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Let $(f_n)_{n\in \omega}$ be a sequence of normal functions (increasing and continuous) from $\omega_1$ to $\omega_1$. I want to show that there is an $\alpha \in \omega_1$ such that $f_n(\alpha)=f_m(\alpha)$ for all $n,m\in\omega$ and I guess it has something to do with Fodor's Lemma but I cannot put my finger on it. Does it use that the range of normal functions are clubs? any other idea how to prove it?

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    $\begingroup$ Try taking the limit of a sequence $\alpha_0\lt\alpha_1\lt\alpha_2\lt\cdots$ of countable ordinals such that $\alpha_{i+1}\gt f_n(\alpha_i)$ for all $i,n\lt\omega.$ $\endgroup$
    – bof
    Commented Nov 4, 2017 at 2:07
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    $\begingroup$ I don't understand why there are votes to close the question. I can't even understand the reason to close the question is `This is not about mathematics'. $\endgroup$
    – Hanul Jeon
    Commented Nov 4, 2017 at 11:06
  • $\begingroup$ IfI got you right @bof after building such sequence (you can find the next term each time because $cf \omega_1=\omega_1$) you proveusing both continuity and monotony ofthe functions that the limit is the fixed point we are looking for? (using that $f_n(lim \alpha_i)=lim f_n(\alpha_i)=lim f_n(\alpha_{i+1})$). $\endgroup$
    – WrabbitW
    Commented Nov 6, 2017 at 18:08

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