Does there exist a strictly increasing and continuous function of ordinal numbers whose smallest critical number (i.e. fixed point) is the smallest non-constructive ordinal number (in the sense of Church and Kleene)? If so, does there exist such a function all of whose critical numbers (i.e. fixed points) are non-constructive and admissible ordinal numbers?
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1$\begingroup$ Doesn't $f(\alpha)=$ first admissible past $\omega+\alpha$ work? $\endgroup$– Andrés E. CaicedoDec 12, 2013 at 19:43
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$\begingroup$ Andres, I take strictly increasing to mean: $\alpha\lt\beta\implies f(\alpha)<f(\beta)$, but your function is constant on intervals. $\endgroup$– Joel David HamkinsDec 12, 2013 at 19:45
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$\begingroup$ @JoelDavidHamkins Ah, strictly, yes. Thanks. So, let's take $f(\alpha)=\alpha$-th admissible ordinal past $\omega$. $\endgroup$– Andrés E. CaicedoDec 12, 2013 at 19:59
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$\begingroup$ But that will jump past $\omega_1^{CK}$, which I thought was to be the fixed point. $\endgroup$– Joel David HamkinsDec 12, 2013 at 20:04
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3$\begingroup$ It’s an easy exercise to show that every ordinal of the form $\omega^\alpha$ with countable cofinality is the least fixed point of some strictly increasing continuous function. Then extend it as Andres says. This does not really have anything to do with admissibility. $\endgroup$– Emil JeřábekDec 12, 2013 at 20:15
1 Answer
Any admissible ordinal is the limit of indecomposable ordinals. Pick an $\omega$-sequence of these ordinals, $\alpha_0<\alpha_1<\dots$ with limit $\omega_1^{CK}$, so the order type of $[\alpha_i,\alpha_{i+1}]$ is $\alpha_{i+1}$. Now define $f$ on $[0,\alpha_0]$ by $f(\beta)=\alpha_0+1+\beta$, on $(\alpha_0,\alpha_1]$ by $f(\beta)=\alpha_1+\beta$, on $(\alpha_1,\alpha_2]$ by $f(\beta)=\alpha_2+\beta$, etc. Then define for $\beta\ge\omega_1^{CK}$, $f(\beta)=\beta$. This $f$ is normal, and has $\omega_1^{CK}$ as its first fixed point. Note that the fact that $\omega_1^{CK}$ is admissible is irrelevant here; in fact, all that matters here is that it has countable cofinality, and is indecomposable.
If instead we want $f(\alpha)$ to be admissible for each $\alpha$, pick a club through the admissible ordinals, and let $f$ be its increasing enumeration. Suppose for instance that $\kappa$ is strongly inaccessible, and note that then there is a club of $\alpha<\kappa$ such that $V_\alpha\prec V_\kappa$, but each such $\alpha$ is of course admissible. We can easily relax the assumptions, to turn this into an argument that there is indeed, provably in $\mathsf{ZF}$, a (proper class) club of admissible ordinals. Note, on the other hand, that a limit of admissible ordinals need not be admissible, so some care is needed.
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$\begingroup$ You don’t need the individual $\alpha_i$ to be indecomposable. Thus defined $f$ is always strictly increasing and continuous, so you only need to ensure that it takes values below the limit, i.e., that the limit is indecomposable. $\endgroup$ Dec 12, 2013 at 20:54
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$\begingroup$ Yes, all I need is that the intervals $(\alpha_i,\alpha_{i+1})$ have length larger than $\alpha_i$. $\endgroup$ Dec 12, 2013 at 20:57
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$\begingroup$ Not even that. The last value assigned in the $i$th interval is $\alpha_i+\alpha_i$, and the first value assigned in the next one is $\alpha_{i+1}+\alpha_i+1\ge\alpha_i+\alpha_i+1>\alpha_i+\alpha_i$. Or to put it simply, $f(\beta)$ is a sum of two numbers, the first of which is nondecreasing, and the second is strictly increasing, hence the result is strictly increasing. $\endgroup$ Dec 12, 2013 at 21:06
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$\begingroup$ Yes, you are right. All that matters is that $\omega_1^{CK}$ itself is (countable and) indecomposable. $\endgroup$ Dec 12, 2013 at 21:08
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$\begingroup$ Thanks alot for all this information. I asked these questions in order to get a clearer intuitive picture of $\endgroup$ Dec 13, 2013 at 20:32