Let $X$ be a topological space and $f:X \to \mathbb R$. Let $a \in X$ and $\mathcal N_a$ be the set of all neighborhoods of $a$. Let $$ \alpha := \liminf_{x \to a} f(x) := \sup _{V \in \mathcal N_a} \inf _{x \in V} f(x). $$
If $(x_d) \subset X$ is a net such that $x_d \to a$ and $f(x_d) \to \beta$, then I proved that $\beta \ge \alpha$.
My question: Is there a net $(x_t) \subset X$ such that $x_t \to a$ and $f(x_t) \to \alpha$?
Update: It seems I have found the required construction.
For each $(V, n) \in \mathcal N_a \times \mathbb N$, by axiom of choice, we pick $x_{V, n}$ such that $$ f(x_{V,n}) < \inf _{x \in V} f(x) + \frac{1}{n}. $$
We define a partial order $\le$ on $\mathcal N_a \times \mathbb N$ by $$ (V_1, n_1) \le (V_2, n_2) \iff (V_1 \supset V_2) \wedge (n_1 \le n_2). $$
Then $(\mathcal N_a \times \mathbb N, \le)$ is a directed set. It's clear that $\{f(x_{V,n}) \mid (V, n) \in \mathcal N_a \times \mathbb N\}$ is a net. Let's prove that $f(x_{V,n}) \to \alpha$. Fix $\varepsilon >0$. Then $(\alpha - \varepsilon, \alpha + \varepsilon)$ is a neighborhood of $\alpha$. There is $V'$ such that $$ \alpha - \varepsilon < \inf _{x \in V'} f(x) < \alpha + \varepsilon. $$
We pick $n'$ such that $\dfrac{1}{n'} < \varepsilon$. Then $$ \alpha - \varepsilon < \inf _{x \in V'} f(x) \le \inf _{x \in V} f(x) \le f(x_{V,n}) < \inf _{x \in V} f(x) + \varepsilon \le \alpha +\varepsilon \quad \forall (V, n) \ge (V', n') $$
This means $f(x_{V,n}) \in (\alpha - \varepsilon, \alpha + \varepsilon)$ for all $(V, n) \ge (V', n')$. This completes the proof.
