Proposition 10

Let $X$ be a set, let $(Y_i)_{i \in I}$ be a family of topological spaces, and for each $i \in I$ let $f_i$ be a mapping of $X$ into $Y_i$. Let $X$ be given the coarsest topology $\mathfrak{T}$ for which the $f_i$ are continuous. Then a necessary and sufficient condition for a filter $\mathfrak{F}$ on $X$ to converge to $a \in X$ is that for each $i \in I$ the filter base $f_i(\mathfrak{F})$ should converge to $f_i(a)$ in $Y_i$.

The condition is necessary since the $f_i$ are continuous (no. 4, Proposition 9, Corollary 1). Conversely, suppose that the condition is satisfied, and let $V$ be an open neighbourhood of $a$ in $X$. By the definition of $\mathfrak{T}$ (§ 2, no. 3, Proposition 4) there is a finite subset $J$ of $I$, and for each $i \in J$ an open subset $U_i$ of $Y_i$, such that $f_i(a) \in U_i$ for $i \in J$ and such that $V$ contains the set

The hypothesis implies that $f_i^{-1}(U_i) \in \mathfrak{F}$ (no. 3, Proposition 7); since $J$ is finite, it follows that belongs to $\mathfrak{F}$, and $M \subset V$. This completes the proof.