show that the correspondence is upper-hemi continuous 
Let $\Gamma_i : X \to Y_i, i =1, ... , k$, be compact valued and upper hemi continuous. Show that $\Gamma : X \to Y = Y_1 \times ... \times Y_i$ defined by
$$\Gamma(x) = \{y \in Y: y = (y_1, ... , y_k), \text{where $y_i \in \Gamma_i(x), i =1, ... ,k$}\},$$
is also compact-valued and upper hemi continuous.

I can show $\Gamma(x)$ is compact. To show upper-hemi continuity, choose any sequence $(x^n) \in X$ converging to $x$, and pick any sequence $(y^n) \in Y$ such that $y^n = \Gamma(x^n)$ for all $n$. I need to show that $(y^n)$ has a subsequence converging to $y \in \Gamma(x)$. Since $\Gamma_i$ is upper-hemi continuous. For each $i$, we can construct $(y_i^{n_{k(i)}})$ converging to $y_i$. I was first thinking that I can collect all $\{n_{k(i)}\}$for each $i$ by taking union $\cup_i \{n_{k(i)}\}$, and then construct a subsequence of $(y^n)$. However, I am not really convinced because for example,  $(y_i^j)$ for $j \in \cup_i\{n_{k(i)}\}$ may not converge to $y_i$ (although we know that $(y_i^{n_{k(i)}})$ converging to $y_i$). How can I construct a subsequence of $(y^n)$ converging to $y \in \Gamma(x)$? I would appreciate if you give some help.
 A: Do you think this prove would hold:
I use the following definition:

Let   $X \subseteq E^{n}$,   $Y \subseteq E^{m}$   and $\Psi: X \rightrightarrows Y$ be a set-valued map.   $\Psi$ is upper
hemicontinuous (u.h.c) at $x_{0} \in X$ if,   for every open set $V \supseteq \Psi (x_{0})$,   there is an open set ${U}$ with $x_{0} \in {U}$ such that \begin{equation} \Psi (x) \subseteq {V} \text{ for every } x \in {U} \cap {X}.    \end{equation} $\Psi$ is upper  hemicontinuous if it is upper hemicontinuous at every $x \in {X}$.

Let thus consider a $x_0 \in X$ and let consider an open set $V$ such that $\Gamma (x_0) \subseteq V$. It is possible to find open sets $V_i$, $1 = 1,\dots,k$, such that $\Gamma_i(x_0) \subseteq V_i$ and that $V_1 \times \dots V_k \subseteq V$.
For all $i = 1, \dots,k$,
by the upper hemicontinuity of $\Gamma_i$, there exists a $U_i$ such that $\Gamma_i (x) \subseteq {V_i} \text{ for every } x \in {U_i} \cap {X}$.
Hence, if one defines $U = U_1 \cap \dots \cap U_k$, one obtains
\begin{equation}
\forall x \in U: \Gamma (x) = \Gamma_1 (x) \times \dots \Gamma_k (x) \subseteq V_1 \times \dots V_k \subseteq V,
\end{equation}
which proves the u.h.c of $\Gamma$.
