Does this topology defined by a map into $\mathbb{R}$ have a name?

Question

Let $f: X \rightarrow \mathbb{R}$ be a function. For $t \in \mathbb{R}_{\pm \infty}$, define the set $X_f(t) = \{x \in X | f(x) > t\}$.

This gives rise to a topology $T(f) := \{X_f(t) | t \in \mathbb{R}_{\pm \infty}\}$ of open sets on $X$, as we can check:

1. Full space: $X = X_f(-\infty) \in T(f)$

2. Arbitrary unions: $\bigcup X_f(t_i) = X_f(\inf\{t_i\})$, since $ x \in X_f(\inf \{t_i\}) \Leftrightarrow f(x) > \inf \{t_i\} \Leftrightarrow \exists i: f(x) > t_i \Leftrightarrow x \in \bigcup X_f(t_i) $

3. Pairwise intersections: $X_f(t) \cap X_f(u) = X_f(\max\{t,u\})$

Is this correct?

Does this topology have a name?

It seems like it is the initial topology for $f$ with respect to the topology $\{(t,\infty) | t \in [-\infty, \infty] \}$ on $\mathbb{R}$.

Also, $T(f)$ is totally ordered, so is the theory of filtrations relevant?

Is there a characterization of the class of functions $g$ that give rise to the same topology as a given $f$, i.e. $T(g) = T(f)$?