Let $D$ be the set of *definable ordinals*. An ordinal *s* is definable if there is a predicate $p$ (in the language of (first-order) set theory), such that $p(x) \iff x=s$ for all $x$. This is definitely a set (not a proper class), since the list of all syntactically valid predicates is countable, and $D$ is $\le$ to that in size. In particular, $D$ is countable (although most of its elements are not).


We can define two ordinals based on $D$:


* The ordinal $\delta$ defined as the least ordinal not in $D$. This ordinal is countable (since $D$ is countable, and so can not contain every countable ordinal), and has the property that an ordinal less than $\delta$ iff it is recursively definable (i.e., every ordinal $x<\delta$ is definable and every $y<x$ is also recursively definable). It is also a limit ordinal, since if $\delta-1$ was an ordinal, it would be definable since it is less than $\delta$, and its definition could be used to define $\delta$, which is a contradiction.
* The ordinal $\Delta$ is the [supremum](https://en.wikipedia.org/wiki/Supremum) of $D$. It is also not a definable ordinal (since if it where, so would $\Delta+1$, which is greater than $\Delta$). This also means its a limit ordinal, since it is the supremum of a set that does not contain it. However, is it definitely not countable. Indeed, its cardinality is greater than any cardinal definable in set theory. It has the property that any ordinal $x$ is $x<\Delta$ iff it is bounded by a definable ordinal.


My question is, which ordinals are $\delta$ and $\Delta$? In particular, have either or both of them been studied in the literature before? If not, can they be defined in linked to ordinals previously studied?



What two ordinals are these (based on definable ordinals)?
In this answer, let me assume as you indicated in the comments that you are working in a second-order set theory with a truth-predicate for first-order truth. Such a theory goes strictly beyond ZFC in consistency strength, but it is provable in Kelley-Morse set theory and indeed, merely in GBC+$\text{ETR}\_\omega$, which is weaker than $GBC+\Pi^1\_1$-comprehension. So, not too strong.


One thing you can say is that $\Delta$, the supremum of the first-order definable ordinals, is a fully correct cardinal, meaning that $V\_\Delta\prec V$. Indeed, $\Delta$ is precisely the smallest correct cardinal. The reason is that $\Delta$ is the supremum of the first $\Sigma\_1$-correct ordinal, and the first $\Sigma\_2$-correct ordinal, and so on, the first $\Sigma\_n$-correct ordinal. So $V\_\Delta$ is the union of the increasingly elementary chain, which is therefore itself fully elementary in $V$. And no smaller cardinal can be correct, because the smaller cardinals are bounded by some definable ordinal. So $\Delta$ is the smallest correct cardinal.


Another way to see that $V\_\Delta\prec V$ is that $V\_\Delta$ satisfies the Tarski-Vaught criterion with respect to $V$, since if $V\models\exists x\ \varphi(x,y)$, where $y\in V\_\Delta$, then since $y\in V\_\theta$ for some definable $\theta$, one can define the least ordinal $\alpha$ that contains such an $x$ in $V\_\alpha$ for all $y\in V\_\theta$ that have such an $x$, and so $\alpha$ will be definable and hence less than $\Delta$, and so the desired witness $x$ can be found in $V\_\Delta$. Thus, the Tarski-Vaught criterion is fulfilled, and so $V\_\Delta\prec V$. 


So there is no first-order way to describe $\Delta$, since any first-order property that $\Delta$ has must be shared also by some ordinals below $\Delta$. Of course, $\Delta$ is second-order definable as the supremum of the definable ordinals, and this seems to be a definition of complexity $\Delta^1\_1$, since the truth predicate is unique when it exists. 


