A quote from section 3.8. of Rudin's functional analysis:
Suppose next that $X$ is a set and $\mathcal{F}$ is a nonempty family of mappings $f: X \to Y_f$, where each $Y_f$ is a topological space (In many important cases $Y_f$ is the same for all $f \in \mathcal{F}$. Let $\tau$ be the collection of all unions of finite intersections of sets $f^{-1}(V)$, with $f \in \mathcal{F}$ and $V$ open in $Y_f$. Then $\tau$ is a topology on $X$, and it is in fact the weakest topology on $X$ that makes every $f \in \mathcal{F}$ continuous.
Why is such topology the weakest?
Also consider in the very same section the following proposition
If $\mathcal{F}$ is a family of mappings $f : X \to Y_f$ where $X$ is a set and each $Y_f$ is a Hausdorff space, and if $\mathcal{F}$ separates points on $X$, then the $\mathcal{F}$-topology of $X$ is a Hausdorff topology.
What is the meaning of the terminology $\mathcal{F}$-topology on $X$ in this context? Is it either 1) a topology on $X$ constructed using the family $\mathcal{F}$ or b) a topology on the set $\mathcal{F}$? From the very short proof below it seems to me is the former, but I wanna be sure.
(1) If you want all the $f$ be continuous, then all the $f^{-1}(V)$ must be open. Also, by definition of topology, all unions of finite intersections of such sets must be open. But this family is a topology, as you can check.
(2) Is the weakest topology that makes continuous all the mappings $f\in\mathcal{F}$.