Order Isomorphism on Totally Ordered Set preserves Total Ordering

Theorem
Let $$\left({S, \preccurlyeq_1}\right)$$ and $$\left({T, \preccurlyeq_2}\right)$$ be posets.

Let $$\phi: \left({S, \preccurlyeq_1}\right) \to \left({T, \preccurlyeq_2}\right)$$ be an order isomorphism.

Then $$\left({S, \preccurlyeq_1}\right)$$ is a totally ordered set iff $$\left({T, \preccurlyeq_2}\right)$$ is also a totally ordered set.

Proof
Let $$\left({S, \preccurlyeq_1}\right)$$ be a totally ordered set

Then by definition $$\preccurlyeq_1$$ is a total ordering.

Let $$x, y \in S$$.

Then either $$x \preccurlyeq_1 y$$ or $$y \preccurlyeq_1 x$$.

From the definition of order isomorphism, either:
 * $$\phi \left({x}\right) \preccurlyeq_2 \phi \left({y}\right)$$

or:
 * $$\phi \left({y}\right) \preccurlyeq_2 \phi \left({x}\right)$$

and so by definition $$\preccurlyeq_2$$ is also a total ordering.

So by definition $$\left({T, \preccurlyeq_2}\right)$$ is also a totally ordered set.

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

So the same technique is used to show that if $$\left({T, \preccurlyeq_2}\right)$$ is a totally ordered set then so is $$\left({S, \preccurlyeq_1}\right)$$.