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• CommentRowNumber1.
• CommentAuthorDavidRoberts
• CommentTimeNov 8th 2012
• (edited Nov 8th 2012)

I’d like to announce the arrival of the pre-preprint $Con(ZF+\neg WISC)$, available here. Abstract:

By considering a variant on forcing using a symmetric model for a proper class-sized group, we show that the very weak choice principle WISC—the statement that there is at most a set of incomparable surjections onto every set—is independent of the rest of the axioms of set theory, in particular those of ZF. Our result applies to any set theory which gives rise to a well-pointed boolean topos with nno. The proof does not rely on the axiom of choice, nor does it make any large cardinal assumptions.

• CommentRowNumber2.
• CommentAuthorTobyBartels
• CommentTimeNov 13th 2012

Maybe this is clear if I read the paper, but this doesn't sound right:

Our result applies to any set theory which gives rise to a well-pointed boolean topos with nno.

Surely ZF + WISC is a set theory that gives rise to a well-pointed boolean topos with NNO, but WISC is not independent of ZF + WISC.

• CommentRowNumber3.
• CommentAuthorTobyBartels
• CommentTimeNov 13th 2012

Well, looking at the first few pages, you seem to mean that given any well-pointed boolean topos $S$ with NNO, you construct a boolean $S$-topos $E$ in which WISC fails (and there's something about local connectedness in there too); but that doesn't mean that given a set theory $T$ and a (or the free?) model $S$ of $T$ that this $E$ is a model of $T$. Whereas, presumably you do want that $E$ is a model of ZF.

So rather than

Our result applies to any set theory which gives rise to a well-pointed boolean topos with nno.

it should be

Our result applies to any well-pointed boolean topos with nno.

(Or so it appears based on reading three pages!)

• CommentRowNumber4.
• CommentAuthorDavidRoberts
• CommentTimeNov 13th 2012

Ok, really it should mean something like ’relatively consistent with’. My aim is to show that $Con(ZF) \Rightarrow Con(ZF + \neg WISC)$, or possibly with $Con(ZFC)$ instead of $Con(ZF)$, but I think the LHS can be replaced with $Con(SEAR[-C])$ with the result being $Con(SEAR + \neg WISC)$, and ditto for $ETCS$ (by the way, what do people call $ETCS$ without AC?)

In fact (thanks to comments from Mike) I know I haven’t quite got the material set theory version yet, but I think the structural version is complete (I haven’t got the structural$\to$material translation done).

So your last comment is correct.

• CommentRowNumber5.
• CommentAuthorTobyBartels
• CommentTimeNov 13th 2012
• (edited Nov 13th 2012)

[never mind]

• CommentRowNumber6.
• CommentAuthorTobyBartels
• CommentTimeNov 13th 2012
• (edited Nov 13th 2012)

by the way, what do people call $ETCS$ without AC?

There's $CETCS$ (constructive ETCS), but that's missing more. Maybe ‘$ETS$’ (^_^)?

• CommentRowNumber7.
• CommentAuthorDavidRoberts
• CommentTimeNov 13th 2012

There’s CETCS (constructive ETCS), but that’s missing more.

I definitely need classical logic for the current proof to work…

• CommentRowNumber8.
• CommentAuthorDavidRoberts
• CommentTimeNov 16th 2012
• (edited Nov 16th 2012)

After Mike’s comments alluded to above, I have resurrected the following material set theory version:

If ZF is consistent, then so is ZFA$+ (\exists \text{ a proper class of atoms}) + \neg WISC$

I’m still working on the independence from $ZF$.

So as it stands, I have only given a pre-Cohen-style independence result, but as a corollary, I now understand the so-called Fraenkel model: it is the material set theory arising from the topos of sets with an action of an open subgroup of $(\mathbb{Z}/2)^\mathbb{N}$ for a certain natural topology on this group. Open subgroups are the finite-index subgroups $\prod_{i\in I} H_i\times (\mathbb{Z}/2)^{\mathbb{N} - I}$ for finite $I\subset \mathbb{N}$ and $H_i \le \mathbb{Z}/2$. Arrows in this topos are allowed to be equivariant for an open (possibly proper) subgroup of the groups acting on the domain and codomain.

I have put the above description at Fraenkel model of ZFA.