(b) Assume towards contradiction existence of well-orders
(A,≤A) and
(B,≤B) such that two or more of the following hold:
A≅B as orders
A is
order-
isomorphic with a
segment of
B
B is
order-
isomorphic with a
segment of
A
Assume (1) and (2) hold. Then exists
b∈B such that
(B,≤B)≅(A,≤B)≅(seg(b),≤B). Hence by composition of
isomorphisms we have an
order-preserving
function f:B→seg(b), which contradicts part (a) of this homework problem. Done!
A similar proof holds if (1) and (3) are true.
Assume (2) and (3) are true. Then exist
a∈A and
b∈B and
functions f,
g so that
(A,≤A) f≅ (seg(b),≤B) inclusion↪ (B,≤B) g≅ (seg(a),≤A)
Composing these
functions gives an
order-preserving map
f:A→seg(a), which contradicts part (a) of this homework problem.