This question arose in the answers to Asaf Karagila's question Does ZFC prove that the universe is linearly orderable?. The answer there was that one can have a ZFC model with no global linear ordering of the universe (and consequently also no global well-ordering). The question here is whether we can separate these two principles from each other.

Question. Is it consistent with ZFC that there is a global linear ordering of the universe but no global well-ordering of the universe?

More specifically, there are two forms of the question, depending on whether one requires the classes to be definable classes as in ZFC, or whether one allows classes in the sense of Gödel-Bernays set theory.

  • If ZFC is consistent, is there a ZFC model with a definable linear ordering of the universe, but no definable (with parameters) well-ordering of the universe?

  • If ZFC is consistent, is there a model of GB+AC with a class linear ordering of the universe, but no class well-ordering of the universe?

The answer to the other question showed that there can be models of ZFC having no definable linear ordering of the universe, because one can make a class forcing extension which adds generic sets in a homogeneous manner, which prevents any definition from ordering them. Can we somehow modify the construction to allow a linear order, but no well-order? I suspect that we can, but I also suspect it will be easier to do this with GB classes than to make them definable.

  • $\begingroup$ This might be a stupid answer, but given that all proper classes have the same cardinality, can't you just put an order on the universe based on a bijection with the ordinals? That is, for sets r and s, r<=s iff o(r)<=o(s)? $\endgroup$
    – user75641
    Jul 2, 2015 at 18:43
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    $\begingroup$ If you define "cardinality" as "the existence of a class bijection", then it is not given that all proper classes have the same cardinality, since this would prove that there is a global well-ordering. This is exactly not the situation. $\endgroup$
    – Asaf Karagila
    Jul 2, 2015 at 18:45
  • $\begingroup$ But obviously all proper classes have the same cardinality. $\endgroup$
    – user75641
    Jul 2, 2015 at 18:54
  • $\begingroup$ You should check this out. $\endgroup$
    – Burak
    Jul 2, 2015 at 19:38
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    $\begingroup$ I'm already a Platonist, and that can affect my opinion about, for example, whether certain bijections exist. But it won't make much difference when the question has the form "Is it consistent with ZFC that $\alpha$?" where $\alpha$ is something that I think is probably false. $\endgroup$ Jul 2, 2015 at 19:49

1 Answer 1


I would like to record here my observation that one natural way of proceeding does not actually succeed.

Specifically, there is a very natural class forcing $\mathbb{P}$ to add a generic linear ordering of the universe, without adding any sets, as follows: conditions in $\mathbb{P}$ are simply set-sized linear orderings of any piece of the universe. We order these conditions by $p\leq q\iff p$ includes $q$ as a suborder. This forcing is $\kappa$-closed for any $\kappa$ and hence adds no new sets. Meanwhile, it is dense for any set to be added to the field of the order, and so the generic filter $G$ provides a generic linear ordering of the universe. Basically, this forcing builds up a linear ordering of the universe by fitting each set generically into the order.

This forcing notion is exactly analogous to the forcing $\mathbb{Q}$ of global AC, where one uses conditions that are well-orderings of any part of the universe, ordered by end-extension. That forcing also adds no new sets, and the generic filter is a global well-ordering. Thus, every model of ZFC can be extended to a model of GBC = GB + global AC without adding any sets. This is why GBC is conservative over ZFC for first order assertions.

Meanwhile, my observation is that adding the global linear order via $\mathbb{P}$ will also create a global well-order. To see this, suppose that $G\subset\mathbb{P}$ is $V$-generic. For any condition $p$, a set-sized linear order, there is some large $\kappa$ such that the field of $p$ does not mention any unbounded subset of $\kappa$. Let $\lhd$ be a well-ordering of the unbounded subsets of $\kappa$, and let $q$ extend $p$ by placing a copy of $\lhd$ above the field of $p$. Thus, $q$ has as a suborder a well-ordering of all the unbounded subsets of $\kappa$. So by density, the generic linear ordering given by $G$ must also have, for arbitrarily large $\kappa$, a suborder that is a well-ordering of all of the unbounded subsets of $\kappa$. This is enough to define a global well-ordering of the universe, since we can say that $X$ preceeds $Y$ if the transitive closure of $X$ has smaller cardinality than $Y$, or they have the same size transitive closure, but for the smallest $\kappa$ having in $G$ a well-ordering of the unbounded subsets of $\kappa$, such that $\kappa$ is large enough to code these sets, that $X$ is coded by an earlier such set than $Y$ is.

So the naive attempt to add a global linear order without adding a global well-ordering doesn't work.

  • $\begingroup$ I would like to add to this observation that if one begins with a model which proves the existence of a global well-ordering (e.g. $V=L$) then by not adding any sets, you could not have affected this statement and therefore could not have violated global choice. $\endgroup$
    – Asaf Karagila
    Nov 3, 2012 at 16:02
  • $\begingroup$ Yes, that's right; the original naive idea was to start with a model having no global well-order, and then add just a linear order, without adding a well-order. But alas, it doesn't work. $\endgroup$ Nov 3, 2012 at 16:39
  • $\begingroup$ This is a complete shot in the dark, as I know very little about forcing, but what if we try to use as $\mathbb{P}$ only dense linear orders, or some sort of linear orders which disallow well-orders to sneek in? $\endgroup$ Nov 3, 2012 at 18:15
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    $\begingroup$ Andrej, forcing with dense orders will give the same result, since every order can be extended to a dense order, and so the collection of dense linear orders is dense in $\mathbb{P}$. In partiular, the generic linear order will be dense, and I could have made the condition $q$ dense, just by adding extra stuff. $\endgroup$ Nov 3, 2012 at 18:18

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