The intention behind this posting is to arrive at a definition of cardinality that can work in $\sf ZF$, and at the same time doesn't exhaust the whole of $V$. The known definition uses Scott's trick, so a cardinal is defined as an equivalence class under bijection of sets of the lowest possible rank. The problem is that this definition exhausts the whole of the universe, that is we have: $$\forall x \exists c: c \text { is a Scott cardinal} \land x \in TC(c)$$. Now, if we have two transitive models $M;N$ of $\sf ZF$, and if the set $\operatorname {Card}^M$ of all Scott cardinals in $M$ is equal to $\operatorname {Card}^N$, then we must have $N=M$. This is a shortcoming in the implementation of Cardinality, since it need not exhaust the whole universe with it.

The usual definition of Cardinality in $\sf ZFC$ after Von Neumann's obviousely doesn't have this shorcoming. Scott cardinals will not enable us to speak of distinct transitive models having the same set of cardinals.

What is needed is a restriction on this definition, we need a predicate $P$ that fulfills the following two sentences: $$\forall x \, \exists c: x \sim c \land P(c) \\ \neg \forall x \exists c: P(c) \land x \in TC(c)$$ where $\sim$ stands for "existence of a bijection".

I think we can take $P$ to be hereditarily_(empty or singleton or a set of empty or singletons) to qualify for the above, formally: $$P(x) \iff \forall y \in TC(\{x\}): \\\operatorname {y=\emptyset \lor singleton}(y) \lor \\\forall z \in y \, (\operatorname {z=\emptyset \lor singleton}(z))$$

I think that would satisfy the above two requirements in $\sf ZF$.

Now we define a modified Scott cardinal as an equivalence class under "bijection" of $P$-sets of the lowest possible rank.

Are there other known examples that would satisfy the requirements here? Especially ones that have less structure or ones that have simpler definition?


1 Answer 1


Your mistake is in misunderstanding $\mathrm{Card}^M=\mathrm{Card}^N$. Indeed, it is easy to take it at face value and simply say that the classes are exactly the same.

However, much like the case with Borel sets, when we move to a different model, what we really mean is the reinterpretation of the Borel code, not the actual set itself.

The way to make sense of this is to say that $\mathrm{Card}^M=\mathrm{Card}^N$ means that:

  1. Every set in $N$ is equipotent to a set in $M$, so no new cardinals are added. (This can happen, e.g. there is a forcing in Cohen's first model which adds an amorphous set, whereas no set in Cohen's model can be amorphous.)

  2. Every $x,y\in M\cap N$ satisfy that $M\models|x|\leq|y|$ if and only if $N\models|x|\leq|y|$. So no cardinals are collapsed, and this prevents something like collapsing $\omega_1$ to be countable without "damaging" the structure as a whole.

  3. Optional(!) Augment the previous condition by replacing $\leq$ by $\leq^*$, i.e. requiring the existence of surjections to be in both or neither. This might be necessary, although it might be redundant or excessive.

Combined, these conditions imply that the cardinal structure of $M$ and $N$ is the same, with the third condition also making sure that $\leq^*$ is preserved.

I will add and say that we know virtually nothing about preservation of cardinals in $\sf ZF$ in this general sense. Beyond some very basic observation in some fairly limited cases, there's no a lot we can say. And we most certainly do not have the tools we have in $\sf ZFC$.

  • $\begingroup$ Yes! But reducing the structure of cardinals would give us more room of freedum, it would enable us to talk about $\operatorname {Card}^M$ at face value. I generally agree with what you said, and thanks for these clarifications, but I still think that modification would have its implementational advantages about cardinality in models. $\endgroup$ Commented Jan 18, 2022 at 20:08
  • $\begingroup$ What do you gain from having the exact same class? You might as well add $M$ as a predicate, and simply require that cardinals of sets in $M$ are the cardinals in $M$. $\endgroup$
    – Asaf Karagila
    Commented Jan 18, 2022 at 20:08
  • $\begingroup$ I didn't get the last comment. Since I'm speaking about Transitive models, then by this definition the cardinals in M are cardinals of sets in M and vise versa $\endgroup$ Commented Jan 18, 2022 at 22:49
  • $\begingroup$ Add a predicate symbol to your language, which is interpreted as $M$ itself, even in larger models. Now in $N$, the larger model, simply state that if $x\in M$, then $|x|$ is the cardinal from $M$, and so, by extension, every cardinal in $N$ of a set from $M$, is the cardinal from $M$. My point here is that if you want to make it clunky, at least make it simple. $\endgroup$
    – Asaf Karagila
    Commented Jan 18, 2022 at 23:24

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