T1 Space is Preserved under Closed Bijection

Theorem
Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.

Let $\phi: T_A \to T_B$ be a closed bijection.

If $T_A$ is a $T_1$ (Fréchet) space, then so is $T_B$.

Proof
Let $T_A$ be a $T_1$ (Fréchet) space.

By definition, all points in $T_A$ are closed.

Let $a \in S_A$.

Then $\set a$ is a closed set.

As $\phi$ is a closed mapping it follows directly that $\phi \sqbrk {\set a}$ is closed.

As $\phi$ is a bijection it follows that every point in $S_B$ is the image under $\phi$ of a single point in $S_A$.

Hence every point in $S_B$ is closed.

That is, $T_B$ is a $T_1$ (Fréchet) space.