Initial Topology with respect to Point-Separating Family of Mappings onto Hausdorff Spaces is Hausdorff

Theorem

Let $X$ be a set.

Let $\family {\struct {Y_i, \tau_i} }_{i \mathop \in I}$ be an indexed family of Hausdorff topological spaces indexed by $I$.

Let $\family {f_i: X \to Y_i}_{i \mathop \in I}$ be an indexed family of mappings indexed by $I$ that separate the points of $X$ in the sense that:

for each $x, y \in X$ with $x \ne y$, there exists $i \in I$ such that $\map {f_i} x \ne \map {f_i} y$.

Let $\tau$ be the initial topology on $X$ with respect to $\family {f_i}_{i \mathop \in I}$.

Then $\struct {X, \tau}$ is Hausdorff.

Let $x, y \in X$.

Then there exists $i \in I$ such that $\map {f_i} x \ne \map {f_i} y$.

Since $Y_i$ is Hausdorff, there exists $U, V \in \tau_i$ such that $U \cap V = \O$ and $\map {f_i} x \in U$, $\map {f_i} y \in V$.

Then we have $x \in f_i^{-1} \sqbrk U$ and $y \in f_i^{-1} \sqbrk V$.

From the definition of the initial topology, $f_i : X \to Y_i$ is $\tuple {\tau, \tau_i}$-continuous.

So $f_i^{-1} \sqbrk U \in \tau$ and $f_i^{-1} \sqbrk V \in \tau$.

We just need to show that $f_i^{-1} \sqbrk U \cap f_i^{-1} \sqbrk V = \O$.

This is immediate since by Preimage of Intersection under Mapping, since we have:

$f_i^{-1} \sqbrk U \cap f_i^{-1} \sqbrk V = f_i^{-1} \sqbrk {U \cap V} = f_i^{-1} \sqbrk \O = \O$

So $f_i^{-1} \sqbrk U$ and $f_i^{-1} \sqbrk V$ are disjoint open sets containing $x$ and $y$ respectively.

So $\struct {X, \tau}$ is Hausdorff.

$\blacksquare$