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.



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


