Holo

@Shinrin-Yoku When I have $C$, I know that for every $a$ there exists some $c\in C$ such that $a\in V_c$. The exact value of $c$ does depends on $a$ indeed, but $C$ does not, we know that $\{V_c\mid c\in C\}$ covers every possible set, without $C$, for each set $a$ we need to find a different $Y$ such that $\{V_c\mid c\in Y\}$ covers $a$, such $Y$ will depend on $a$

@Shinrin-Yoku as I said to Eric, to "pick any least well ordered set ..." You need ZFC2+second order symbol+second order global choice, it is not something you can 'just' do

@Shinrin-Yoku you literally can't express quantification over equivalence class of this relation in $\sf ZF(C)$, each equivalence class is a proper class (apart from 0), you can't quantify over proper classes. The method of making this possible in $\sf ZF(C)$ is Scott's trick, which uses $C$

@Shinrin-Yoku with $C$ you can just do induction over $C$: $V_{∅}=∅$, $V_{A+1}=\mathcal{P}(V_A)$ and for limit points take the union. By replacement and foundation, this hierarchy is covers all of the sets (regardless what $C$ is, as long as it is well-ordered and set-like), and then you just take the minimal $A∈C$ that $a\in V_A$. Eric gave a good point that if $a$ is fixed, we can replace $C$ with $\aleph'(2^{\operatorname{trcl}(a)})$, but the problem with this approach is that it depends on $a$, unlike $C$ that works for all $a$ simultaneously

@EricWofsey You are correct that I overlooked $\aleph'(2^{\operatorname{trcl}(a)})=Y$, but my point was less about $Y$ itself, and more about how to get to $Y$ from $X$. Your second idea is more along the lines of what I envision, but like you pointed out you get into a problem without Scott trick, but Global Choice doesn't solve the problem. Global choice let you have a choice class. Without Scotts trick the situation we have here is that we have a conglomerates of classes, this "conglomerates choice" is not expressible even in $\sf ZFC2$ (you need $\sf ZFC2$+second order function symbol)

[Cont.] Well given a well-ordering $Z$ we can define the limit stages of $Z$, so we can continue till $\aleph'^{\operatorname{otp}(Z)}(X)$, but "most of the times" (in some precises meaning of "most of the time"), finding $Z$ such that $\aleph'^{\operatorname{otp}(Z)}(X)$ reaches $Y$ requires you to find well-ordering greater than $Y$, which is what we tried to find. Having a class $C$ gives all of this to us, because you can define the induction over all of $C$ (and when we assume replacement, this will result in every possible stage)

[Cont.] A possible idea to find such $Y$ is to repeat Hartogs theorem a lot, but what is "a lot"? You start with $X$, look at $\aleph'(X)$ ($\aleph'(X)$ is the well-ordering defined in Hartogs theorem, it is not the Hartogs number, it is not an ordinal), then look at $\aleph'(\aleph'(X))=\aleph'^2(X)$ and then on $\aleph'^3(X)$ and so on, but it is possible that for each natural number $n$ we have that $a\notin V_{\aleph'^n(X)}$. So how do we continue? The instinct is to say "at limit stages we take union/supremum", but what is a "limit stage" without ordinals? [Cont.]

@Shinrin-Yoku (re: second comment) No. You can define $V_X$ for any well-order $X$, and $V_X=V_{\operatorname{otp}(X)}$, but it will miss some important stuff: First of all, you won't be able to quantify over ranks, as a rank of a set will be a proper class of order-isomorphic well-ordering. Second of all, it won't uniformally cover all of the sets, given $a\notin V_X$, there exists some well-ordering $Y$ such that $a\in V_Y$, furthermore, $X$ embeds into $Y$, but how do we find such $Y$? The only methods I can think of is using the class $C$. [Cont.]

@Shinrin-Yoku (re: first comment) When you have a fixed set $X$, you can usually replace the ordinals with some well-ordering using Hartogs, this is done in e.g. $\sf Z$, but it less convenient in a lot of cases. For example, say you have a sets $X,Y$ and you for some reason need the Hartogs numbers of $X$ and $Y$, with ordinals you can easily look at $\max(\aleph(X),\aleph(Y))$ and to have some induction that exhaust both $X,Y$. Without ordinals you can still do this, but it requires you to explain what $\max$ means here, and why it behaves like we want. It is possible, but less convenient

@LittleCheese (3) See (2)

@LittleCheese (2) The idea is that $V$ is the literal universe, there are no way to have more stuff than $V$, but using set theory with classes, e.g. NBG, we can extend the universe from $V$ to $V^*$ by making some proper classes into sets

@LittleCheese (1) It is possible to do forcing on $V$ itself to get $V^*$, it is called "proper-class forcing", but the "basic" method is by starting with some model $M$

The cofinality is still an ordinal (or rather, a well orderable cardinal), so we know that it exists an unique, even for non well orderable cardinals

(Also, this definition doesn't extends the cofinality for ordinals, only cardinals)

While this interpretation works only for well orderable cardinals, it can be extended to not well orderables by "switching" the unbounded part to use sums, that is: $\operatorname{cf}(κ)$ is the minimal $α$ such that there is a partition of $κ$ into $α$ parts such that each is smaller than $κ$. This definition agrees with your definition for well orderable cardinals, and extends to non-well orderables. For example, $\operatorname{cf}(κ)=2$ for all infinite Dedekind-finite cardinals

(Continue the last comment) I am not sure if $κ<κ^{\operatorname{cf}(κ)}$ is still true in this case, but then $\operatorname{cf}$ can be finite (well, it can be 2)

Using this definition implicitly assume order to the set, for non-well orderable sets we don't have some "canonical" order like we have for well orderable sets (Sadly)

How do you define $\operatorname{cf}$ without AC? And in no interpretation of $\operatorname{cf}$ I see how it is obvious that $κ<κ^{\operatorname{cf}(κ)}$ is false (think about an infinite Dedekind finite set, then as long as ${\operatorname{cf}(κ)}$ is not $1$ we have $κ<κ^{\operatorname{cf}(κ)}$)

@TheSimpliFire you can always bugging me

@TheSimpliFire hi, sorry, a lot of stuff at work need attention because of Covid

I don't have time today @TheSimpliFire , emergency at work

I will finish the 2 digits

Which is annoying

About 3 digits. It should be the same but with 4 cases instead of 2

It is consistent throughout the proof, because it is "carry" vs "no carry"

Hmmm

Oh that

Other then considering the 2 possible cases?

b[1]=(2k+1)/3=(2(3s+1)+1)/3=(6s+2+1)/3=(6s+3)/3=2s+1

No, 2s+1, 3s+1=k

Those is the only solutions

From this we get that b[2]=s+1

First lets finish what we already have

But do you see where in your proof in overleafwhat I wrote above helps?

I linked the wrong thingy

But after we finish the proof we must carefully go over it and explain where exacty we use that

In this interval we find why it fails to s<3

@TheSimpliFire To here

@TheSimpliFire from here

Lemme find the message

I did figure it out

1 digits: a=2, we proved that b<k/2 so a+b=ab implies a=b=2

So we do need 1/2/3 digits

Oh :/

Did we assume that q-1≠2?

Wait no

So only need to consider 1 and 2 digits

?