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Consistently with $\mathsf{ZFC}$ there is a forcing which preserves cardinals but whose square does not always preserve cardinals - that is, some $\mathbb{P}$ such that for every $\mathbb{P}$-generic $G$ we have $\mathrm{Card}^{V}=\mathrm{Card}^{V[G]}$ but for some $\mathbb{P}^2$-generic $H=\langle H_0,H_1\rangle$ we have $\mathrm{Card}^{V}\not=\mathrm{Card}^{V[H]}$.

However, the only way I know how to get this is via a bit of a cheat: find two different forcings which are individually "good" but have "bad" product, and then look at their lottery sum. This construction has the drawback that the square of the resulting forcing doesn't always collapse cardinals - we're only guaranteed "bad" behavior in the extension if the two coordinates of our generic lie on different "sides" of the original lottery sum.

I recall$^*$ seeing a stronger example of this phenomenon, but I can't track it down or reconstruct it on my own:

Is it consistent with $\mathsf{ZFC}$ that there is a forcing $\mathbb{P}$ such that $\mathbb{P}$ preserves cardinals but $\Vdash_{\mathbb{P}^2}\mathrm{Card}^V\not=\mathrm{Card}^{V[\langle H_0,H_1\rangle]}$?


$^*$Actually my original memory was that even the weaker phenomenon can't happen, but after it was pointed out to me that it can, I now remember differently. I'm sure eventually my memory will have been right. :P

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  • $\begingroup$ Consistently? Sure. Suslin trees are ccc. $\endgroup$ – Asaf Karagila Sep 6 '20 at 23:09
  • $\begingroup$ @AsafKaragila Does forcing with the square of a Suslin tree (always) collapse cardinals? I wasn't aware of that. $\endgroup$ – Noah Schweber Sep 6 '20 at 23:57
  • $\begingroup$ I believe so, because $T\times T$ will collapse $\omega_1$. But now I'm not sure. But I'm pretty sure that it is at least consistently the case. $\endgroup$ – Asaf Karagila Sep 7 '20 at 0:02
  • $\begingroup$ @AsafKaragila Yeah, I don't see that $T\times T$ need collapse $\omega_1$. I wouldn't be surprised if it were consistently the case, but I don't see it. $\endgroup$ – Noah Schweber Sep 7 '20 at 0:03
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    $\begingroup$ It seems obligatory to give the example of Todocevic square: if $\mathbb{S}$ adds a generic $\square(\kappa)$ for $\kappa$ regular, and $\mathbb{T}$ is the threading forcing, then $\mathbb{S} \ast \mathbb{T}$ is equivalent to the $\kappa$-Cohen forcing while $\mathbb{S} \ast (\mathbb{T}^2)$ collapses $\kappa$ to be countable. $\endgroup$ – Yair Hayut Sep 7 '20 at 8:37
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A self-specializing Souslin tree gives you a ccc notion of forcing whose square collapses $\omega_1$ (See, e.g., the answer to Ultrafilters preserved by $\mathbb{P}$ but not by products?). Such trees exist under $\diamondsuit$, and so are consistent with ZFC.

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One can indeed prove more: starting with $V=L$, there exists a tame and cardinal preserving class forcing notion $\mathbb{P}$ such that forcing with $\mathbb{P} \times \mathbb{P}$ collapses all uncountable cardinals. See

  • Adam Figura, Collapsing algebras and Suslin trees, Fundamenta Mathematicae 114 (1981) 141-147, doi:10.4064/fm-114-2-141-147.
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  • $\begingroup$ Very cool! (I've made a minor clarifying edit.) $\endgroup$ – Noah Schweber Sep 7 '20 at 7:22

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