Order Isomorphism between Linearly Ordered Spaces is Homeomorphism

Theorem
Let $\left({S_1, \le_1, \tau_1}\right)$ and $\left({S_2, \le_2, \tau_2}\right)$ be totally ordered spaces.

Let $\phi: S_1 \to S_2$ be an order isomorphism from $\left({S_1, \le_1}\right)$ to $\left({S_2, \le_2}\right)$.

Then $\phi$ is a homeomorphism from $\left({S_1, \tau_1}\right)$ to $\left({S_2, \tau_2}\right)$.

Proof
By the definition of order isomorphism, $\phi$ is a bijection.

Thus to show $\phi$ is a homeomorphism it remains to be shown that:
 * $\phi$ is continuous

and:
 * $\phi^{-1}$ is continuous.

First it is shown that $\phi^{-1}$ is continuous.

By Order Isomorphism Preserves Initial Segments and its dual, $\phi$ maps open rays in $\left({S_1, \le_1}\right)$ to open rays in $\left({S_2, \le_2}\right)$.

By Continuity Test using Sub-Basis, $\phi^{-1}$ is continuous.

Next it is shown that $\phi^{-1}$ is continuous.

By Inverse of Order Isomorphism is Order Isomorphism, $\phi^{-1}$ is an order isomorphism.

Applying the above shows that $\phi = \left({\phi^{-1}}\right)^{-1}$ is continuous.

Both $\phi$ and $\phi^{-1}$ have been shown to be continuous.

Thus, by definition, $\phi$ is a homeomorphism.