# Domain of Continuous Injection to Hausdorff Space is Hausdorff

## Theorem

Let $T_\alpha = \left({S_\alpha, \tau_\alpha}\right)$ and $T_\beta = \left({S_\beta, \tau_\beta}\right)$ be topological spaces.

Let $f: S_\alpha \to S_\beta$ be a continuous mapping which is an injection.

If $T_\beta$ is a $T_2$ (Hausdorff) space, then $T_\alpha$ is also a $T_2$ (Hausdorff) space.

## Proof

Let $x, y \in S_\alpha$ be distinct points.

We want to find disjoint open sets $U, V \in \tau_\alpha$ containing $x$ and $y$ respectively.

Since $f$ is injective the points $f \left({ x }\right), f \left({ y }\right) \in S_\beta$ are distinct.

By assumption $T_\beta$ is Hausdorff.

Therefore we can choose disjoint open sets $U', V'$ of $T_\beta$ such that $f\left({ x }\right) \in U'$ and $f\left({ y }\right) \in V'$.

Since $f$ is continuous, the sets $U = f^{-1}\left({ U' }\right)$ and $V = f^{-1}\left({ V' }\right)$ are open.

Moreover we have $x \in U$ and $y \in V$.

Finally we show that the sets $U$ and $V$ are disjoint.

Suppose $z \in U \cap V$.

Then by the definition of $U$, $V$ we have $f \left({ z }\right) \in U'$ and $f\left({ z }\right) \in V'$.

This is a contradiction, since $U' \cap V' = \varnothing$.

$\blacksquare$