What two ordinals are these (based on definable ordinals)? 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 $D$:


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*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 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?
 A: There are several subtle issues with your post. 
It is not in general possible to express the notion of "definable", because it leads to contradictions. For example, the class $D$ is not definable in the language of set theory (or whatever language you are using), since if it were, then the least ordinal not in $D$ would be definable, but not in $D$, which is a contradiction. 
Meanwhile, there is the phenomenon in set theory of pointwise definable models, which are models of ZFC in which every set is definable without parameters. For example, we discuss this issue at length in my paper:  


*

*Hamkins, Joel David; Linetsky, David; Reitz, Jonas, Pointwise definable models of set theory, J. Symb. Log. 78, No. 1, 139-156 (2013). (dx.doi.org/10.2178/jsl.7801090) ZBL1270.03101.


In a pointwise definable model, there is no least ordinal that is not definable, and there is no ordinal larger than every definable ordinal. In the article, we point out the following, regarding the extent to which definability is first-order expressible:

Let us now turn to the question of the extent to which
  definability is first-order expressible, by presenting a
  number of examples that illustrate the range of
  possibility. We have already observed that the property of
  a model being pointwise definable is not first order
  expressible, since it is not preserved by nontrivial
  elementary extensions. Since pointwise definability is a
  strong generalization of the axiom V=HOD, it is tempting
  to introduce such notation as V=D or V=HD to express
  that a model is pointwise definable, thereby maintaining a
  parallel to the classical V=HOD notation while
  emphasizing that the definitions need no parameters. We
  hesitate to adopt this notation, however, because we fear
  it would incorrectly suggest that the concept is
  first-order expressible, which isn't the case.
(i) There is no uniform definition of the class of
  definable elements. Specifically, there is no formula
  $\mathop{\rm df}(x)$ in the language of set theory that is
  satisfied in any model $M\models\text{ZFC}$ exactly by the
  definable elements. The reason is that if $M_0$ is
  pointwise definable and $M_0\prec M$ is a nontrivial
  elementary extension, then the definable elements of $M_0$
  and $M$ are precisely the elements of $M_0$, and so $M_0$
  should satisfy $\forall x\,\mathop{\rm df}(x)$ but $M$
  would satisfy $\exists x\,\neg \mathop{\rm df}(x)$,
  contrary to $M_0\prec M$.
(ii) The class of definable elements can form a
  definable class. Although there is no uniform definition
  of the class of definable elements, it can sometimes happen
  that a model enjoys a certain structure that allows it to
  see its collection of definable elements as a definable
  class. For example, in a pointwise definable model, the
  class of definable elements includes every object and is
  therefore defined by the formula $x=x$. See also (iv) and
  (v) below.
(iii) The collection of definable elements might
       not form a class. Consider any model
       $M\models\text{ZFC}$, and let $N$ be an ultrapower of
       $M$ by an ultrafilter on the natural numbers. The
       parameter-free definable elements of $N$ are
       necessarily contained in the range of the
       ultrapower map, and in particular, do not include
       any of the newly added nonstandard natural
       numbers. Thus, the class of definable elements of
       $N$ is not amenable to $N$, for it would reveal
       that its natural number are not well-founded.
(iv) The definable elements can form a definable
       class in a model having no class function $r\mapsto\psi_r$
       mapping definable elements to definitions. Suppose
       that $M$ is a pointwise definable model of \ZFC.
       The definable elements of $M$ are all of $M$,
       which is certainly a definable class in $M$. But
       $M$ cannot have a function $r\mapsto\psi_r$
       associating to each element $r$ of $M$, or even to
       each real of $M$, a defining formula $\psi_r$,
       since such a map would reveal to $M$ that it has
       only countably many reals.
(v) The definable elements can be a set in a model
  that does have a definability map $r\mapsto\psi_r$.
  Suppose that $\kappa$ is an inaccessible cardinal (this
  hypothesis can be reduced), and observe by a
  Lowenheim-Skolem argument that there are numerous
  $\gamma<\kappa$ with $V_\gamma\prec
V_\kappa\models\text{ZFC}$. It follows that the definable
  elements of $V_\kappa$ are all in $V_\gamma$ and satisfy
  the same definitions there as in $V_\kappa$. Since
  $V_\gamma$ is a set in $V_\kappa$, we may construct in
  $V_\kappa$     the function $r\mapsto \psi_r$ that maps
  every     definable element $r$ of $V_\gamma$ to the
  smallest definition $\psi_r$ of it, and because
  $V_\gamma\prec V_\kappa$, this function has the same
  property with respect to $V_\kappa$, as desired. The large
  cardinal hypothesis can be reduced; it is sufficient to
  have an $\omega$-model $M$ with some $M_0\in M$ having
  $M_0\prec M$.
(vi) No model can have a definable definability
  map $r\mapsto\psi_r$. If such a map were definable, then
  since there are only countably many definitions $\psi_r$,
  we could easily diagonalize against it to produce a
  definable real not in the domain of the map. In (v), the
  map is definable from parameter $\gamma$.

Meanwhile, the universal definition shows that there is a single definition that can in principle define any desired object. For example, any set at all, including any ordinal, can be made definable in a forcing extension of the universe. 
A: 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. 
