Inverse of Order Isomorphism is Order Isomorphism

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Theorem

Let $\left({S, \preceq_1}\right)$ and $\left({T, \preceq_2}\right)$ be ordered sets.

Let $\phi$ be a bijection from $\left({S, \preceq_1}\right)$ to $\left({T, \preceq_2}\right)$.


Then:

$\phi: \left({S, \preceq_1}\right) \to \left({T, \preceq_2}\right)$

is an order isomorphism if and only if:

$\phi^{-1}: \left({T, \preceq_2}\right) \to \left({S, \preceq_1}\right)$

is also an order isomorphism.


Proof

Follows directly from the definition of order isomorphism.

Let $\phi: \left({S, \preceq_1}\right) \to \left({T, \preceq_2}\right)$ be an order isomorphism.

Then:

$\forall x, y, \in S: x \preceq_1 y \implies \phi \left({x}\right) \preceq_2 \phi \left({y}\right)$

and:

$\forall \phi \left({x}\right), \phi \left({y}\right) \in T: \phi \left({x}\right) \preceq_2 \phi \left({y}\right) \implies \phi^{-1} \left({\phi \left({x}\right)}\right) \preceq_1 \phi^{-1} \left({\phi \left({y}\right)}\right)$

That is:

$\forall \phi \left({x}\right), \phi \left({y}\right) \in T: \phi \left({x}\right) \preceq_2 \phi \left({y}\right) \implies x \preceq_1 y$

Hence the result.

$\blacksquare$


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