show/hide this revision's text 2 added 400 characters in body

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 $L[G]$.
show/hide this revision's text 1

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.