In this question, the notation $P^x(\alpha)$ denotes a situation where a particular OTM-program $P$ performs a computation on input $x$ with an ordinal parameter $\alpha$, assuming that $x$ is written on the initial segment of length $\omega$ (the smallest limit ordinal) of the tape of $P$ at time $0$. That is, $x$ is the input for $P$ written in cells indexed by finite ordinals $(0, 1, 2, \ldots)$ before the start of computation, yet all cells indexed by all ordinals greater than or equal to $\omega$ are initially blank, except one cell indexed by $\alpha$ (this cell is marked by a non-zero symbol.)

Let $\beta$ denote the smallest ordinal such that for any pair of an OTM-program $P$ and a real $x$ (that is, $P$ quantifies over all programs and $x$ quantifies over all reals) exactly one of the following statements is true:

  1. There does not exist an (uncountable or countable) ordinal $\alpha$ such that $P^x(\alpha)$ halts;

  2. If there exists at least one (uncountable or countable) ordinal $\alpha$ such that $P^x(\alpha)$ halts, then, assuming that $\alpha_0$ is the smallest such ordinal, $\alpha_0 < \beta.$

How large is $\beta$?

  • $\begingroup$ So your question is what is the least non-OTM-computable with a real parameter, am I correct? $\endgroup$
    – Hanul Jeon
    Sep 19, 2020 at 8:12
  • $\begingroup$ @HanulJeon I didn't understand the question. Regarding what you wrote (just to be sure) did you mean: "what is the least non-OTM-computable ordinal with any arbitrary real parameter allowed"? Admittedly, I don't understand the second part of your answer though (w.r.t. upper-bound). For V=L, my personal reasoning goes as follows: Given any arbitrary ordinal parameters less than a countable $\alpha$ the sup of values clocked (with parameters $< \alpha$) can be shown to be countable. And hence the upper-bound follows (because any real can be computed with some countable ordinal parameter). $\endgroup$
    – SSequence
    Sep 19, 2020 at 10:48
  • $\begingroup$ @SSequence: "I didn't understand the question" — can you please specify which part of the question is unclear? $\endgroup$ Sep 19, 2020 at 11:07
  • $\begingroup$ @SSequence: "You could just have $\omega$ as a parameter and a certain machine would halt regardless of what real input was placed on it" — yes, of course, a particular program $P_1$ will halt. But a particular program $P_2$ will not (with the same input.) I have emphasized that we take into account all programs, all inputs and all ordinal parameters (assuming that the parameter is minimal, as is written in the question). $\endgroup$ Sep 19, 2020 at 11:44
  • $\begingroup$ @SSequence: [1/2]: yes, of course, but these facts do not affect the definition of $\beta$ at all. Consider the following game. I pick an arbitrary ordinal $\tau_0$. You pick an arbitrary OTM $P$ and an arbitrary real $x$, then write $x$ on the cells indexed by finite ordinals. $\endgroup$ Sep 19, 2020 at 12:13

1 Answer 1


Since there is disputation on how to interpret the problem, I think it would be better to clarify my interpretation:

Let $P(x,\alpha)$ be a program, which takes a binary sequence $x\in 2^\mathbb{N}$ (also called a real, which is standard terminology in set theory) and an ordinal $\alpha$. Consider the set $$H = \{\alpha\mid \text{$\alpha$ is the least ordinal such that $P(x,\alpha) $ halts for some $x$, $P$} \}.$$ Then $H$ is a set. What is the value of $\sup H$?

If I understand your problem correctly, then the answer is $\omega_1$. Please feel free to comment if there is an error in my proof.

For the lower bound, we will find an OTM-program with a parameter $x\in 2^\mathbb{N}$ that computes a countable ordinal. Assume that $x$ codes a well-order over $\omega$ whose order-type is $\alpha$. Consider the following procedure: decode $x$ and enumerate ordinals less than the order-type of $x$ by brute force. (This is possible since there are only countably many members in $x$ and we have infinite time.) In this way, we can compute $\alpha$ from $x$. Now take $P(\beta)$ as follows: if $\beta=\alpha$, it halts. If not, it does not halt.

For the upper bound, assume that we have a program $P$ of real parameter $x$. By Lemma 2.6 of Koepke's Turing Computations on Ordinals, the ordinal computation by $P$ is absolute between $V$ and $L[x]$. Assume that $P$ halts with an input $\alpha_0$, and $\alpha_0$ is the smallest such an ordinal. Moreover, assume that we take time $\theta$ to compute $P(\alpha_0)$.

Now consider the Skolem hull $M$ of sufficiently large $L_\gamma[x]$ generated by $\{\theta,\alpha_0,x\}$. By condensation, there is an isomorphism $\pi:M\to L_\beta[x]$ for some countable $\beta$. Then $L_\beta[x]$ thinks $P$ halts with an input $\pi(\alpha_0)$ and does not halt if we plug in ordinals smaller than $\pi(\alpha_0)$. By $\pi(\alpha_0)\le \alpha_0$, Lemma 2.6 of Koepke and minimality of $\alpha_0$, we have $\pi(\alpha_0)=\alpha_0$. Hence $\alpha_0$ is countable.

  • $\begingroup$ [1/2]: I am not sure what you mean by "real parameter $x$" and why you wrote "$x$ codes a well-order" and "$P$ halts with an input $\alpha_0$", so in order to avoid ambiguity, I have to clarify that $x$ is not a parameter, $x$ is an input: it is an arbitrary real (the word "real" here implies an infinite binary sequence) written on all cells indexed by natural numbers. Note that since $x$ is arbitrary, it is not required to code a well-order. $\endgroup$ Sep 19, 2020 at 10:31
  • $\begingroup$ [2/2]: Then $\alpha_0$ is not an input, it is always the smallest ordinal greater than or equal to $\omega$ such that a program $P$ halts given $x$ as the input and a single "1" written on the $\alpha_0$-th cell of the tape. Do these clarifications affect the answer? $\endgroup$ Sep 19, 2020 at 10:32
  • $\begingroup$ @lyricallywicked The second half of your question is difficult to understand. You don't seem to specify any relation between $\alpha$ and $\beta$ in your possibilities-(1),(2) in the second half of your question. The answer given is for the following question: "what is the least ordinal not reachable by any OTM (with no ordinal parameters and any arbitrary real input allowed)". $\endgroup$
    – SSequence
    Sep 19, 2020 at 10:59
  • $\begingroup$ @SSequence: [1/2] "You don't seem to specify any relation between $\alpha$ and $\beta$ in your possibilities-(1),(2)" — why? I have provided the full definition. If there exists at least one ordinal parameter such that an arbitrary OTM halts with $\alpha$ as the parameter and $x$ as the input, then there is the smallest such parameter, denoted by $\alpha_0$. Then $\beta$ is the smallest ordinal greater than any $\alpha_0$ under the assumption that $x$ is an arbitrary real. $\endgroup$ Sep 19, 2020 at 11:21
  • $\begingroup$ @SSequence: [2/2] Regarding [The answer given is for the following question: "what is the least ordinal not reachable by any OTM (with no ordinal parameters and any arbitrary real input allowed)"] — if this is so, then no, this is not what this question is about. Definition of $\beta$ is not related to being (non-)reachable, (non-)computable or (non-)writable. It is related to ordinal parameters. Note that ordinal parameters for OTMs may be uncountable (for example, $0^{\sharp}$ is recognizable from $\omega_1$, if $0^{\sharp}$ exists, but it is not recognizable from any countable ordinal.) $\endgroup$ Sep 19, 2020 at 11:27

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