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Exercise 6.1.4
For every infinite subset of , is isomorphic to .
Answers
Proof. Let be an infinite subset of . Then (where is the standard well-ordering of ) is a well-ordering since any is also a subset of and therefore has a least element. Hence by Theorem 6.1.3 either:
- 1.
- and are isomorphic,
- 2.
- An initial segment of is isomorphic to , or
- 3.
- is isomorphic to an initial segment of
We show that they must be isomorphic (1) by showing that (2) and (3) lead to contradictions.
Suppose (2), i.e. that an initial segment of is isomorphic to . Then since it follows from Lemma ?? that is finite. But since this is isomorphic to it means that , which is a contradiction!
Now suppose (3) so that is isomorphic to an initial segment of . Again Lemma ?? tells us that is finite whereas is infinite. But since they are isomorphic this implies that , which is again a contradiction!
Hence it has to be that and are isomorphic. □