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 that $\kappa$ has cofinality $>\omega$.

For an ordinal $\alpha$, $\theta_\alpha=\alpha$ is equivalent to the inability to define $\alpha$ using first-order formulas and finitely many parameters from ordinals $<\alpha$.

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.

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.

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 that $\kappa$ has cofinality $>\omega$.

For an ordinal $\alpha$, $\theta_\alpha=\alpha$ is equivalent to the inability to define $\alpha$ using first-order formulas and finitely many parameters from ordinals $<\alpha$.

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. 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 neccesary for such a cardinal (if any other than ZF)?

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

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 that $\kappa$ has cofinality $>\omega$.

For an ordinal $\alpha$, $\theta_\alpha=\alpha$ is equivalent to the inability to define $\alpha$ using first-order formulas and finitely many parameters from ordinals $<\alpha$.

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.