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Math Help - Decreasing sequence Aleph's

  1. #1
    Senior Member Dinkydoe's Avatar
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    Lightbulb Decreasing sequence Aleph's

    I need to show: there does not exist a sequence:

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

    Supposing there exists such a sequence:
    We can observe that for any 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 X_n \prec X_k for all 0\leq k < n. In particular, for any n we have \mathcal{P}^n(X_n)\preceq X_0.

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

    From this I conclude we get a strictly decreasing sequence Aleph's: Since \aleph(X_n) \preceq \mathcal{P}(\mathcal{P}(\mathcal{P}(X_n)))\preceq X_{n-3}\prec \aleph(X_{n-3}). Thus the subsequence \left\langle X_{3n}:n\in \omega\right\rangle is a sequence such that \aleph(X_n) \prec \aleph(X_{n-1}) and \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: \left\langle \aleph(X_n):n\in \omega \right\rangle should contain a smallest ordinal. But can't we simply define that one as \bigcap_{n\in \omega} \aleph(X_n)?
    Last edited by Dinkydoe; December 24th 2009 at 02:11 PM.
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  2. #2
    Senior Member
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    Hi

    Yep you can find a contradiction when considering a strictly decreasing sequence of ordinals:

    Assume (\alpha_n)_{n\in\omega} is a sequence of ordinals such that \alpha_n>\alpha_{n+1} for any n\in\omega.

    What about X=\bigcup\{\alpha_n\ ;\ n\in\omega\} ?
    Take x\in X, there is a k\in\omega such that x=\alpha_k and \alpha_{k+1} is a strictly lower element of X:\ X, a set of ordinals, has no minimum. Contradiction.
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