBN: 978-0-8176-4847-3, page 38.), this concept of continuity is extended to set-valued maps as:

Let $F$ be a set-valued map from $X$ to $Y$. $F$ is said to be upper semicontinuous at $x_0 \in X$ if for every neighborhood $N_Y$ of $F(x_0)\subset Y$, there exists a neighborhood $N_X$ of $x_0$ such that $F(N_X) = \bigcup_{x \in N_X}F(x) \subset N_Y$.

What bothers me is the possibility of $N_Y = F(x_0)$ when $F(x_0)$ is open. For example, if we let $X = Y = \mathbb{R}$, $F(x) = (0,x^2+1)$, and $x_0 \in \mathbb{R}$, then $F(x_0)$ itself is a neighborhood of $F(x_0)$ and $F(N_X) \not\subset F(x_0)$ for all neighborhood $N_X$ of $x_0$. Therefore, $F$ is not upper semicontinuous at $x_0$. However, it does not make sense for me; $F$ should be continuous, isn't it? Would you give me any reason why people do not put the additional condition like "$N_Y$ is a neighborhood of the closure of $F(x_0)$" on the definition?

• One reason is that, in your example, that condition wouldn't help anyway. – Alex Degtyarev Sep 22 '16 at 10:29
• @AlexDegtyarev Thanks for pointing my mistake. I changed the condition. – flyingwith Sep 22 '16 at 10:34
• At least in my experience, it is more normal to define these properties in the context of functions whose values are closed sets. I'd have to go to the library to check, but as far as I recall this is what is done in Kuratowski's "Topology". – Ian Morris Sep 22 '16 at 10:46
• @IanMorris Thanks for your comment but my references don't assume the closed values in the definition. – flyingwith Sep 22 '16 at 11:07
• One possible reason I can come up with is that if $F(x_0)$ is not closed, then there could be a single-valued function $f(x) \in F(x)$ that converges to the boundary point of $F(x_0)$ at $x_0$ such that $f$ is discontinuous at $x_0$. So, it might be preferable to say $F$ is not upper semicontinuous. However, this reason does not work in the general case because $F \equiv (0,1)$ is upper semicontinuous. – flyingwith Sep 22 '16 at 11:13

I think there is not much more to say than that most interesting results about upper semicontinuous set-valued maps involve closed-valued or even compact-valued maps. Indeed some authors choose to define upper semicontinutiy only for such maps. But there are good reasons not to. It helps to rephrase continuity notions for set-valued maps in terms of preimages of open sets, and here it turns out that there is more than one natural notion of a preimage.

Let $F:X\to 2^X$ be a set valued map and let $W\subseteq Y$. The upper inverse of $W$ under $F$ is

$$F^U(W)=\{x\in X:F(X)\subseteq W\}$$

and the lower inverse is

$$F^L=\{x\in X:F(x)\cap W\neq\emptyset\}.$$ Then one can define upper semicontinuity by the requirement that the upper inverse of an open set is again open and lower semicontinuity by the requirement that the lower inverse of an open set is again open. If one identifies functions with set-valued maps whose values are singletons, both notions coincide with ordinary continuity there. Indeed, one often talks about hemicontinuity instead of semicontinuity nowadays exactly because a, say, upper semicontinuous function is in general not upper semicontinuous as a set-valued map.

Now these two notions of continuity for set-valued maps represents different aspects of continuity. The example you gave does satisfy lower hemicontinuity so it does satisfy an idea of continuity, just not the one you used.

A very accessible reference for results on set-valued maps is Chapter 17 of Infinite Dimensional Analysis by Aliprantis and Border.

This map is single valued map. $f$ is upper semicontinuous at $x_0$ if for every positive epsilon there exist a nbhd $U$ of $x_0$ such that $f(x) \le f(x_0)+\varepsilon$.

• For some basic information about writing math at this site see e.g. here, here, here and here. – Martin Sleziak Sep 25 '17 at 12:32