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I suppose that one way to make Gödel's constructible hierarchy as "concrete" as possible is to define it by saying that the sets of $L_{\alpha+1}$ for any ordinal $\alpha$ in $ON$ (the class of all ordinals) are exactly those that can be written on the form $\{x\in L_\alpha|\phi(x)\}$ where $\phi(x)$ is build up from variables ranging over $L_\alpha$, terms of the form $\{x\in L_\beta|\phi(x)\}$ with a $\beta<\alpha$, connectives, and quantifiers; and letting $L_0$ be $\emptyset$ and taking unions at limit ordinals. That means that my title question reduces to "Can every ordinal in $ON$ be described in language?". Now, if $ON$ is understood in a naive way, the answer is clearly "no". But is it consistent with ZF that $ON$ is so small that the answer is "yes"?

I would like to elaborate a little on why I asked that last question, but I will not be able to formulate myself with perfect precision - hopefully, I will be able to get across what it is I'm trying to get information about anyway.

The constructible hierarchy is defined using an already given $ON$. Could $ON$ not itself be determined somehow by the constructive hierarchy? In particular, might there not just be countable many (as seen from the outside) countable ordinals (as seen from the inside)? And could that not contribute to keeping $ON$ small?

If the answer to my (overall) question is negative, I'm also interested in whether at least all the real numbers of this model of ZF can be described in language.

Casper
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    Are you basically asking this question? –  Apr 02 '17 at 00:49
  • Also, *constructible hierarchy*, not "constructive hierarchy". :) – Asaf Karagila Apr 02 '17 at 06:20
  • I don't understand your question but I can answer part of it: 'might there not just be countable many (as seen from the outside) countable ordinals (as seen from the inside)?' Yes, certainly. This is just the statement that $\omega_1^L$ is countable in $V$. This, for example, is true if $0^\sharp$ exists. – Stefan Mesken Apr 02 '17 at 12:32

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It is - very surprisingly! - possible to have a model $M$ of ZFC+V=L such that every element of $M$ (not just the ordinals in $M$!) is definable in $M$ without parameters. That is, for each $a\in M$ there is a single formula $\varphi(x)$ in the language $\{\in\}$ such that $$\{c\in M: M\models\varphi(c)\}=\{a\}.$$ In such a model, every ordinal (indeed, every set) is defined by a single formula.

Meanwhile, we could always have a forcing extension $V\subset V[G]$ such that $V[G]$ thinks that there are only countably many countable ordinals in the sense of $V$ (so $V$ is internal, $V[G]$ is external). Note that unlike above, $L$ plays no role here; this is a coarse fact. And finally, you may find this interesting: it is consistent with ZF that the set of countable ordinals is a countable union of countable sets.

Noah Schweber
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  • Thanks, that was very helpful! But after reading the paper you link to and also this paper (http://www.ams.org/journals/bull/1963-69-04/S0002-9904-1963-10989-1/home.html) by Cohen, I'm still wondering what an $ON$ where all the cardinals are nameable would look like. – Casper Apr 03 '17 at 21:47