1. Infinite exponents
Last lecture we showed that for any Later, we will see that for any and any there is such that On the other hand (recall we are assuming choice), there are no nontrivial positive partition relations with infinite.
(To avoid vacuously true statements, we are always assuming implicitly that requires for at least one )
Proof: It is enough to show that In effect, if holds and we may assume that is regular, so any countable subset of is bounded, and the ordinal is still strictly below
For let denote the subset of consisting of its first many members.
Now, given let be the function If is homogeneous for and of size then is homogeneous for In effect, let so If then by homogeneity of
This shows that for infinite implies that there is some such that so it is enough to show that the latter never holds. In fact, we cannot even have (Recall our convention that denotes the order type and the size.)
For this, let be an arbitrary infinite cardinal, and let be a well-ordering of Define by iff is the -least member of and otherwise.
If is homogeneous for and is the -least member of then so is -homogeneous. Now consider any infinite sequence of infinite subsets of Since for all we must have that contradicting that is a well-order.
In the presence of the axiom of choice, Theorem 1 completely settles the question of what infinite exponent partition relations hold. Without choice, the situation is different.