A basic doubt on upper semi-continuity of set-valued maps Upper  Semi-Continuity for set valued maps have two definitions 
$h:\Bbb R^d \to 2^{\Bbb R^d}$ is upper semi-continuous if 
1) Sequential definition : $x_n \to x$, $y_n \to y$ and $y_n \in h(x_n)$ then $y \in h(x)$
2) For every open set $ V \in \Bbb R^d$ the set $\{x:h(x) \subset V\}$ is open in $\Bbb R^d$
I want to prove 1 -> 2
So, assume 2) is false. So, there exist an open set $V$ s.t. $\{x:h(x) \subset V\}$ is not open. That means we can find a $p$ s.t. $h(p) \subset V$ and a sequence $x_n \to p$ s.t. $h(x_n)\not\subset V$. So, we have $y_n \in h(x_n)$ and $y_n \notin V$. But, there is no gurantee that $y_n$ converges, or one of its subsequence converges. How to prove then ?
 A: It seems the following. 
Counterexample 1. Let $h_0(x):\Bbb R^d\to\Bbb R^d$ be any map. Define a map $h: \Bbb R^d\to 2^{\Bbb R^d}$ by putting $h(x)=\{h_0(x)\}$.  Then the map $h$ cannot satisfies Condition 2 iff the map $h_0$ is continuous. From the other side, assume that $\{x_n\}$ and $\{y_n\}$ are sequences of points of the space $\Bbb R^d$ such that the sequence $\{x_n\}$ converges to a point $x\in \Bbb R^d$, the sequence $\{y_n\}$ converges to a point $y\in \Bbb R^d $, and $y_n \in h(x_n)$ for each index $n$. That means that $y_n=x_n$, therefore $y=x\in \{x\}=h(x)$.
Counterexample 2. Let $d=2$ and $h(u,v)=\{(u’,v’)\in\Bbb R^2:v’\le v\}$ for each point $(u,v)\in\Bbb R^2$. Put $V=\{(u,v)\in\Bbb R^2: v< e^u\}$. Then $V$ is an open subset of the space $\Bbb R^d$ but $h^{-1}(V)=\{(u,v)\in\Bbb R^2: v\le 0\}$ is not open. From the other side, assume that there are sequences $\{x_n=(u_n, v_n)\}$ and $\{y_n=(u’_n,v’_n)\}$ of points of the space $\Bbb R^d $ such that the sequence $\{x_n\}$ converges to a point $x=(u,v)\in \Bbb R^d$, the sequence $\{y_n\}$ converges to a point $y=(u’,v’)\in \Bbb R^d$, and $y_n \in h(x_n)$ for each index $n$. Then $v’_n\le v_n$ for each $n$, so $v’\le v$ and thus $y \in h(x)$.
