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?

closedsets. 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". $\endgroup$ – Ian Morris Sep 22 '16 at 10:46