Topological Space Separated by Mappings is Hausdorff
Theorem
Let $X$ be a topological space.
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.
Proof
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.
$\blacksquare$