There is no sequence $X_n$ such that $\forall n(\mathscr P(X_{n+1})\preceq X_n)$.

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:

  1. $\aleph(X)<\aleph(\mathscr P(\mathscr P(\mathscr P(X))))$
  2. There is no sequence $\langle X_n: n \in \omega\rangle$ such that $\forall n(\mathscr P(X_{n+1})\preceq X_n)$.
  3. 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

Solutions Collecting From Web of "There is no sequence $X_n$ such that $\forall n(\mathscr P(X_{n+1})\preceq X_n)$."

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.