I need to show: there doesnotexist a sequence:

$\displaystyle \left\langle X_n:n\in \omega \right\rangle$ such that $\displaystyle (\forall n)(\mathcal{P}(X_{n+1})\preceq X_n)$.

Supposing there exists such a sequence:

We can observe that for any $\displaystyle n\in\omega: X_n \prec \mathcal{P}(X_n)\preceq X_{n-1}\prec \cdots \prec \mathcal{P}(X_1)\preceq X_0$. Thus we have $\displaystyle X_n \prec X_k$ for all $\displaystyle 0\leq k < n$. In particular, for any $\displaystyle n$ we have $\displaystyle \mathcal{P}^n(X_n)\preceq X_0$.

In a previous excercise I had to show that (without the axiom of choice): $\displaystyle \aleph(X) \preceq \mathcal{P}(\mathcal{P}(\mathcal{P}(X)))$. And since $\displaystyle \aleph(X)$ is the smallest ordinal $\displaystyle \alpha$ such that $\displaystyle \alpha\not\preceq X$ we obtain: $\displaystyle \aleph(X)\prec \aleph(\mathcal{P}(\mathcal{P}(\mathcal{P}(X))))$.

From this I conclude we get a strictly decreasing sequence Aleph's: Since $\displaystyle \aleph(X_n) \preceq \mathcal{P}(\mathcal{P}(\mathcal{P}(X_n)))\preceq X_{n-3}\prec \aleph(X_{n-3})$. Thus the subsequence $\displaystyle \left\langle X_{3n}:n\in \omega\right\rangle$ is a sequence such that $\displaystyle \aleph(X_n) \prec \aleph(X_{n-1})$ and $\displaystyle \aleph(X_0) = \bigcup_{n\in\omega} \aleph(X_n)$.

Those were some of my conclusions: However I don't see where the contradiction kicks in. Can anyone offer me some suggestions? Probably there can not exist a infinite decreasing sequence of ordinals, intuitively that might sound logical but I can't proof it.

Since the set of ordinals are wellordered, our sequence: $\displaystyle \left\langle \aleph(X_n):n\in \omega \right\rangle$ should contain a smallest ordinal. But can't we simply define that one as $\displaystyle \bigcap_{n\in \omega} \aleph(X_n)$?