Exercise 6.2.6

An ordinal α is a natural number if and only if every nonempty subset of α has a greatest element.

Answers

Proof. ( ) Suppose that n is a natural number and consider any nonempty subset A of n . Since A n it follows that | A | | n | = n so that A is finite. Thus A is a finite set of natural numbers and so has a greatest element. This can be proven by a trivial inductive argument.

( ) We show this by contrapositive. Suppose that α is an ordinal such that α 𝑵 . Then α ω = 𝑵 so that α ω , from which it follows that α ω Hence α = ω or α > ω , in which case ω α so that ω α since α is transitive. Thus in either case 𝑵 = ω α . Clearly 𝑵 has no greatest element (since if n were such a greatest element then n + 1 𝑵 but n < n + 1 ). Thus there is a nonempty subset A of α such that A has no greatest element. □

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2024-07-15 11:42
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