# Can $\omega_1$ be supercompact?

Is "ZF + $\omega_1$ is supercompact" consistent relative to "ZFC + there is a supercompact cardinal"?

In particular, if $\delta$ is supercompact, does it remain so in $V(\mathbb{R} \cap V[G])$ where $G \subset Col(\omega,<\delta)$ is $V$-generic? This seems to be the case for measurability but I am having trouble proving it for supercompactness. It seems likely that someone else has tried this, so I though I'd ask here.

The appropriate definition of supercompactness in ZF is the one in terms of normal fine measures, where normality is defined using diagonal intersections.

I am aware that $\omega_1$ has some amount of supercompactness under AD. I am interested in a more direct proof using forcing, which I hope will give (full) supercompactness.

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– Asaf Karagila Aug 3 '12 at 20:31
Ah, I thought I remembered a question like that on here but I couldn't find it. Sorry for posting a similar question. – Trevor Wilson Aug 3 '12 at 20:50
Trevor, I think that my question may be a bit overly broad. It is just like the time I asked about the ability to destroy weak choice principles (or add some of them) via generic extensions (unlike full choice), and later Stefan Geschke asked about a concrete example. It's a reasonable question when trying to tackle a broad question, I think. – Asaf Karagila Aug 3 '12 at 20:54
The Jech construction preserves supercompactness. This is Lemma 1.3 in Apter-Henle, Large cardinal structures below $\aleph_{\omega}$. – tci Aug 13 '12 at 12:50
@TanmayInamdar: Great! Could you post that as an answer? (Maybe including the statement of the relevant theorem?) – Trevor Wilson Aug 13 '12 at 15:43

Thanks, that mentions a way to extend an ultrapower of Ord taken under ZF to an ultrapower of $V$ in a class generic extension. This would be a useful thing to do if $\omega_1$ were supercompact. It looks like the only example given of supercompactness without choice given in the article comes from AD though. – Trevor Wilson Aug 3 '12 at 20:45
Trevor, I had several chats over a cup of coffee with Magidor on the problem of an inner model for supercompactness. If you think about Solovay's model it has an $L(\mathbb R)$ sort of construction, and in a sense if we make $\aleph_1$ measurable we essentially take some inner model of measurability. However supercompactness does not yet have such canonical inner model, so I don't think that it's that easy to construct a model of $\aleph_1$ being supercompact for all $\lambda$. – Asaf Karagila Aug 3 '12 at 20:51
I was hoping that the inner model would just be $V(\mathbb{R} \cap V[G])$. Doesn't this work for measurability? Every subset of $\delta$ in this model is added by some proper initial segment of $G$. So we have a measure on $\delta$ given by the union of the small-forcing extensions of the original measure. This is countably complete, again because every countable sequence is added by a proper initial segment of the forcing. But if there is a countably complete measure on $\delta$ then there is a normal measure. – Trevor Wilson Aug 3 '12 at 21:00
...This argument fails to adapt to supercompactness in two ways: first, the set being measured is not added by any proper initial segment of the forcing, and second, normality does not come for free with $\mathcal{P}_\kappa(\lambda)$. Could you explain how the question relates to the inner model problem for supercompact cardinals? – Trevor Wilson Aug 3 '12 at 21:03