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.


(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.

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    $\begingroup$ And one can add that ZFC (along with most large cardinal axioms) is not suitable for proving these sort of statements. For example it cannot even prove that the universe is linearly ordered. $\endgroup$ – Asaf Karagila Aug 18 at 13:04
  • $\begingroup$ @AsafKaragila Do you mean it is natural to add some axioms to obtain that ability? (I mean except the trivial variant that states that "universes are linearly ordered") Or to use some kind of higher order logic? $\endgroup$ – georgy_d Aug 18 at 14:40
  • $\begingroup$ @georgy_d: Well, for example $V=L[E]$ or $V=\rm HOD$ are such "natural axioms" (here $E$ is "some set"). $\endgroup$ – Asaf Karagila Aug 18 at 14:44
  • $\begingroup$ The definitions for the fist example can be found in wiki on Constructible universe, for the second: en.wikipedia.org/wiki/Ordinal_definable_set , Thanks a lot! $\endgroup$ – georgy_d Aug 18 at 16:26

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