Topological Space Separated by Mappings is Hausdorff

Theorem

Let $\family {Y_i}_{i \mathop \in I}$ be an indexed family of Hausdorff spaces for some indexing set $I$.

Let $\family {f_i : X \to Y_i}_{i \mathop \in I}$ be an indexed family of continuous mappings.

Suppose $\family {f_i : X \to Y_i}_{i \mathop \in I}$ separates the points of $X$.

Then, $X$ is Hausdorff.

Let $x \ne y$ be elements of $X$.

By definition of separating points, there exists some $i \in I$ such that:

$\map {f_i} x \ne \map {f_i} y$

As $Y_i$ is Hausdorff, there exist open sets $U, V \subseteq Y_i$ such that:

$\map {f_i} x \in U$

$\map {f_i} y \in V$

$U \cap V = \O$

By definition of continuous mapping, we have that:

$f_i^{-1} \sqbrk U$ and $f_i^{-1} \sqbrk V$ are open sets of $X$

By the definition of preimage, it is immediate that:

$x \in f_i^{-1} \sqbrk U$

$y \in f_i^{-1} \sqbrk V$

It remains to show that $f_i^{-1} \sqbrk U$ and $f_i^{-1} \sqbrk V$ are disjoint.

Aiming for a contradiction, suppose there is some $z \in X$ such that:

$z \in f_i^{-1} \sqbrk U$

$z \in f_i^{-1} \sqbrk V$

Then, by definition of preimage:

$\map {f_i} z \in U$

$\map {f_i} z \in V$

contradicting the fact that $U$ and $V$ are themselves disjoint.

By Proof by Contradiction, it follows that $f_i^{-1} \sqbrk U$ and $f_i^{-1} \sqbrk V$ are disjoint.

As $x \ne y$ were arbitrary, it follows that $X$ is Hausdorff by definition.

$\blacksquare$