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Feb 20, 2021 at 6:20 comment added Trevor Wilson Regarding SWVP versus WVP, there is another such principle that is equivalent to "Ord is Woodin", namely: for every Ord-length inverse limit system of graphs (or whatever) there is an ordinal $\alpha$ such that for all ordinals $\beta > \alpha$ the canonical homomorphism from the inverse limit up to $\beta$ to the inverse limit up to $\alpha$ is a retraction. Perhaps requiring the morphisms to commute is more natural than requiring them to be unique.
Feb 20, 2021 at 6:15 comment added Trevor Wilson ...so the inaccessible weak Vopenka cardinals are the Woodin cardinals, but there can also be smaller weak Vopenka cardinals. In particular, if the Proper Forcing Axiom holds then $\aleph_2$ is a weak Vopenka cardinal -- I wonder whether there is any application of this fact to category theory.
Feb 20, 2021 at 6:03 comment added Trevor Wilson @TimCampion Yes, "Ord is Woodin" sounds a bit awkward. Maybe this point in the large cardinal hierarchy will eventually be called just "WVP" with the equivalence to a large cardinal principle being understood implicitly, the same way "VP" is often used now in set theory to refer to any large cardinal principle equivalent to VP. By the way, the analogy WVP : Woodin cardinals :: VP : Vopenka cardinals has an interesting hole in it, namely that the natural definition of "weak Vopenka cardinal" doesn't imply inaccessibility like any natural definition of "Vopenka cardinal" does...
Feb 14, 2021 at 21:00 comment added Tim Campion I'm also just now appreciating that I actually like SWVP better than WVP -- it's related to the formulation of VP which says "For every ORD-indexed sequence of objects $(A_\alpha)_{\alpha<ORD}$, there is a morphisms $A_\alpha \to A_\beta$ for some $\alpha<\beta$" in the same way as WVP is related to the discrete-subcategory formulation of VP. I like the formulation leading to SWVP better because it doesn't require messing around with rigid objects or ensuring that various morphisms are unique -- which ends up feeling kind of unnatural category-theoretically.
Feb 14, 2021 at 20:55 comment added Tim Campion Somehow when I first read this post, the formulation "ORD is Woodin" sounded a bit weird and technical to me. I'm only just now noticing the obvious fact that a "basically equivalent" way of saying this is that "A weak-Vopenka cardinal is precisely a Woodin cardinal". The intrusion of "ORD is" in the formulation is 100% parallel to the fact that Vopenka's principle basically says "ORD is a Vopenka cardinal". So from this perspective -- seeing that WVP is to Woodin cardinals as VP is to Vopenka cardinals -- it's clear that WVP is an extremely canonical point in the large cardinal hierarchy
Jul 2, 2019 at 6:24 history edited Trevor Wilson CC BY-SA 4.0
typo
Jul 2, 2019 at 6:02 history edited Trevor Wilson CC BY-SA 4.0
Added link to ArXiv; removed outline of obsolete argument
Apr 25, 2019 at 16:24 comment added Trevor Wilson @Victoria Thanks! I will let him know.
Apr 25, 2019 at 12:33 comment added Victoria Gitman Trevor, this is great! If my understanding is correct, the question has been open for some time and was of interest to many people. You should write to Joan Bagaria to let him know that it has finally been answered.
Apr 23, 2019 at 3:23 comment added Trevor Wilson @TimCampion Thanks! I hope to have time to do that soon.
Apr 23, 2019 at 2:44 comment added Tim Campion This is awesome! I hope you write it up in a paper!
Apr 22, 2019 at 13:43 vote accept Tim Campion
Apr 22, 2019 at 6:12 history edited Trevor Wilson CC BY-SA 4.0
Statement S is called SVWP
Apr 22, 2019 at 0:43 history edited Trevor Wilson CC BY-SA 4.0
Rewrote the proof of the lemma, clarifying some things and fixing a minor error
Apr 21, 2019 at 23:10 comment added Trevor Wilson Also, I think a refinement of the argument should prove that WVP for $\mathbf{\Pi}_1$ classes and WVP for $\mathbf{\Sigma}_2$ classes are both equivalent to the existence of a proper class of strong cardinals. Perhaps one can obtain analogous results higher in the Levy hierarchy, like Bagaria did for VP. I don't know what the higher definable analogs of strong cardinals are.
Apr 21, 2019 at 23:07 comment added Trevor Wilson It doesn't seem to make much difference to the proof, so it's best to prove it for arbitrary classes in GBC. Then we obtain the following consequences as special cases: (1) In ZFC, WVP for definable classes is equivalent to "Ord is definably Woodin", and (2) for every inaccessible cardinal $\kappa$, WVP in $V_\kappa$ (for all classes that are elements of $V_{\kappa+1}$) is equivalent to "$\kappa$ is a Woodin cardinal."
Apr 21, 2019 at 23:02 history edited Trevor Wilson CC BY-SA 4.0
Added proof of converse implication
Apr 21, 2019 at 22:38 comment added Asaf Karagila Now, when you say class, I'm guessing you mean definable, or are we talking about a hierarchy of strength here by stronger and stronger second-order systems?
Apr 21, 2019 at 20:13 history edited Trevor Wilson CC BY-SA 4.0
deleted 243 characters in body
Apr 21, 2019 at 20:06 history answered Trevor Wilson CC BY-SA 4.0