Order Isomorphism on Totally Ordered Set preserves Total Ordering

From ProofWiki
Jump to navigation Jump to search

Theorem

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

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 if and only if $\left({T, \preccurlyeq_2}\right)$ is also a totally ordered set.


Proof

Necessary Condition

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.

$\Box$


Sufficient Condition

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

Applying the other implication (proven above) to $\phi^{-1}$, it follows that if $\left({T, \preccurlyeq_2}\right)$ is a totally ordered set then so is $\left({S, \preccurlyeq_1}\right)$.

$\blacksquare$


Sources