Order Isomorphism Preserves Initial Segments

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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 $\left({A_1, \prec_1}\right)$ and $\left({A_2, \prec_2}\right)$.

Suppose $x \in A_1$.


Then $\phi$ maps the $\prec_1$-initial segment of $x$ to the $\prec_2$-initial segment of $\phi \left({x}\right)$.


Proof

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

\(\displaystyle \phi \left[{\left\{y \in A: y \prec_1 x \right\} }\right]\) \(=\) \(\displaystyle \phi \left[{\left\{y \in A: \phi \left({y}\right) \prec_2 \phi \left({x}\right) \right\} }\right]\) by the definition of initial segment and order isomorphism
\(\displaystyle \) \(=\) \(\displaystyle \left\{ \phi\left({y}\right) \in \phi \left[{A}\right]: \phi \left({y}\right) \prec_2 \phi \left({x}\right) \right\}\) by the definition of order isomorphism and image


This is the $\prec_2$-initial segment of $\phi \left({x}\right)$.

$\blacksquare$


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