Let $\theta_\alpha$ be the smallest ordinal such that there is no first-order formula $\phi$ such that:

$$\exists\beta_0,\beta_1...\beta_n\in\alpha\forall x(\phi(x,\theta_{\beta_0},\theta_{\beta_1}...\theta_{\beta_n})\Leftrightarrow x=\theta_\alpha)$$

Unless $\alpha=0$, in which case $\theta_0$ is the smallest ordinal with no first-order formula $\phi$ such that:

$$\forall x(\phi(x)\Leftrightarrow x=\theta_0)$$

With this definition, $\alpha\leq\theta_\alpha<|\alpha|^+$ (and thus $|\theta_\alpha|=|\alpha|$). These two arguments are shown separately from one another. ($\alpha\leq\theta_\alpha$ is because $\theta_\alpha$ is greater than or equal to a certain normal function at every point, and $\theta_\alpha<|\alpha|$ because for the set of all first-order formulas $\phi(v)$, which has cardinality $\aleph_0$, it can be shown that $\theta_\alpha$ has an injection onto $\aleph_0\cdot{}^{<\omega}|\alpha|$.)

However, because $\theta$ itself is not a normal function, it is currently unknown whether or not $\theta$ has any fixed points at all. If $\theta_\kappa=\kappa$ for an initial ordinal $\kappa$, then $\kappa$ is an $\aleph$-fixed point, a fixed point of the enumeration of those, etc.

It is also known (using a simple argument from Klev's $\mathcal{O}^{++}$) that $\theta_0>\zeta$ where $\zeta$ is the supremum of all eventually writable ordinals. This should give one an idea of how large these ordinals are. In fact, $\theta_0$ is the supremum of all first-order defined ordinal notation supremums. This information is not very pertinent to the question at hand, however.

Let $\Theta_0$ be the smallest $\kappa$ such that $\theta_\kappa=\kappa$. If $\Theta_0$ is inaccessible, it is the $\Theta_0$-th inaccessible. If $\Theta_0$ is Mahlo, it is the $\Theta_0$-th Mahlo. If $\Theta_0$ is Measurable, it is the $\Theta_0$-th Measurable. In general, if $\phi(\Theta_0)$ holds for a first-order formula $\phi$, then $\Theta_0$ is the $\Theta_0$-th ordinal such that $\phi(\kappa)$ holds.

My question is this: Are there any initial ordinals $\kappa$ such that $\theta_\kappa=\kappa$? If so, what is the minimum of them? What axiomatic requirements are necessary for such a cardinal (if any other than ZF)?

A bonus question which is completely unnecessary and possibly more difficult is generalizing this to $n$-th order logic, $\Sigma_n$, $\Pi_n$, or $\Delta_n$ formulas.

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    $\begingroup$ Do you restrict the complexity of $\phi$ in some way? Otherwise, it is possible that all ordinals are definable without parameters. $\endgroup$
    – Yair Hayut
    Oct 11, 2017 at 8:52
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    $\begingroup$ Following up on Yair's comment: arxiv.org/abs/1105.4597 $\endgroup$ Oct 11, 2017 at 13:29
  • $\begingroup$ @Yair I restrict $\phi$ to be finite and first-order. I also restrict you to only look at "Tall" models (models which contain every ordinal in $V$) $\endgroup$
    – Zetapology
    Oct 11, 2017 at 14:30
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    $\begingroup$ I feel like this is quite like your previous question mathoverflow.net/q/281377/1058. As was explained there, by Yair, it is not clear without further assumptions that even $\theta_0$ exists, so could you spell out exactly what you have in mind? Do you want your models to have access to a truth predicate, for example? $\endgroup$ Oct 11, 2017 at 16:27
  • $\begingroup$ I would simply like to see if any models of ZFC have $\theta_0$ and if so what axioms are required to restrict ZFC to said models. $\endgroup$
    – Zetapology
    Oct 11, 2017 at 17:03

1 Answer 1


Under reasonable assumptions (to avoid such things as pointwise-definable models), there is indeed a $\kappa$ with $\theta_\kappa=\kappa$ and $cf(\kappa)=\omega$.

To see this, let $M(\alpha)$ be the supremum of the ordinals definable with parameters $<\alpha$. Note that $M(\alpha)$ is not necessarily defined in an arbitrary model of ZFC - e.g. in a pointwise definable model, $M(0)$ already doesn't exist. Now we define a sequence of ordinals as follows:

  • $\beta_0=0$.

  • $\beta_{n+1}=\vert M(\theta_{\beta_n})\vert^+$.

Let $\kappa=\sup\beta_i$. Clearly $\kappa$ is an initial ordinal with cofinality $\omega$, and $\theta_\kappa=\kappa$ follows immediately from the fact that $\kappa>M(\beta)$ for all $\beta<\kappa$. And in fact, by extending the $\beta$-sequence through $Ord$ in the obvious way, we get a club of $\theta$-fixed points (since $\theta_{\beta_\lambda}=\beta_\lambda$ whenever $\lambda$ is a nonzero limit ordinal).

The point is that while the function $\theta(-)$ itself need not be continuous everywhere, its composition with a sufficiently fast-growing function will be - we "smooth out" $\theta(-)$ with $\vert M(-)\vert^+$. And note that this trick will work if we replace first-order logic with an arbitrary logic, so long as the appropriate suprema for that logic always exist.

It is not immediately clear to me whether the $\kappa=\beta_\omega$ above is (in appropriate models anyways) the least initial ordinal which is a $\theta$-fixed point.

  • $\begingroup$ So this does solve the second question in the comments, but the question of what axiomatic requirements are necessary to warrant the existence of $\Theta_0$ are there; for example, does the existence of a weakly inaccessible imply the existence of $\Theta_0$? I still accept this question and commend you for your answer however. $\endgroup$
    – Zetapology
    Oct 12, 2017 at 3:10
  • $\begingroup$ @Zetapology See Theorem 4 and Observation 5 of this paper on pointwise definable models. If $P$ is a large cardinal compatible with V=HOD (and this includes basically everything), then we can have pointwise definable models realizing $P$ - and in a pointwise definable model, even $\theta_0$ doesn't exist. So you won't get such a simple sufficient condition. (That paper is well worth a read by the way, especially given your interests.) $\endgroup$ Oct 12, 2017 at 3:26
  • $\begingroup$ Wow... I guess this cardinal is simply a really strong, yet somehow small type of cardinal. $\endgroup$
    – Zetapology
    Oct 12, 2017 at 4:00
  • $\begingroup$ Does the existence of a critical point of some nontrivial elementary embedding from $j:\mathrm{HOD}\rightarrow\mathrm{HOD}$ imply the existence of at least $\theta_0$? I know every Reinhardt cardinal is a $\theta$-Fixed point. $\endgroup$
    – Zetapology
    Oct 12, 2017 at 4:06
  • $\begingroup$ In fact, if $M$ is a transitive inner model of ZFC such that for every first order $\phi$ and every $v$ a finite sequence of ordinals in $M$, $\phi(v)\Leftrightarrow M\models\phi(v)$ (i.e. $M$ is almost "complete" in a sense), then the critical point of an elementary embedding $j:V\rightarrow M$ is always a theta fixed point. $\endgroup$
    – Zetapology
    Oct 12, 2017 at 4:19

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