Let E and F be partially ordered classes, and let g : E → F be an isomorphism.
Prove that for arbitrary x$\in E$,
g($S_x$)=$S_{g(x)}$
Attempted proof v1.0
By hypothesis g is an isomorphism,g is bijective.By Theorem 4.1.14every element g($S_x$) in E has a corresponding element $S_{g(x)}$ in F(So I have to create such a 1-1 map)
But first Image of g is defined as:
g[E]={g(x)|x $\in $E}
Now my mapping
x $\mapsto$ g(x)
$S_x \mapsto $g($S_x$)
g($S_x$) $\mapsto $ $S_{g(x)}$
Since x,y $\in E $ then $S_x, S_y$ are two initial segments of class X.
if x < y, then$S_x$ is an initial segment of $S_y$ and$S_x$ ⊂$S_y$;
Since g is in increasing and by theorem 4.1.1 x <y $\iff$ $S_x < S_y \iff$g($S_x$)< $S_{g(x)}$
Help .