Given a correspondence $f:X\rightarrow 2^X$, suppose X is a closed simplex in $\mathbb{R}^n$, and $f$ is compact-valued. We say $f$ is upper hemicontinuous if, $\forall x\in X$ and every open subset $V$ of $X$ such that $f(x)\subset V$, there exists a neighborhood $U$ of $x$ such that, $\forall x'\in U$, $f(x')\subset V$.
However, there is a sequential characterization of upper hemicontinuity, which is useful in proofs, given below.
$f$ is upper hemicontinuous if and only if, $\forall x\in X$, $\{x_n\}\subset X$ converging to $x$, $y_n\in f(x_n)$, and $y_n\rightarrow y$ all together imply that $y\in f(x)$.
I have tried to prove the equivalence between the definition of upper hemicontinuity and its sequential characterization for a week, but still got no clue. Is there any trick?
I don't know whether the following method is correct or not, just a personal opinion.
Consider the open set, for $\forall m>0$, $V=N_{\frac{1}{2m}}(f(x))$, obviously, $f(x)\subset V $
Since $f$ is UHC, $\exists \delta>0$, s.t $f(N_{\delta}(x))\subset V$
And $\lim x_{n}=x$, so $\exists N_1(m)>0$, s.t $n>N_1$, $x_n\in N_{\delta} (a)$
Hence $f(x_n)\subset V$, notice that $y_n\in f(x_n)$, $y_n\in N_{\frac{1}{2m}}(f(x))$, which implies that $\exists \alpha \in f(x)$, s.t $y_n\in N_{\frac{1}{2m}}(\alpha)$
Similarly, $\lim y_n=y$, $\exists N_2(m)>0$, s.t $n>N_2$, $y_n\in N_{\frac{1}{2m}}(y)$
Just pick the $n> \max\{N_1,N_2\}$, then $y\in N_{\frac{1}{m}}(\alpha)$
we know that
$y=\alpha$, then $y\in f(x)$
$y\neq \alpha$, notice that $m$ is an arbitrary value. Therefore, every neighborhood of $y$ contains a point $\alpha \in f(x)$, which implies that $y$ is a limit point of $f(x)$, since $f(x)$ is closed set which contains all its limit points, we know that $y\in f(x)$