I was reading about the fixed-point lemma for normal functions and I was wondering if we could weaken strict monotonicity to non-decreasing. I see why it's necessary for the proof, but is there a continuous non-decreasing ordinal operation without a fixed point? I'm probably missing something obvious but I can't think of any.
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asked Mar 31, 2018 at 22:07
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2 Answers
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You don't need that $f$ is strictly increasing to have at least one fixed point. Consider the sequence $\alpha_n = f^n(0)$. By induction it follows that this is non-decreasing. If $\alpha_n=\alpha_{n+1} $ for some $n$ then we are done as this means that $\alpha_n = f(\alpha_n)$. Else consider the sup of this sequence and follow by continuity that this is a fixed point.
Of course you cannot get arbitrarily large fixed points as witnessed by the constant $0$ function.
answered Apr 1, 2018 at 10:08
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With $\ell(\alpha)=\min\{\,\lambda\mid \lambda\text{ is limit ordinal}, \lambda>\alpha\,\}$, let $$ f(\alpha)=\begin{cases}\ell(\alpha)&\text{if }\alpha\text{ is limit ordinal}&\\\ell(\ell(\alpha))&\text{otherwise}\end{cases}$$ Then $f$ is continuous, non-decreasing, and has no fixed point.
answered Mar 31, 2018 at 22:21
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1$\begingroup$ Why is this continuous? It looks discontinuous at $\omega\cdot\omega$, or am I missing something? $\endgroup$ Apr 1, 2018 at 9:45
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$\begingroup$ @Jonathan: It actually seems discontinuous at every limit ordinal, for example every finite $\alpha$ gives $f(\alpha)=\omega$, but $f(\omega)=\omega+\omega$. $\endgroup$– Asaf Karagila ♦Apr 1, 2018 at 10:19
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$\begingroup$ @Asaf finite successors $\alpha$ give $\omega +\omega$, the next next limit. $\endgroup$ Apr 1, 2018 at 13:12
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$\begingroup$ @Jonathan: Ah, you're right. I misread. Thanks. $\endgroup$– Asaf Karagila ♦Apr 1, 2018 at 13:13