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This question is motivated by the papers http://arxiv.org/abs/1405.7456 and http://arxiv.org/abs/1410.1224.


Say that a set $X$ is "generically presentable" over $V\models ZF$ if there is some forcing $\mathbb{P}\in V$ such that $X\in V[G]$ whenever $G$ is $\mathbb{P}$-generic over $V$. More precisely, we say that a $\mathbb{P}$-name $\nu$ is a "generically presentable set" if $$\Vdash_{\mathbb{P}^2}\nu[G_0]=\nu[G_1].$$

It's natural to ask: "What are the generically presentable sets?" It turns out this has a very simple answer: the only generically presentable sets are those in $V$ already. (Specifically, if $\nu\in V^\mathbb{P}$ is a generically presentable set, then there is some $A\in V$ such that $\Vdash_\mathbb{P}A=\nu[G]$.) This was proved by Solovay (see Who proved "sets in every generic are already in the ground model?").

Of course, this only answers the question of generically presentable sets up to equality. In general, we could talk about "generically presentable [things] up to [equivalence relation]." For example, we could talk about countable structures which are generically presentable up to isomorphism: $\nu\in V^\mathbb{P}$ is a generically presentable structure if $$\Vdash_\mathbb{P}\mbox{"$\nu$ is a structure with domain $\omega$"}\quad\mbox{and}\quad\Vdash_{\mathbb{P}^2}\nu[G_0]\cong\nu[G_1].$$ Now things get more interesting: in general, just because a structure is generically presentable, it does not follow that the structure must have a copy in $V$. (This is proved in each of the two papers cited above.)

My question is about "generically presentable sets up to equinumerosity," that is, "generically presentable cardinalities." It is a basic fact about forcing that every cardinal in $V[G]$ is a cardinal in $V$, so forcing cannot add new cardinalities. However, this breaks down if $V$ is not a model of choice: we may have a set in $V[G]$ with no bijection (in $V[G]$) with any set in $V$.

This raises a general question: "How nice can the cardinalities added by forcing be?" In particular, are there any "definable" ones?

More precisely, fix a forcing notion $\mathbb{P}$ and say that a $\mathbb{P}$-name $\nu$ is a "generically presentable cardinality" if $$\Vdash_{\mathbb{P}^2}\nu[G_0]\equiv\nu[G_1].$$ (Here "$\equiv$" means "is equinumerous with.") We now ask:

Is there a $V\models ZF$, a $\mathbb{P}\in V$, and a generically presentable cardinality $\nu\in V^\mathbb{P}$, such that for every $X\in V$, $$\mathbb{P}\Vdash X\not\equiv\nu[G]?$$

EDIT: in the other direction, what are some axioms - weaker than $AC$, such as $SVC$ - which imply that this can't happen?


Technical observation: "sets up to equality" and "countable structures up to isomorphism" are both better behaved than "sets up to equinumerosity." In particular, forcing can make two non-equinumerous sets have the same size. This means that, while in the first two cases the specific forcing used to present an object doesn't matter too much, we really should say "$\mathbb{P}$-generically presentable cardinality" to be perfectly clear.

In particular, we have an interesting side question:

Fix $V\models ZF$. Is there a snappy characterization of those forcing notions $\mathbb{P}\in V$ such that there is a generically presentable cardinality $\nu\in V^\mathbb{P}$ with $\forall X\in V(\mathbb{P}\Vdash \nu\not\equiv X)?$

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  • $\begingroup$ This is just a "half-baked" thought: if $V$ satisfies SVC then one could force AC and that would make the generic cardinality "uninteresting". So, I guess, if there is such a thing then AC must fail "really badly" in $V$. $\endgroup$ – François G. Dorais Jan 7 '15 at 2:40
  • $\begingroup$ It's always possible for the generic cardinality to be made "uninteresting": just collapse it to $\omega$. (This is what I'm getting at in the paragraph mentioning bad behavior of equinumerosity.) I'm not sure we can conclude even that no model of SVC admits a "nonexistent generic cardinality," but maybe I'm missing something. $\endgroup$ – Noah Schweber Jan 7 '15 at 2:41
  • $\begingroup$ Yeah, that's the "half-baked" part. The thing is that with SVC you can force AC for reasons totally unrelated to the "generic cardinal" and that seems like an issue. I haven't thought this through much so I'm not even convincing myself with this idea but it's not outright ridiculous... $\endgroup$ – François G. Dorais Jan 7 '15 at 2:58
  • $\begingroup$ @François: But do you have to force back the axiom of choice? No, you don't have to force it back. $\endgroup$ – Asaf Karagila Jan 15 '15 at 19:32

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