Cardinal numbers and Well ordered Set (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.
 A: I dislike being so blunt, but so far as I can tell, you are completely lost. Your argument for (1) makes no sense: (1) is not a statement about ordinals in the set $A$. And even if it were, it is not at all clear what $P(h^*(A),\alpha)$ is intended to be. Your argument for (2) is completely incomprehensible. In your argument for (3) you seem to be trying to prove something about ordinals that are known to be less than $h^*(A)$, when in fact you are supposed to be proving that a certain ordinal is less than $h^*(A)$. And your argument for (4) is simply incoherent.
I’m going to do the first three and let you take another crack at the fourth one; spaceisdarkgreen’s comment should at least point you in the right general direction.
(1) Suppose that $h^*(A)\le\alpha$; then the function
$$f:\alpha\to h^*(A):\xi\mapsto\begin{cases}
\xi,&\text{if }\xi<h^*(A)\\
0,&\text{otherwise}
\end{cases}$$
is a surjection. If $g:A\to\alpha$ is a surjection, $f\circ g$ is a surjection from $A$ onto $h^*(A)$, contradicting the definition of $h^*(A)$, so there is no surjection of $A$ onto $\alpha$, and therefore $\alpha\not\le^* A$.
(2) Since $h^*(A)$ is an ordinal, there is a cardinal $\kappa\le h^*(A)$ such that there is a bijection $f:\kappa\to h^*(A)$. If $g:A\to\kappa$ is a surjection, then $f\circ g$ is a surjection of $A$ onto $h^*(A)$, which is impossible. Thus, there is no surjection from $A$ onto $\kappa$. By definition $h^*(A)$ is the smallest ordinal that is not a surjective image of $A$, so $h^*(A)\le\kappa$, and hence $h^*(A)=\kappa$ is a cardinal.
(3) Suppose that $h(A)>h^*(A)$; $h(A)$ is the smallest ordinal $\alpha$ such that there is no injection from $\alpha$ into $A$, so there is an injection $f:h^*(A)\to A$. Let $R=\operatorname{ran}f$, so that $f^{-1}$ is a bijection from $R$ to $h^*(A)$. Then the function
$$g:A\to h^*(A):\xi\mapsto\begin{cases}
f^{-1}(\xi),&\text{if }\xi\in R\\
0,&\text{otherwise}
\end{cases}$$
is a surjection from $A$ onto $h^*(A)$, which is impossible, and therefore $h(A)\le h^*(A)$. (I am assuming here that $h$ is the Hartogs function; it certainly isn’t a choice function.)
