Is the class of all sets well ordered?(in broader sense) I saw the question Well ordered Proper Classes. and I want to ask the following.
Is the class of all sets linearly ordered? I mean, let us assume we use ZFC set theory. (Or ZFC + Tarski axiom. (1) By the way, does such system contain known inconsistencies?).
Every universe is well ordered by Zermelo's theorem.
(2) But does exist a class that is a bijection between Ord and Set?
I think that class of universes is linearly ordered.
We can preserve an order on the lower universe and add an order of the set-theoretic difference between current universe and previous one. (Which is also a set because it belongs to the next universe.)
(3) Are my statements valid?
(4) How to continue them or prove well-ordering of Set the other way?
All I want is to somehow prove that there exists a "minimal" element of every proper class.
 A: (1) Almost all set theorists believe in the consistency of ZFC and ZFC + Tarski's axiom (or equivalently, ZFC with a proper class of inaccessible cardinals.) Of course, we cannot prove its consistency due to Gödel's incompleteness theorem if they are consistent.
(3) In fact, the collection of all (Tarski-Grothendieck) universes are well-ordered: they are of the form $V_\kappa$ for some inaccessible $\kappa$, and the class of all inaccessible are a subclass of the class of all ordinals. Hence they are well-ordered.
(Note that if you mean a universe mere a model of ZFC, then they are not linearly ordered.)
However, we cannot prove the class of all sets $V$ is well-ordered from this fact, even if we have Tarski's axiom. You have to choose a well-order in each step, and it needs a proper class many choices, which is not justifiable unless we have the axiom of Global choice.
(2) The class of all ordinal-definable sets $\mathrm{OD}$ is a bijective image of the class of ordinals $\mathrm{Ord}$. In fact, if $X$ is a class which is a bijective image of $\mathrm{Ord}$ under a definable bijective class function, then $X\subseteq \mathrm{OD}$. Hence if $V\neq \mathrm{OD}$, then there is no definable bijection between $\mathrm{Ord}$ and $V$.
Even if we drop the definability, there is no reason to assume there is a bijection between $\mathrm{Ord}$ and $V$. See the relevant answer on Mathoverflow.
(4) It is known that they are equivalent:

*

*$V$ has a well-ordering,

*There is a bijection from $\mathrm{Ord}$ to $V$, and

*The axiom of Global Choice.

There are some axioms that imply the axiom of Global choice: for example, the axiom of constructibility proves there is a canonical global well-ordering. However, the mere ZFC does not prove the axiom of Global Choice, even if we assume Tarski's axiom. Hence there is no way to prove Global choice from your theories.
