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Note: For the purposes of this question, I’m considering only set universes whose ordinals are well-founded, as discussed here and here

It seems like no $\mathsf{ZF}$ universe can ever really be said to contain "every conceivable ordinal" (more or less by definition, given the way the concept of von Neumann ordinal is defined). That is, no matter what conditions/properties you stipulate about a $\mathsf{ZF}$ set universe $\mathcal{V}$, it seems I can always define some kind of new universe $\mathcal{W}$ that has additional larger ordinals.

One particularly simple-minded example: given whatever universe $\mathcal{V}$ whose existence you want to stipulate, I can always define a different universe $\mathcal{W}$ by starting with $\mathcal{V}$ and then just shoving some additional ordinals in there:

  • Introduce the new ordinal $\beta = Ord^{\mathcal{V}}$ (i.e. $\beta$'s members are all the ordinals of $\mathcal{V}$)
  • And then you can introduce $\beta + 1$ (i.e. $\beta \cup$ {$\beta$}), $\beta + 2$, and so on.

Obviously a $\mathcal{W}$ so simplistically defined would be a pretty disappointing set universe, in that it would fail to satisfy many of our desired axioms. But the point is to illustrate my basic conclusion that (as far as I can tell), for any given $\mathsf{ZF}$ set universe $\mathcal{V}$, one can always define some new universe $\mathcal{W}$ which contains all the ordinals of $\mathcal{V}$ plus some additional larger ones.

Is this basic conclusion correct? Or am I missing something?

NikS
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    What is the definition of a conceivable ordinal? You might want to read about set theoretic potentialism. – Asaf Karagila Sep 08 '23 at 06:10
  • Are you looking for this kind of theorem : For any model $M$ of ZFC, there is model $N$ such that $M \subsetneq N$ and $\text{Ord}^{M} \subsetneq \text{Ord}^{N}$ ? – user700974 Sep 08 '23 at 11:39
  • @AsafKaragila - To frame this more specifically: Is it true that for any given $\mathsf{ZF}$ set universe $\mathcal{V}$, one can always define some new universe $\mathcal{W}$ which contains all the ordinals of $\mathcal{V}$ plus some additional larger ones? (text of question modified to make this more clear) – NikS Sep 11 '23 at 06:27
  • This paper by Hamkins and Linnebo seems to imply that, according to "one commonly held view amongst set theorists", the answer to my question is "yes". But is there a school of thought that would say the answer is "no"? I can't think of what the argument for that view would even look like. What property of $\mathcal{V}$ could one possibly stipulate that, in effect, "blocks" my ability to add more ordinals to it? That sounds like a basic logical contradiction... – NikS Sep 11 '23 at 06:28
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    The question of "can" is a question of your assumption, and in this case also what is a "universe". If a universe is a proper class containing all the ordinals, then in what sense can we add new ordinals? This isn't necessarily a bad question, but it requires a lot of carefully set up context to be answered in a satisfactory way, and unfortunately, most of it is going to be philosophically subjective anyway. – Asaf Karagila Sep 11 '23 at 07:44
  • If you’re considering well-founded set models as synonymous with “universes”, then it’s consistent that the answer is “no” in the sense that (for instance) there may be exactly one possible height of a transitive model of ZF. – spaceisdarkgreen Sep 11 '23 at 14:25
  • @AsafKaragila : Sounds like this is more complicated than what can reasonably be written up in a StackExchange post. Would you happen to know of some resource (book, paper, etc.) that describes the gist of the philosophical perspectives for and against (i.e. supporting vs. not supporting the notion that “for any given $\mathcal{V}$ there’s always some bigger universe $\mathcal{W}$ with additional larger ordinals”)? – NikS Sep 12 '23 at 04:46

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