Reference no: EM133281284

Question 1: Why is the ordering < of (2) in Example 140 a total ordering?

Question 2: We'll work with the natural numbers, w, but rather than using their natural ordering, we define some alternative well-orderings, which we denote by <.
(1) 1, 2, 3, ...0;
(2) 0, 2, 4, ..., 1,3,5,

The idea with (1) is that we put 0 at the end of the ordering. So for every n such that n ≠ 0, we have n < 0 and we keep the ordinary ordering for the rest.

To verify that (2) gives us a well-ordering, let X ⊆ ω. Then either X contains an even number or it doesn't. Suppose it does. Then there is some <-least even number that it contains, say m. Then is clear that m is the <-least even number; and m is < all of the odd numbers by definition. Suppose X doesn't contain any even numbers; i.e., it only contains odd numbers. Then there is a <-least odd odd number, n, in X which is also the <-least odd number. And since the odd numbers are all > the even numbers, n is the <-least of any of the numbers in X. Thus in either, case X has a <-minimal element. Hence, < is well-founded.

Question 3: Provide an example of a bijective homomorphism that is not an isomorphism (you'll need to use relation symbols). Show that a surjective order preserving map between two (strict) total orderings is an isomorphism.

Question 4: Show that if α and β are ordinals, then a α ∩ β is an ordinal.

Question 5: Show that x is transitive if x ⊆ P(x).