Characterization lower semi continuity

Question

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.

Answer 1

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 $x_0=0.$