# Rudin functional analysis, section 3.8 (Topological preliminaries)

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 $$f\in\mathcal{F}$$.