The smallest initial ordinal which is not defined using first-order formulas with parameters of smaller ordinals?

Question

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.

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.

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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.