# Appealing axioms incompatible with large cardinal axioms

I'm interested to know what are some 'appealing' axioms that are inconsistent with ZFC plus some large cardinal axiom. I saw the question On the contradictory nature of large cardinals & choice-like axioms, which sparked my curiosity. One reason is that in most mathematical practice all elements of a power-set that we construct are in fact in Godel's $L$, where GCH holds. Thus arguably whatever intuitive understanding of the set-theoretic universe $V$ that one may claim to have is actually more like an understanding of $L$, simply because we cannot actually conceive of an uncountable number of things except via proxy, namely the description of the phenomenon of being not countable. GCH ($κ \le λ \le 2^κ \rightarrow κ = λ \lor λ = 2^κ$ for any infinite cardinals $κ,λ$) is inconsistent with ZFC+PFA, and hence it suggests that there are other 'appealing' axiom pairs besides (PFA,GCH) and (measurable, $V=L$) that are mutually inconsistent over ZFC.

Many people have mentioned that it seems unusual that there is a linear ordering among the large cardinal axioms so far considered, but since at some point they contradict natural extensions of AC, I'm curious as to whether this linear ordering is perhaps due to the fact that these axioms are all of roughly the same kind. I had also seen this post on situations in which 'choice' principles fail, and I was told that it is still not known whether Berkeley cardinals are consistent with ZF (without AC). If they are consistent then there would be at least two incompatible kinds of 'appealing' set-theoretic universes, one with strong 'choice' principles and one having 'all sorts' of inner models.

But surely there ought to be a far richer logical landscape? There are infinitely many incomparable theories, so if it is mere consistency ($Π_1$ arithmetical sentences) we are talking about then there should be infinitely many pairwise incomparable or incompatible extensions of ZFC, just like for PRA, where PA (plus names for all the p.r. functions) and PRA+TI($ε_0$) are mutually incomparable extensions but both of which have significant consequences (induction for all formulae, Con(PA) respectively). Also, both of them are in turn interpretable in Z$_2$. In this case the two theories are not incompatible simply because we already had the standard model of $\mathbb{N}$ in mind when we constructed both of them. However, in set theory it seems to me that there is no standard model describable in even natural language, so I see no obstruction to having a wide variety of incompatible extensions of ZFC that are intuitively appealing.

So what other pairs are known? I don't know the literature so I'd appreciate references too. Thanks!

• Isn't there an axiom of limitation or something like that, that basically says if you can't guarantee or build a set from ZFC axioms then it doesn't exist? – Jacob Wakem Aug 8 '16 at 16:56
• I don't know what you're talking about. V=L says that the universe is constructible, which I've already mentioned in my question. – user21820 Aug 8 '16 at 17:39

My personal favorite - although it seems not to be very popular - is the inner model hypothesis (IMH) and its variants. IMH states that anything that can happen in an inner model, does: specifically, $$\mbox{Any parameter-free \varphi which holds in an inner model of some outer model of V,}$$ $$\mbox{already holds in some inner model of V.}$$
• @user21820 I didn't mean a specific outer model, I meant any outer model. That is, if there is some outer model $W\supseteq V$ such that $W\models \varphi$, then there is an inner model $U\subseteq V$ such that $U\models \varphi$. (An inner model is a transitive proper class satisfying ZFC; an outer model is, well, hard to define without using a more powerful theory than ZFC, but basically it's an object of which $V$ is an inner model; think forcing extension, if you're familiar with those.) – Noah Schweber Jul 1 '16 at 11:19
• I did say "some" in my phrasing, but it seems significantly different from what you stated in your comment. What do you mean by $W \supseteq V$? – user21820 Jul 1 '16 at 11:21
• @user21820 Not really. The problem is it doesn't really make sense to talk about $V$ satisfying IMH, inside $V$, since we need to talk about sets that don't exist. Here's how I'd phrase it: a set model $V$ of ZFC also satisfies IMH if, whenever $W$ is a set model of ZFC with the same ordinals as $V$ and containing $V$, and $U$ is an inner model of $W$ satisfying $\varphi$, then $V$ has an inner model satisfying $\varphi$ (for $\varphi$ parameter-free). (cont'd) – Noah Schweber Jul 1 '16 at 11:28
• Now, internal versions of IMH can be phrased by considering "definable constructions" of outer models - for example, forcing. Call the small IMH the scheme "If there is some set forcing $\mathbb{P}$ such that $\Vdash_\mathbb{P}$"$\varphi$ holds in an inner model," then there is some inner model of $V$ which satisfies $\varphi$." (Note that even this scheme requires something like MK on the face of it, because of the quantification over inner models!) I think Friedman's paper jstor.org/stable/4093051 goes into this in more detail. – Noah Schweber Jul 1 '16 at 11:29