Let $X$ a topological space satisfying the first countability axiom. I want to prove the following result:
$\varphi : X \rightarrow R $ is lower semicontinuous (this mean that $\varphi^{-1} (a, + \infty)$ is open for all $a \in R$) if and only if $ \varphi (\hat{u}) \leq \liminf \varphi (u_n) $ (for all sequence $u_n$ converging to $x$).
For the first part my proof is :
Let $u_n \subset X$ with $u_n \rightarrow \hat{u}$. Write $a = \varphi (\hat{u})$. Let $\epsilon >0$. We have $\hat{u} \in \varphi^{-1}(a - \epsilon , + \infty)$ . This set is open in X, then exists $n_0$ a natural number such that
$$u_n \in \varphi^{-1}(a - \epsilon , + \infty), \forall n \geq n_0$$ then
$$ \varphi (\hat{u}) \leq \inf_{n > n_0} \varphi(u_n) + \epsilon \leq \inf_{n > k} \varphi(u_n) + \epsilon \ \forall k \geq n_0$$
then $ \varphi (\hat{u}) \leq \liminf_{n > n_0} \varphi(u_n) + \epsilon $ Then follow the first part.
The proof of the first part is correct? Someone can give me a help with the other part?