Exercise 6.1.6

Show that the lexicographic product ( 𝑵 × 𝑵 , < ) (see Lemma 4.6 in Chapter 4) is isomorphic to ω ω .

Answers

Proof. Suppose that is the lexicographic ordering of 𝑵 × 𝑵 . Now we define f : 𝑵 × 𝑵 ω ω by

f ( n , m ) = ω n + m

for any ( n , m ) 𝑵 × 𝑵 . Clearly f ( n , m ) ω × ω .

First we show that f is surjective. So consider any k ω ω so that there are n , m 𝑵 where k = ω n + m . Then we clearly have that f ( n , m ) = ω n + m = k . Since clearly ( n , m ) 𝑵 × 𝑵 it follows that f is surjective.

Now we show that f is an increasing function. To this end consider any ( n 1 , m 1 ) , ( n 2 , m 2 ) 𝑵 × 𝑵 where ( n 1 , m 1 ) ( n 2 , m 2 ) .

Case: n 1 = n 2 . Then since ( n 1 , m 1 ) ( n 2 , m 2 ) it must be that m 1 < m 2 . Hence we have that f ( n 1 , m 1 ) = ω n 1 + m 1 = ω n 2 + m 1 < ω n 2 + m 2 = f ( n 2 , m 2 ) .

Case: n 1 n 2 . Then since ( n 1 , m 1 ) ( n 2 , m 2 ) it must be that n 1 < n 2 . Hence we have that f ( n 1 , m 1 ) = ω n 1 + m 1 < ω n 2 ω n 2 + m 2 = f ( n 2 , m 2 ) .

Thus in all cases f ( n 1 , m 1 ) < f ( n 2 , m 2 ) so that f is increasing. It then follows that f is injective and isomorphic. Hence ( 𝑵 × 𝑵 , ) is isomorphic to ω ω . □

User profile picture
2024-07-15 11:42
Comments