Corollary 1

Let $X$ be a set, let $(Y_i)_{i \in I}$ be a family of Hausdorff topological spaces, and for each $i \in I$ let $f_i$ be a mapping of $X$ into $Y_i$. Let $X$ carry the coarsest topology $\mathfrak{T}$ for which the $f_i$ are continuous. Then a necessary and sufficient condition for $X$ to be Hausdorff is that for each pair of distinct points $x, y$ of $X$ we have $f_i(x) \neq f_i(y)$ for some index $i \in I$.

The condition is sufficient by reason of Proposition 5 of no. 1. Conversely, suppose $X$ is Hausdorff; let $Y = \prod_{i \in I} Y_i$ and let $f = (f_i)_{i \in I}$ be the mapping $x \to (f_i(x))$. By Proposition 7 above $Y$ is Hausdorff, and by Proposition 3 of § 4, no. 1, $\mathfrak{T}$ is the inverse image under $f$ of the topology of $Y$. If $f(x) = f(y)$ for two distinct points $x, y$ of $X$ it is clear that every open set (in the topology $\mathfrak{T}$) which contains $x$ also contains $y$, contrary to the hypothesis that $X$ is Hausdorff.