Define a ranking function $\cal R$ as:

$\mathcal{R}: V \to ON; \,\mathcal {R}(x)= \min \alpha \, \forall y \in x: \alpha > \mathcal {R}(y) $

Now the constructible rank $\mathcal R^c$ of a set $X$ in $L$ is the ordinal index of the first constructible stage $X$ appears as an element of.

Accordingly for some set $X$ we may have $\mathcal {R^c}(X) > \mathcal {R} (X) +1$

Now define: $concordant(X) \equiv_{df} \mathcal {R^c}(X) = \mathcal {R}(X) + 1$; otherwise, $X$ is discordant.

Let $ [Cc]$ be the class of all concordant sets. Let $ L`[Cc]$ be defined as $ L`[Cc] = \bigcup L_\alpha`[Cc]$

Where each $L_{\alpha+1} `[Cc] $ is the set of all concordant subsets of $L_\alpha `[Cc]$ that are definable after formulas with parameters and quantifiers restricted $L_\alpha `[Cc]$; with limit stages being unions of all prior stages.

Now $L`[Cc]$ is a proper subclass of $L$ so it cannot be a model of $\sf ZFC$.

Now what sentences $L`[Cc]$ satisfy? In particualr does $L`[Cc]$ satisfy set union?

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    $\begingroup$ It seems you can always pump up the rank artificially simply by adding all the ordinals as elements, and then working as usual with non-ordinal sets. In this sense, you won't really be missing much. $\endgroup$ Apr 23, 2021 at 11:20
  • $\begingroup$ @JoelDavidHamkins, yes possibly the height of the hierarchy won't change (not sure of that) but, what about sets contained in each stage, for example all subsets of a stage must belong to the next stage, they cannot appear later, since this would be discordant. It is expected that we'd find hard times with maintaining impredicative constructability statements (with respect to type theory (which can be captured with constructible ranks)) since those are responsible for subsets of lower stages appearing high up in the hierarchy. I'd like to see if in particular set unions can be preserved. $\endgroup$ Apr 23, 2021 at 12:02
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    $\begingroup$ My point is that they can in effect appear later, provided that they are accompanied by a suitable ordinal at that stage. So the entirety of L will be there, in this weird coded manner, where every set also contains the ordinals up to the stage it was added. $\endgroup$ Apr 23, 2021 at 12:33
  • $\begingroup$ @JoelDavidHamkins, hmmm..., that's nice. I would like to know the details of that trick. $\endgroup$ Apr 23, 2021 at 13:03
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    $\begingroup$ Your notation "$L_\alpha[Cc]$" clashes with the existing relative constructibility notation (according to which we trivially have $L_\alpha[Cc]\supseteq L_\alpha$ and $L[Cc]=L$); I suggest a different notation. $\endgroup$ Apr 23, 2021 at 16:00

1 Answer 1


We assume we are working in ZF. We assume that the definition of concordant is $concordant(X) \equiv_{df} \mathcal {R^c}(X) = \mathcal {R}(X)$+1(without the "+1", no set is concordant).

We will show that union does not hold in πΏβ€˜[𝐢𝑐]. For simplicity we will use Quine ordered pairs so that if x and y are in πΏπ›Όβ€˜[𝐢𝑐], and 𝛼 is infinite, then (x,y) is in πΏπ›Όβ€˜[𝐢𝑐]. We will construe

formulas as elements of Ο‰. We can express that a formula holds in πΏπ›Όβ€˜[𝐢𝑐], in 𝐿(𝛼+1)β€˜[𝐢𝑐]. Let QX be the one element subsets of X. Let A be the set of elements of 𝐿(Ο‰+1)β€˜[𝐢𝑐], given by

(f,p)∈A<-->("f is a formula with two free variable holds" ,p∈𝐿(Ο‰+1)β€˜[𝐢𝑐], and "f(p,(f,p)) does not hold in 𝐿(Ο‰+1)β€˜[𝐢𝑐]"). We note that there is an infinte p and an f such that

"f is a formula with 2 free variables" holds, and (f,p)∈A.(For instance let f be the formula corresponding to r=r∧sβ‰ s and let p=Ο‰. Then p=p∧(f,p)β‰ (f,p) does not hold in 𝐿(Ο‰+1)β€˜[𝐢𝑐].)

Then Aβˆ‰πΏ(Ο‰+2)β€˜[𝐢𝑐], but QQA∈𝐿(Ο‰+4)β€˜[𝐢𝑐]. If QA∈𝐿(Ο‰+3)β€˜[𝐢𝑐], then QA is an example of a set whose union does not exist in πΏβ€˜[𝐢𝑐]. If QAβˆ‰πΏ(Ο‰+3)β€˜[𝐢𝑐], then QQA is an example of a set whose union does not exist in πΏβ€˜[𝐢𝑐].

  • $\begingroup$ I didn't finish examining your argument, but I've noticed that your proof contains a non-stratified piece of information that is f(p,(f,p)). Do you have a stratified version of your argument? $\endgroup$ Apr 27, 2021 at 16:02
  • $\begingroup$ No............. $\endgroup$ Apr 27, 2021 at 16:57

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