(ZF) For every set A we denote with h*(A) the smallest ordinal α for which there is no surjective function from A to α (α is nonzero):
h∗(A) = µα[¬(α ≤∗ A)]. Prove that for every set A it is true that:
- h∗(A) ≤ α ⇒ ¬(α ≤∗ A);
- h∗(A) is a cardinal number;
- h(A) ≤ h∗(A);
- If A can be well ordered set, then it follows that h(A) = h∗(A).
Before I start, I have to say that I am a liitle bit confused of the problem itself. The reason is in the notation of h(A). In previous cases, we have named h as a function of choice for A,but here the meaning is different, so I stick to what is written above,since we are in ZF. However,
My ideas are: To use the following
Lemma: .For every set A there exists an ordinal α such that there is no surjective function from A to α: (∀A)(∃α)[¬(α ≤∗ A). /1/ Since we have picked up the smallest such ordinal, I will try to use Transfinite recursion to prove the statement. Namely, Let P(h∗(A),α) be a property defined over all ordinals α in A,satisfying the condition we want. Then:(∃ h∗(A))P(h∗(A),α,u)->(∃ h(A))(P(h(A),α,u) and for every β(β<h(A)->¬(P(h(A),β,u)). This directly proves 1). /2/ Since for every β(β<h*(A)->¬(P(h*(A),β,u)), a.k.a ¬(Limit(β)) and ord(h*(A)) => Nat(h*(A)). But this means card(h*(A)).
To prove (3) I will use the fact that h*(A) is the min cardinal for which there is no surjection. Hence for an arbitrary cardinal number h(A)<h*(A) there is an image with Dom in h(A) and Rng in A.
For (4) I think that the direct application of the Definition for well ordered set which states that : if for every subset A' of A there is a smallest element (<), then A is well ordered. But then since we have assumed that h*(A) is the min cardinal satisfying this and h(A)<h*(A), they should be equal.
I will be extremely thankful if someone help to verify the correctness of my sttempts to solve it. Thank you.