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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.

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

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  • $\begingroup$ For (2). Why is such topology the weakest? $\endgroup$ – user8469759 Dec 10 '18 at 11:22
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    $\begingroup$ @user8469759 Because any topology that has those maps be continuous must have those sets as open sets, hence their finite intersections be open, and unions of open sets must also be open. So any topology that has these open sets must contain the topology they generate, so by definition, it is the weakest topology making those maps continuous. $\endgroup$ – Aaron Dec 10 '18 at 11:26

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