Exercise 6.1.4

For every infinite subset A of 𝑵 , ( A , < ) is isomorphic to ( 𝑵 , < ) .

Answers

Proof. Let A be an infinite subset of 𝑵 . Then ( A , < ) (where < is the standard well-ordering of 𝑵 ) is a well-ordering since any B A is also a subset of 𝑵 and therefore has a least element. Hence by Theorem 6.1.3 either:

1.
( A , < ) and ( 𝑵 , < ) are isomorphic,
2.
An initial segment of ( A , < ) is isomorphic to ( 𝑵 , < ) , or
3.
( A , < ) is isomorphic to an initial segment of ( 𝑵 , < )

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 S of ( A , < ) is isomorphic to ( 𝑵 , < ) . Then since A 𝑵 it follows from Lemma ?? that S is finite. But since this is isomorphic to 𝑵 it means that | S | = | 𝑵 | = 0 , which is a contradiction!

Now suppose (3) so that ( A , < ) is isomorphic to an initial segment S of ( 𝑵 , < ) . Again Lemma ?? tells us that S is finite whereas A is infinite. But since they are isomorphic this implies that | S | = | A | = 0 , which is again a contradiction!

Hence it has to be that ( A , < ) and ( 𝑵 , < ) are isomorphic. □

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2024-07-15 11:42
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