Homepage › Solution manuals › Karel Hrbáček › Introduction to Set Theory › Exercise 6.1.7
Exercise 6.1.7
Let be a well-ordered set, and let . Extend to by making greater than all . Then has a smaller order type than .
Answers
This problem is looking ahead to future sections where order types and how to compare them are defined.
Proof. Suppose that is the order type of and that is the isomorphism. Now let . We then claim that is isomorphic to . So define a by
for . Clearly we have that for any .
Now consider any . If then set so that . If then so set so that then . We then have that . Therefore is surjective.
Now consider any where
Case: . Then since is an isomorphism and we have that .
Case: and . Then and so that . Hence so that by the definition of we have that .
Note that these cases are exhaustive since it can’t be that since (and therefore ). It also cannot be that but since then it would be that since is the greatest element of , which contradicts . Thus in all cases so that is increasing, and therefore injective and an isomorphism.
Hence is the order type of , is the order type of , and since . □