Composite of Order Isomorphisms is Order Isomorphism

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\left({S_1, \preceq_1}\right)$, $\left({S_2, \preceq_2}\right)$ and $\left({S_3, \preceq_3}\right)$ be ordered sets.

Let:

$\phi: \left({S_1, \preceq_1}\right) \to \left({S_2, \preceq_2}\right)$

and:

$\psi: \left({S_2, \preceq_2}\right) \to \left({S_3, \preceq_3}\right)$

be order isomorphisms.


Then $\psi \circ \phi: \left({S_1, \preceq_1}\right) \to \left({S_3, \preceq_3}\right)$ is also an order isomorphism.


Proof

From Composite of Bijections is Bijection, $\psi \circ \phi$ is a bijection, as, by definition, an order isomorphism is also a bijection.


By definition of composition of mappings, $\psi \circ \phi \left({x}\right) = \psi \left({\phi \left({x}\right)}\right)$.


As $\phi$ is an order isomorphism, we have:

$\forall x_1, y_1 \in S_1: x_1 \preceq_1 y_1 \implies \phi \left({x_1}\right) \preceq_2 \phi \left({y_1}\right)$


As $\psi$ is an order isomorphism, we have:

$\forall x_2, y_2 \in S_2: x_2 \preceq_2 y_2 \implies \psi \left({x_2}\right) \preceq_3 \psi \left({y_2}\right)$


By setting $x_2 = \phi \left({x_1}\right), y_2 = \phi \left({y_1}\right)$, it follows that:

$\forall x_1, y_1 \in S_1: x_1 \preceq_1 y_1 \implies \psi \left({\phi \left({x_1}\right)}\right) \preceq_3 \psi \left({\phi \left({y_1}\right)}\right)$


Similarly we can show that:

$\forall x_3, y_3 \in S_3: x_3 \preceq_3 y_3 \implies \phi^{-1} \left({\psi^{-1} \left({x_3}\right)}\right) \preceq_1 \phi^{-1} \left({\psi^{-1} \left({y_3}\right)}\right)$

$\blacksquare$


Sources