Formulae of standard set theory interpreted in suitable segments of L In a letter dated 16 October Georg Kreisel asked me the question that follows. He agreed to my idea of posting the question here.  
I’d be grateful if you could recommend an exposition of formulae of standard set theory interpreted in suitable segments $L_\alpha$ of $L$. (a) It is commonly said that each set of $L$ is definable by such a formula with symbols for specific ordinals, usually without specifying suitable $L_\alpha$ where the formula is interpreted. (Of course, if a set is defined at all, there is a bound for the $\alpha$ beyond which the definition is stable.) (b) Are there definitions without any additional symbols interpreted in $L$ which define (in $V$) uncountable, let alone, strongly inaccessible ordinals at all? 
 A: I'm not sure I have the intended sense of the question, as there seem to be several ways to interpret it,
so let me give several replies. Please clarify if I've misunderstood.


*

*Perhaps you intend to ask: is there a formula
$\varphi(x)$ in the language of set theory such that
$L_\alpha\models\varphi(\kappa)$ if and only if $\kappa$ is
an uncountable cardinal in $V$? And is there a formula
$\psi(x)$ such that $L_\alpha\models\psi(\kappa)$ if and
only if $\kappa$ is inaccessible in $V$?


In this case, the answer is no. One way to see that we
should expect this is that by forcing we may make any set
countable (and hence also non-inacessible), but forcing
does not change the satisfaction relation of formulas in
$L$ or in $L_\alpha$, and so we may change the truth of the
right-hand-side of these proposed equivalences by moving to
a forcing extension, without changing the left-hand-side.
Thus, they cannot always be equivalent. But another direct
argument is that whenever $L_\alpha\models\varphi(\kappa)$,
then by the downward Lowenheim-Skolem theorem, there is in
$L$ a countable elementary substructure $X\prec L_\alpha$
containing $\kappa$, and the Mostowski collapse of $X$ is
isomorphic by the condensation principle to some $L_\beta$,
which will satisfy $\varphi(\kappa_0)$ for the image of
$\kappa$ under the collapse. Thus,
$L_\beta\models\varphi(\kappa_0)$, but $\kappa_0$ is
countable in $L$, violating the desired property.


*

*Perhaps you intend to ask: is there a formula
$\varphi(x)$ such that $L\models\varphi(\kappa)$ if and
only if $\kappa$ is uncountable in $V$ (and another formula
for inaccessibility).


In this case, the answer is that it depends on $V$. On the
one hand, if $V=L$, then there are such a formula, because
the property, $\kappa$ is uncountable,'' is expressible
in the first-order language of set theory, as the assertion
that there is no surjective function from $\omega$ to
$\kappa$. Similarly the property of being inaccessible is
expressible. The point is that it is consistent with ZFC
that the concepts of uncountable and inaccessible are in
agreement between $L$ and $V$.
But meanwhile, it is also consistent that there are no such
formulas. For example, one quick way to see this is that if
$0^\sharp$ exists, then all the Silver indiscernible
ordinals have the same first-order properties in $L$, and
so from the perspective of $L$, the cardinal $\aleph_1^V$
satisfies the same formulas as many countable ordinals.
But one needn't make the $0^\sharp$ assumption, and one can
do it equiconsistently with ZFC. The reason is that it is
equiconsistent with ZFC that there is a cardinal $\delta$
with $L_\delta\prec L$, expressed as a scheme in the
language with a constant for $\delta$. In such a model, we
may move to the forcing extension $L[G]$ collapsing
$\delta$ to $\omega$. In $L[G]$, all the uncountable
ordinals are above $\delta$, but by our $L_\delta\prec L$
hypothesis, for any ordinal above $\delta$ with a certain
property in $L$, there will be ordinals below $\delta$ with
that same property. But since these will all be countable
in $L[G]$, it violates the desired feature.
François pointed out in the comments below that you may have intended the question:


*

*Is it consistent that every ordinal that is definable in $L$ is countable in $V$? 


The answer here is yes, since in the model $L[G]$ above, where $L_\delta\prec L$ and $G$ collapses $\delta$ to $\omega$, we have that every definable object of $L$ is in $L_\delta$, and these are all countable in $L[G]$.
