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I’m working on the following exercise from Kunen:

Define, in ZF without the axiom of regularity, $\aleph(X)=\{\alpha:

> \exists f \in \, ^\alpha X(f \text{ is } 1-1\}$. Show:

- $\aleph(X)<\aleph(\mathscr P(\mathscr P(\mathscr P(X))))$
- There is no sequence $\langle X_n: n \in \omega\rangle$ such that $\forall n(\mathscr P(X_{n+1})\preceq X_n)$.
- The axiom of choice implies that $\aleph(X)=|X|^+$ whenever $X$ is infinite.

I have already proved 1. and 3., but I don’t know what to do with 2. Thinking without AC is hard :p

- Why does $\omega$ have the same cardinality in every (transitive) model of ZF?
- How can I construct a sequence of injections $\langle f_\alpha\colon\alpha\to\omega\mid\alpha<\omega_1\rangle$ with a particular coherence property?
- Axiomatic Set Theory (ZFC): Intersection
- What are some natural arithmetical statements independent of ZFC?
- Finite choice without AC
- Why does the set of all singleton sets not exist?

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- Defining cardinality in the absence of choice
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- Is there any generalization of the hyperarithmetical hierarchy using the analytical hierarchy to formulas belonging to third-order logic and above?
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- How many positive numbers need to be added together to ensure that the sum is infinite?
- Formulations of Singular Cardinals Hypothesis
- Where does TREE(n) sit on the Fast Growing Hierarchy?

**HINT:** Use the first part to show that from such sequence you get a decreasing sequence of ordinals.

Let $\kappa_n=\aleph(X_n)$, then you can note that $\kappa_{n+3}<\kappa_n$ by that first part, and the trivial observation that $X\preceq Y\implies\aleph(X)\leq\aleph(Y)$.

First, notice that if $X \preceq Y$, then $\aleph(X) \leq \aleph(Y)$: Since $\aleph(Y) \not \preceq Y$, it follows that $\aleph(Y) \not \preceq X$. Since $\aleph (X)$ is the smallest ordinal $\alpha$ satisfying $\alpha \not \preceq X$, it follows that $\aleph(X) \leq \aleph(Y)$

Now, suppose such a sequence does exists. given $n \in \omega$, notice that $\aleph(X_n) \geq \aleph(\mathscr P(X_{n+1}))\geq\aleph(\mathscr P^2(X_{n+2}))\geq \aleph(\mathscr P^3(X_{n+3}))>\aleph(X_{n+3})$. So the sequence $\langle \aleph(X_{3n}): n \in \omega\rangle$ is a strictly decreasing sequence.

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