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Math Help - Regressive problem with ordinals regarding countability

  1. #1
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    Unanswered - Regressive problem with ordinals regarding countability

    I've spent a lot of time thinking about this problem, but I'm running out of time and need a skeleton of the process to solve it if someone wants to tackle it...

    Prove:
    If  f: \omega_{1} \rightarrow \omega_{1} \wedge \forall \alpha \in \omega_{1} (f(\alpha)<\alpha \vee f(\alpha)=0) \exists \alpha_{0} \in \omega_{1} such that the set  \{ \alpha \in \omega_{1}: f(\alpha)= \alpha_{0} \} is uncountable.

    I've been trying to do this by finding a counterexample to the following sentence

    let  f: \omega_{1} \rightarrow \omega_{1} \wedge \forall \alpha \in \omega_{1} (f(\alpha)<\alpha \vee f(\alpha)=0), \Rightarrow \forall \alpha_{0} \in \omega_{1} the set  \{ \alpha \in \omega_{1}: f(\alpha)= \alpha_{0} \} is countable.

    If I can find a single counterexample, then the original problem is true. I've looked at unions to see if I can get high enough in the ordinals so that uncountably many things drop to \alpha when applying the function, but that would mean \forall \gamma \in \omega_{1} \exists \alpha \in \omega_{1} (\cup \gamma = \alpha, \alpha < \gamma) is true. Which, according to my professor, it is not.

    Thus, I need help.... please.
    Last edited by tcRom; November 16th 2009 at 07:46 PM.
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  2. #2
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    is omega a subset of the reals? or just any set. Are there any other conditions on f?
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  3. #3
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    In this case, \omega_{1} is the first uncountable ordinal. f is just a function that maps  f:\omega_{1} \rightarrow \omega_{1} where the output is less than the input or equal to 0.

    For example, if we were using the naturals the following could be possible:
    f(0) = 0
    f(1) = 0
    f(2) = 0 or 1
    f(3) = 0 or 1 or 2
    .
    :
    f(n) = anything less than n

    So, f is not actually defined in the problem, that's what I am trying to do; define f so that it maps an uncountable number of things to \alpha_{0}:\alpha_{0} \in \omega_{1}. I hope that clarifies things.
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  4. #4
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    I'm pretty sure to prove that statement you have to show that any function that satisfies the criteria will have a a_0 so that the preimage of a_0 is uncountable. I'll think about this and see if I can come up with anything, my intuition says that to show uncountablility you may have to use cantor's trick: Cantor's diagonal argument - Wikipedia, the free encyclopedia this is just my intuition though, I'll think about it.
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  5. #5
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    What about doing this with limit ordinals? Say we choose an \alpha_{0} and look at the limit of it's inverse image, then we do the same for everything in \alpha_{0}. So, the next thing we choose above \alpha_{0} must come from above the highest limit ordinal that anything in \alpha_{0} mapped to. If we keep going with this, we'll eventually find a \beta that maps to itself, a contradiction. Thus, the assumption that everything in our set is countable is false, making the original statement true.

    Any thoughts?
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