I'm having trouble understanding a couple of things when studying well orderings and ordinals.
I know that given a well ordering $(A,<)$ there is no $a\in A$ s.t. $(A,<)\cong (A_a,<)$ where $A_a$ is the proper initial segment of $A$ determined by $a$.
Now I have to prove that given two well orderings $(A,<)$ and $(B,\triangleleft)$ exactly one of the following holds:
- $(A_a,<)\cong (B,\triangleleft)$ for some $a\in A$;
- $(A,<)\cong (B,\triangleleft)$;
- $(A,<)\cong (B_b,\triangleleft)$ for some $b\in B$.
I have a proof of this theorem, but there are some points that are left to the reader and I struggle understanding them. Firstly I can't get a contradiction when 1. and 3. simultaneously hold, I think it should contradict the result I stated above but I can't get it. Next, during the proof we defined a function $f=\{(x,y)\in A\times B : (A_x,<)\cong (B_y,\triangleleft)\}$. We proved that this is a function, we shown what are its domain and codomain, but I don't get why it is an order isomorphism, i.e. why given $a<a'\in dom(f)$ we must have $f(a)\triangleleft f(a')$.
Any help is very much appreciated :)
As mentioned in the comments, 1 and 3 contradict each other since if $A$ were isomorphic to an initial segment of $B$ and $B$ were isomorphic to an initial segment of $A,$ then these isomorphisms would compose to give an isomorphism of $A$ with an initial segment of itself, which is impossible. Note the same argument shows that 1 and 2 are contradictory, and same for 3 and 2.
If we have $A_a\cong B_b$ and $A_{a'}\cong B_{b'}$ and $a<a'$, then since $A_a$ is an initial segment of $A_{a',}$ $B_b$ must be isomorphic to an initial segment of $B_{b'}.$ Then by similar logic as the first part, we can't have $b'\le b$ since then we'd either have $B_{b'}$ equal to $B_b$ or it is an initial segment of $B_b$ and then $B_b$ would be isomorphic to an initial segment of itself.