Equivalence of Definitions of Homeomorphic Topological Spaces

Theorem
Let $T_\alpha = \struct {S_\alpha, \tau_\alpha}$ and $T_\beta = \struct {S_\beta, \tau_\beta}$ be topological spaces.

Let $f: T_\alpha \to T_\beta$ be a bijection.

Definition 1 iff Definition 2
Let $f$ and $f^{-1}$ both be continuous.

As $f$ is continuous, then by definition:
 * $V \in \tau_\beta \implies f^{-1} \sqbrk V \in \tau_\alpha$

and as $f^{-1}$ is continuous, then by definition:
 * $U \in \tau_\alpha \implies \paren {f^{-1} }^{-1} \sqbrk U = f \sqbrk U \in \tau_\beta$

That is:
 * $\forall U \subseteq S_\alpha: U \in \tau_\alpha \iff f \sqbrk U \in \tau_\beta$

Conversely, suppose that:
 * $\forall U \subseteq S_\alpha: U \in \tau_\alpha \iff f \sqbrk U \in \tau_\beta$

By definition:
 * $U \in \tau_\alpha \implies f \sqbrk U \in \tau_\beta$ $f^{-1}$ is continuous.

Also by definition:
 * $f \sqbrk U \in \tau_\beta \implies U = f^{-1} \sqbrk {f \sqbrk U} \in \tau_\alpha$ $f$ is continuous.

That is, both $f$ and $f^{-1}$ are continuous.

Definition 1 iff Definition 3
Let $f$ and $f^{-1}$ both be continuous.

Then by Bijection is Open iff Inverse is Continuous, $f$ is open.

Thus $f$ is both open and continuous.

Conversely, let $f$ be both open and continuous.

Then also by Bijection is Open iff Inverse is Continuous, $f^{-1}$ is continuous.

Thus both $f$ and $f^{-1}$ are continuous.

Definition 3 iff Definition 4
Let $f$ be both open and continuous.

Then by Bijection is Open iff Closed it follows that $f$ is both closed and continuous.

Conversely, let $f$ be both both closed and continuous.

Then also by Bijection is Open iff Closed it follows that $f$ is both open and continuous.