Let (X,d) be a metric space. $F: X\to \overline{\mathbb{R}}$. $F$ is sequentially lower semicontinuous iff $F$ is lower semicontinuous
The definitions that I have to use are:
(1) $F$ is sequentially lower semicontinuous if for all sequences $(x_n)_n \subseteq X$ s.t $ x_n\to x$ , $\liminf_n F(x_n)\ge F(x)$
(2) $F$ is lower semicontinuous if $\forall x \in X$ and $t \in \mathbb{R} $ with $F(x)>t$, there exists $r>0$ s.t $F(y)>t$ for all $y\in B(x,r)$
I have already proven $(1) \implies (2)$ by contradiction
I am having trouble proving $(2) \implies (1)$ I tried using contradiction as well , but I got nowhere:
If by absurd (1) is false, it means it's negation holds:
$\exists x_n \to x$ s.t $\liminf_n F(x_n)< F(x)$.
Since between two distint real numbers there's always a third one:$\exists t \in \mathbb{R}$ s.t
$\liminf_n F(x_n)< t <F(x)$.
So using (2), $\exists r>0,$ s.t $F(y)>t$ for all $y\in B(x,r)$
$\liminf_n F(x_n)< t <F(y) $,for all $y\in B(x,r)$.
Then I tried using that the last inequality implies $F(x_n)< t <F(y)$, for infinitely many $n$'s . I also tried using the definition of $\liminf_n$, but nothing. How should I proceed? Maybe contradiction is not the way to go here?
$F(x) >t$ so there exists $r>0$ s.t $F(z)>t$ for all $z\in B(x,r)$. We can take $z=x_n$ for $n$ sufficiently large. So $F(x_n) >t$ for such $n$ which implies $\lim \inf F(x_n) \geq t$, a contradiction.