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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?

WrabbitW
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    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 $i,n\lt\omega.$ – bof Nov 04 '17 at 02:07
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    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'. – Hanul Jeon Nov 04 '17 at 11:06
  • 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 $f_n(lim \alpha_i)=lim f_n(\alpha_i)=lim f_n(\alpha_{i+1})$). – WrabbitW Nov 06 '17 at 18:08

1 Answers1

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This answer assumes the (rather weak) Choice Principle that a countable union of countable sets is again countable, where by "countable," I mean "injectable into $\omega_0$". As a result of this, we have that $\operatorname{cf}(\omega_1)=\omega_1,$ which is critical.

A key observation is the following. If $f:\omega_1\to\omega_1$ is monotone nondecreasing, then the following are equivalent:

  • The function $f$ is continuous on $\omega_1$ with respect to the order topology.
  • For all limit ordinals $\gamma<\omega_1,$ $\sup f[\gamma] = f(\sup\gamma).$

(Interestingly, the above is not true if we don't require $f$ to be monotone nondecreasing.)

Another key fact is that if $f:\beta\to\mathbf{ON}$ for some ordinal $\beta$ and $f$ is increasing, then $\alpha\leq f(\alpha)$ for all $\alpha<\beta.$

Now on to the proof.

We recursively define a sequence $\alpha:\omega_0\to\omega_1$ as follows. Let $\alpha_0=0.$ Given $k<\omega_0$ such that $\alpha_k$ has been defined, we have that $\{f_n(\alpha_k):n<\omega_0\}$ is a countable subset of $\omega_1$ (and so not cofinal in $\omega_1$), and we let $\alpha_{k+1}$ be the least ordinal in $\omega_1$ greater than all $f_n(\alpha_k).$ Then for each $k<\omega_0$ and each $n<\omega_0,$ we have $\alpha_k\leq f_n(\alpha_k)<\alpha_{k+1}.$

Since $\{\alpha_k:k<\omega_0\}$ is countable (and so not cofinal in $\omega_1$), then letting $\gamma:=\sup_{k<\omega_0}\alpha_k,$ we have $\gamma\in\omega_1.$

Fixing any $n<\omega_0,$ we have $\alpha_k\leq f_n(\alpha_k)<\alpha_{k+1}$ for each $k<\omega_0,$ so $k\mapsto\alpha_k$ is increasing, so $\alpha_k<\gamma$ for all $k<\omega_0,$ and so $\alpha_{k+1}<\gamma$ for all $k<\omega_0.$

For any $k<\omega_0$ and any $n<\omega_0,$ $$\alpha_k\leq f_n(\alpha_k)<\alpha_{k+1}<\gamma,$$ so $$\alpha_k\leq f_n(\alpha_k)<\gamma.$$ Thus, for any $n<\omega_0,$ we have $$\gamma=\sup_{k<\omega_0}\alpha_k\leq\sup_{k<\omega_0}f_n(\alpha_k)\leq\gamma,$$ and so $$\sup_{k<\omega_0}f_n(\alpha_k)=\gamma.$$

Note that that $k\mapsto\alpha_k$ is a cofinal mapping $\omega_0\hookrightarrow\gamma,$ so it follows for any $n<\omega_0$ that $$\sup f_n[\gamma]=\sup_{\beta<\gamma}f_n(\beta)=\sup_{k<\omega_0}f_n(\alpha_k)=\gamma.$$ Thus, observing that $\gamma$ is necessarily a limit ordinal, we have by continuity of $f_n$ that $$f_n(\gamma)=f_n(\sup\gamma)=\sup f_n[\gamma]=\gamma$$ for any $n<\omega_0,$ whence for all $m<\omega_0$ and $n<\omega_0,$ we have $f_m(\gamma)=f_n(\gamma),$ as desired.

Cameron Buie
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