Characterization lower semi continuity

I want to show the characterization of semi-continuity in a general setting, as stated below.

Let $$X$$ a topological space, for $$x \in X$$ let me denote $$\mathcal V(x)$$ its neighborhoods. Let $$f : X \longrightarrow [-\infty , + \infty]$$, the following are equivalent:

1. $$\forall x_0 \in X,\quad \forall \epsilon > 0 ,\quad \exists V \in \mathcal V(x_0),\quad \forall x \in V,\quad f(x) \geq f(x_0) - \epsilon$$

2. Forall $$\lambda \in \mathbb R$$, $$\{ f\leq \lambda \}$$ is closed.

3. The epigraph of $$f$$ defined by $$epi(f) = \{ (x,t) \in X \times \mathbb R : f(x) \leq t \}$$ is closed.

I manage to prove that $$1 \Longrightarrow 2 \Longrightarrow 3$$ but for $$3 \Longrightarrow 1$$ I am stuck when $$f(x_0) = + \infty$$.

My attempt for $$3 \Longrightarrow 1$$:

Assume the epigraph is closed, let $$x_0 \in X$$ and $$\epsilon > 0$$. If $$f(x_0) = -\infty$$ any neighborhood of $$x_0$$ does the job in 1. If $$f(x_0)$$ is real then $$(x_0,f(x_0)-\epsilon) \in (epi(f))^c$$ that is open so there exists $$V \in \mathcal V(x_0)$$ and $$\delta > 0$$ such that $$\forall x \in V,\quad \forall t \in (f(x_0)-\epsilon -\delta, f(x_0)-\epsilon +\delta),\quad f(x) > t.$$ Then forall $$x \in V$$ and $$t = f(x_0) - \epsilon$$ we get $$f(x) > f(x_0)-\epsilon.$$ In the case $$f(x_0) = + \infty$$ I don't know what to do.

Also I know that there is a characterization with sequences for first countable spaces, if anyone could provide an example of $$X$$ and $$f$$ such that $$f$$ is sequentially lower semi continuous and not lower semi continuous I would be glade.

• (3) implies that $\{f>c\}$ is open for all $c$; that will be equivalent to saying that $f$ is l.s.c. (1) I think should be replaced by the condition: for every $x_0\in X$ and $\alpha\in\mathbb{R}$, if $\alpha<f(x_0)$ then there is a neighborhood $V(x_0)$ such that $f(x)>\alpha$ whenever $x\in V(x_0)$. Commented May 6, 2022 at 22:03

I think what you are trying to prove is false. Consider the function $$f:x\in[0,+\infty)\mapsto 1/x\in [-\infty,\infty],$$ where $$1/0:=\infty$$. Then the epigraph is closed but $$f$$ fails to satisfy your point (1) for $$x_0=0.$$