You can even replace "for all limit ordinals" with "for all aleph-numbers" by using the sequence $λ_0=α$ and $λ_{n+1}=\aleph(V_{λ_n})$

@JoelDavidHamkins Did you find a counterexample yet? You had plenty enough time to think

@ZuhairAl-Johar it should be $M\cap Ord = N\cap Ord$ in their comment

@ZuhairAl-Johar you are right, I missed the part in your previous question that stated that ZFC+Classes is still in finitary logic, and assumed everything moved to infinitary logic. Sorry

To name drop a relevant foundation that can go beyond "small, big, very big", there is the "Tarski–Grothendieck set theory", which is basically "ZFC + every set is inside of a universe" (note that the TG set theory uses Tarski-classes, and not Grothendieck-universes, the main difference between the 2 is that Tarski-classes implies AC, so ZF+every set is inside a universe does not prove AC while ZF+every set is inside of a Tarski-class does implies choice)

Re:the first sentence, In cases you didn't read it already, the paper What do we want a foundation to do? Comparing set-theoretic, category-theoretic, and univalent approaches by Penelope Maddy is a great read about the different values each foundational approach gives to the table

@TimothyChow did you just assumed LEM in this server??

"I’m finding myself taking the consistency of ZFC for granted as a ‘religious belief’.", As Gödel (in)famously showed, every theory we would want to consider will require some "leap of faith" (Ignoring the ""boring theories""). Calling the belief of Con(ZFC) "religious" means that Con(ZFC) dictates the way we act, for mathematicians this is a weird statement think about, as even if someone believes in a platonic universe of sets that satisfy ZFC it does not prevent them from from exploring the vast amount of incompatible set theories/weaker set theories.

@shabunc something you can do to better develop this intuition is as such: every time you do something without the axiom of choice, check if you repeat some kind of choice infinitely many times. If you do repeat it infinitely many times, you need to justify why you can do that, if you are only doing it finitely many times (for a fix finite number), then you don't need any justifications

If there exists a cofinite block there is no such sequence of bijections

Unless $j$ is defined over transitive set of course

I am confused, by your definition of $j^z$, $j^2[x]=\{j(y)\mid y\in x\}$ so $x⊆Dom(j)$, how $_n\mathsf{F/B}_j(S)$ is well defined for any general $S$? (That question extend to any $z\ne 1,-1$)

>Proper classes are so large, that they don't really have a cardinality $$$$ Fun fact 1: In $ZF$, this statement (when state correctly) is equivalent to AC. $$$$ Fun fact 2: In $ZF$, even tho classes can be "not big", there exists a (class) function $f:V→Ord$ such that a class $C$ is proper if and only if $f''C$ is a proper class (An example of such $f$: the rank function) $$$$ Fun fact 3: It is consistent with $ZF$-regularity that there is no function like the above $f$