Order Isomorphism Preserves Initial Segments

Theorem

Let $A_1$ and $A_2$ be classes.

Let $\prec_1$ and $\prec_2$ be strict orderings.

Let $\phi: A_1 \to A_2$ create an order isomorphism between $\struct {A_1, \prec_1}$ and $\struct {A_2, \prec_2}$.

Suppose $x \in A_1$.

Then $\phi$ maps the $\prec_1$-initial segment of $x$ to the $\prec_2$-initial segment of $\map \phi x$.

$\phi$ maps the $\prec_1$-initial segment of $x$ to:

$\ds \phi \sqbrk {\set {y \in A: y \prec_1 x} }$

$=$

$\ds \phi \sqbrk {\set {y \in A: \map \phi y \prec_2 \map \phi x} }$

Definition of Initial Segment and Definition of Order Isomorphism

$\ds $

$=$

$\ds \set {\map \phi y \in \phi \sqbrk A: \map \phi y \prec_2 \map \phi x}$

This is the $\prec_2$-initial segment of $\map \phi x$.

$\blacksquare$