Minimal foundations for Cardinal Arithmetic I would like to develop a theory of cardinal numbers that relies on as weak a basis as possible. Therefore, I would like to know if there is a way to even define a cardinal number for every set without assuming Foundation or Choice. Thank you for your answers.
 A: It turns out that in a precise sense the answer is no! Let $T$ be $ZF$ minus Foundation. Then there is a model $M$ of $T$ such that there is no definable function $f: M\rightarrow M$ (note that $f$ will be a class function) such that $$f(a)=f(b)\iff\mbox{ there is a bijection between $a$ and $b$.}$$ Thus, if we want "the cardinality of [thing]" to be appropriately definable, we either need a bit of Foundation, or Choice, or we need some other axiom not already present in ZFC.
See Andres Caicedo's answer to Scott's trick without the Axiom of Regularity.

Meanwhile, even in ZF, it is consistent that we cannot assign cardinality representatives: see Asaf Karagila's answer to Defining cardinality in the absence of choice.
A: Well. The actual question is what you mean by developing cardinal arithmetic. This can be understood in several ways.

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*You want to have a formula $\varphi_\sim(x,y)$ which is true if and only if there exists a bijection between $x$ and $y$, and a formula $\varphi_+(x,y,z)$ which is true if and only if there exists a bijection between $x\times\{0\}\cup y\times\{1\}$ and $z$; and similarly for multiplication.
This setting gives you very rudimentary cardinal arithmetics, and you can probably get them from a very weak fragment of $\sf ZF$ which certainly does not include choice or foundations.


*You want to have a definable class of sets, $\Gamma$ with definable $\sim,\leq,+,\cdot$ class relation and operators on $\Gamma$, such that for every $x$ in the universe you can identify a unique $\gamma_x\in\Gamma$, and whenever there is an injection from $x$ into $y$, then $\gamma_x\leq\gamma_y$, and so on.
For this you can do with choice and without foundations (using ordinals), or with foundations and without choice (using Scott's trick). But if you give up both, it's not at all clear how you can do it. It is possible to resort into "local" versions, so you can extend the rudimentary definitions from the previous case to "set many at a time", but in order to ensure it works, I don't know if you can give up on too many instances of replacement and separation. Because "set many at a time" means that you need stronger axioms if your sets are more complicated in some way.


*You could argue that cardinal arithmetic involves cardinal representatives. Namely, you want a class of sets $\Gamma$, again with $\sim,\leq,+,\cdot$ such that this time $\gamma_x$ has a bijection with $x$. So if in the previous meaning $\gamma_x$ was allowed to be some random set, this time we require that it will actually be of the same cardinality of $x$.
This requires both foundations and some choice. Namely, without either one you can't really talk about $\Gamma$ as I pointed out above, and without some sort of choice you cannot choose a $\gamma_x$ from each equivalence class in the relation defined by "there is a bijection between the two sets". You do not need the full power of the axiom of choice here, but there is no "nice" choice principle to guarantee you this choice of representatives.
Let me finish by saying something about the axiom of foundations. You can talk about $\sf ZFA$ (set theory with atoms) as being a two-sorted theory, or $\sf ZF$ with weaken extensionality, or $\sf ZF$ with weakened regularity (your atoms will be sets of the form $x=\{x\}$). If we consider the last version, then one can claim that foundations is not needed since you can effectively run Scott's trick in $\sf ZFA$.
So in all the instances of foundations that I refer to, I really mean that there is some set $A$ that using iteration of power sets and unions you obtain the entire universe from $A$.
