Exercise 6.1.7

Let ( W , < ) be a well-ordered set, and let a W . Extend < to W = W { a } by making a greater than all x W . Then W has a smaller order type than W .

Answers

This problem is looking ahead to future sections where order types and how to compare them are defined.

Proof. Suppose that α is the order type of W and that f : W α is the isomorphism. Now let β = S ( α ) = α { α } . We then claim that W is isomorphic to β . So define a g : W β by

g ( w ) = { f ( w ) w W α w W

for w W . Clearly we have that g ( w ) β for any w W .

Now consider any x β . If x = α then set w = a W so that g ( w ) = α = x . If x α then x α so set w = f 1 ( x ) so that then w W . We then have that g ( w ) = f ( w ) = f ( f 1 ( x ) ) = x . Therefore g is surjective.

Now consider any w 1 , w 2 W where w 1 < w 2

Case: w 1 , w 2 W . Then since f is an isomorphism and w 1 < w 2 we have that g ( w 1 ) = f ( w 1 ) < f ( w 2 ) = g ( w 2 ) .

Case: w 1 W and w 2 = a . Then g ( w 1 ) = f ( w 1 ) α and w 2 W so that g ( w 2 ) = α . Hence g ( w 1 ) g ( w 2 ) so that by the definition of < we have that g ( w 1 ) < g ( w 2 ) .

Note that these cases are exhaustive since it can’t be that w 1 = w 2 = α since w 1 < w 2 (and therefore w 1 w 2 ). It also cannot be that w 2 W but w 1 = a since then it would be that w 2 w 1 since a is the greatest element of W , which contradicts w 1 < w 2 . Thus in all cases g ( w 1 ) < g ( w 2 ) so that g is increasing, and therefore injective and an isomorphism.

Hence β is the order type of W , α is the order type of W , and α < β since α β . □

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2024-07-15 11:42
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