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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?

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I don't understand the intended role of the $L_\alpha$'s? The ordinal $\alpha$ can often be used to simulate an ordinal parameter in the defining formula. For example, if $\kappa$ is any cardinal in $V$ (or even just in $L$), then $\kappa$ is the unique solution to the formula "$x$ is the largest cardinal" in $L_\alpha$ where $\alpha = (\kappa^+)^L$. So every cardinal number is definable without parameters in some $L_\alpha$... –  François G. Dorais Oct 18 '11 at 22:15
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And hence also every ordinal, since $\kappa=\aleph_\beta$ for some $\beta$. Thus, every object in $L$ is definable without parameters in some $L_\alpha$. –  Joel David Hamkins Oct 18 '11 at 22:29
    
And moreover: every object $x$ in $L$ is definable without parameters in some $L_\alpha$ satisfying any desired fixed finite fragment $\text{ZFC}^*$ of ZFC, because otherwise the $L$-least such $x$ not so definable is definable in $L$ itself (by this very description), and so by reflection it is definable in such an $L_\alpha$, a contradiction. –  Joel David Hamkins Oct 19 '11 at 12:57

1 Answer 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.

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 $L[G]$.

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There is a third, slightly different, interpretation. Part (b) seems to talk about definability without parameters (in $L$). This leads me to read the question as: is it consistent that every ordinal definable without parameters in $L$ is countable? This also depends on $V$, but in a very different way... –  François G. Dorais Oct 18 '11 at 21:40
    
Ah, yes, I can see that reading of the question now. But the answer seems to be the same as in my second answer. Namely, if $V=L$, then we can define $\omega_1^L$, which is uncountable in $V$. But if $L_\delta\prec L$, then all definable ordinals of $L$ are below $\delta$, but in the model $L[G]$ collapsing $\delta$ to $\omega$, these are all countable. So in $V=L[G]$, every ordinal definable in $L$ is countable in $V$. –  Joel David Hamkins Oct 18 '11 at 21:47

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