3
$\begingroup$

Assume ZFC and a supercompact cardinal $κ$. Is it consistent that there is a weak extender model $N⊨\text{ZF}$ for supercompactness of $κ$ such that the axiom of choice and well-ordering of $P_κ(λ)^N$ for large enough $λ$ fails in every set generic extension of $N$? If yes, do such models still absorb large cardinal structure in $V$ above $κ$, analogously to ZFC weak extender models for supercompactness?

Recall that $N$ is a weak extender model for supercompactness of $κ$ iff for every $λ>κ$ there is a normal fine ultrafilter $U$ on $P_κ(λ)$ such that $N∩P_κ(λ)∈U$ and $N∩U∈N$. I do not require $N⊨\text{ZFC}$ (but AC holds in $V$). Thus, perhaps one should also require an elementary embedding formulation of supercompactness in $N$ (agreeing with $N∩U$), and I am fine with an answer either way.

For short extenders, inner model theory can be based on the extenders on $V$. However, supercompact cardinals use long extenders, and $N∩P_κ(λ)∈U$ requires so many sequences of ordinals to be present that there does not appear to be a canonical way to well-order them. In particular, $N$ must satisfy the $κ$-covering property: $∀s∈P_κ(λ) \, ∃t∈P_κ(λ)∩N \, s⊆t$. Thus, I do not expect there are canonical ZFC + V=HOD weak extender models for supercompactness (but this is controversial; see Woodin's HOD Conjecture and Ultimate L Conjecture for a contrary view).

However, canonical models of ZF are not so limited. For example, assuming AD in $L(ℝ)$, the structure of $L(ℝ)$ is canonical even though $L(ℝ)$ includes arbitrary reals and thus does not know how to well-order them. Similarly, without choice, there might be some way to put enough sequences of ordinals into a canonical model $N$ to make $N∩P_κ(λ)∈U$ if we allow that in every set generic extension of $N$, for a large enough $λ$, $N∩P_κ(λ)$ cannot be well-ordered. The axiom of choice is basic to current inner model theory (for short extenders), but large cardinals beyond choice suggest that a new approach will eventually be required. We cannot yet rule out that the axiom of choice will fail in canonical inner models for supercompact cardinals (with supercompacts in ZFC still obtainable through class generic extensions).

Improve this question
$\endgroup$
7
  • $\begingroup$ I don't know the answer to your question, but I think your intuition is misguided. Solovay proved that if $U$ is a supercompactness measure, then there is a set $A \in U$ such that the function $x \mapsto \sup(x)$ is injective on $A$. Therefore there is a canonical way to well-order enough sequences of ordinals to get a degree of supercompactness into a relatively small model. Just take $L(A)[U]$ for such an $A,U$. Now, is it "canonical"? The consensus is no. $\endgroup$
    – Monroe Eskew
    Commented Apr 19 at 3:00
  • $\begingroup$ @MonroeEskew Adding one subset of ordinals per ordinal does not mean that it is a small model, but you are right that the issue is not the cardinality of sets in $U$ but what those sets must include. Solovay's construction uses a partition of the set of cofinality $ω$ ordinals $<λ$ into $λ$ disjoint stationary sets, and in turn shows that under large cardinal axioms, such partitions must capture much of $V$. $\endgroup$
    – Dmytro Taranovsky
    Commented Apr 19 at 4:08
  • $\begingroup$ I still contend that your assertion, "$N \cap P_\kappa(\lambda) \in U$ requires so many sequences of ordinals to be present that there does not appear to be a canonical way to well-order them," is simply wrong. Despite the background structures in $V$ that enable this, we only add a certain $A \in U$ which is definably well-ordered. Any $B \subseteq A$ also in $U$ will suffice! And then when we add $U$ as a predicate, we get a supercompactness measure! So there appears to be a canonical way to well-order the sequences at a local level. $\endgroup$
    – Monroe Eskew
    Commented Apr 19 at 4:46
  • $\begingroup$ Supercompactness without choice cannot, and should not, be defined via ultrafilters. If Łoś theorem fails, which it bounds to with AC failing this hard, then ultrafilters, and certainly ultrafilters over ordinals, are practically meaningless. What you want to talk about is embeddings. $\endgroup$
    – Asaf Karagila ♦
    Commented Apr 19 at 12:07
  • $\begingroup$ @MonroeEskew A well-ordering can be defined using $A$ (and if needed, $U$) as a parameter, but $A$ itself might not be definable (neither in $V$ nor in $L(A)[U]$). The $κ$-covering property precludes $N$ from being too small. $\endgroup$
    – Dmytro Taranovsky
    Commented Apr 19 at 18:19

0

Reset to default

You must log in to answer this question.