# What is the large cardinal strength of the assertion that every $\kappa$-complete filter on $\kappa$ extends to a $\kappa$-complete ultrafilter?

It is well-known that an uncountable regular cardinal $\kappa$ is strongly compact if and only if every $\kappa$-complete filter on any set extends to a $\kappa$-complete ultrafilter on that set. The usual proof of this, in the one direction, uses $\theta$-strong compactness to handle filters of size $\theta$, even when those filters concentrate on base sets of size less than $\theta$. My question concerns the nature of the limitation of the size of the base set in this equivalence, and in particular, what is the strength of the assumption in the case of filters on $\kappa$ itself.

Question. What is the large cardinal strength of the assumption that $\kappa$ is an uncountable regular cardinal for which every $\kappa$-complete filter on $\kappa$ extends to a $\kappa$-complete ultrafilter on $\kappa$?

For a lower bound, every such $\kappa$ is easily seen to be a measurable cardinal, since the club filter on $\kappa$ is $\kappa$-complete and so the property gives a measure on $\kappa$.

For an upper bound, if $\kappa$ is $2^\kappa$-strongly compact, then the property holds by the usual characterization of strong compactness I alluded to above. Namely, if $F$ is a $\kappa$-complete filter on $\kappa$, then let $j:V\to M$ be a $2^\kappa$-strong compactness embedding. Since $F$ has size at most $2^\kappa$, the strong compactness cover property ensures that there is some $s\in M$ with $j^{\prime\prime}F\subset s$ and $|s|^M<j(\kappa)$. We may assume $s\subset j(F)$, and so $\bigcap s\in j(F)$ by $j(\kappa)$-completeness in $M$. Pick any $\alpha\in \bigcap s$, and it follows that $F\subset\mu$ where $X\in\mu\leftrightarrow\alpha\in j(X)$, which the standard arguments show is a $\kappa$-complete ultrafilter on $\kappa$.

So the property is trapped between $\kappa$ being measurable and $\kappa$ being $2^\kappa$-strongly compact.

Further refined questions would be:

• If there is a cardinal $\kappa$ with the property, then can one undertake the construction of inner models with stronger than measurable cardinals? For example, can one construct an inner model with a Woodin cardinal?

• Does the measurable cardinal in the canonical inner model $L[\mu]$ fail to have the property?

• Can one force an instance of a measurable cardinal without the property?

• Can one force a cardinal $\kappa$ to have the property, but not be $2^\kappa$-strongly compact?

I suspect that the answers to all these questions is affirmative.

More generally,

Question. What is the strength of the assumption that every $\kappa$-complete filter on a set of size $\theta$ extends to a $\kappa$-complete ultrafilter on that set?

As above, any $2^\theta$-strongly compact cardinal has this property, and this property implies that there are uniform $\kappa$-complete ultrafilters on every regular cardinal up to and including $\theta$, which by a result of Ketonen implies that $\kappa$ is strongly compact up to that degree. So when $\theta$ is regular, this property is trapped between $\theta$-strong compactness and $2^\theta$-strong compactness.

This question grew out of an issue arising in an answer by Noah S to a previous question on MO.

• Is it obvious that this is not just measurability? – Asaf Karagila Jun 27 '14 at 1:43
• No, it is conceivable it is just measurability, although I suspect it is stronger than this. – Joel David Hamkins Jun 27 '14 at 1:50
• Let me rephrase that. Why do you suspect it's stronger than measurability? – Asaf Karagila Jun 27 '14 at 1:56
• Just because the only way I know how to prove it is from $2^\kappa$-strong compactness, which is very strong. – Joel David Hamkins Jun 27 '14 at 2:07
• In particular, the lower bound of measurability is within the realm of the inner model theory, but the upper bound of $2^\kappa$-strong compactness is to my knowledge quite exceeding that realm. So the exact strength of the hypothesis is somewhat unsettled. – Joel David Hamkins Jun 27 '14 at 3:03

## 2 Answers

Let us call a cardinal $\kappa, \kappa-$compact, if every $\kappa-$complete filter on $\kappa$ extends to a $\kappa-$complete ultrafilter.

The following is proved in Gitik's paper On measurable cardinals violating the continuum hypothesis:

Theorem. If there exists a $\kappa-$compact cardinal, then there exists an inner model with a strong cardinal.

At the end of paper, it is asked as an open question, if the existence of a strong cardinal is enough?

Addition: Recently Gitik has improved his results. For example:

If there exists a $\kappa-$compact cardinal, then there exists an inner model with a Woodin cardinal.

See his recent paper "On $\kappa$-compact cardinals".

• Thanks very much! The main part of section 2 of that paper, entitled, "On the strength of $\kappa$-compactness," is devoted to this question. He proves the result you mention, and then asks several additional questions: whether a strong cardinal is enough for $\kappa$-compactness; what is the strength of assuming only that every complete filter extending the club filter extends to an $\kappa$-complete ultrafilter or that every stationary set is in a $\kappa$-complete ultrafilter containing also the club filter. – Joel David Hamkins Jun 27 '14 at 13:00
• Have any of those questions (or my other questions) been answered since Gitik's paper? – Joel David Hamkins Jun 27 '14 at 13:19
• I don't know about the progress on questions, but it seems that the main problems are still open. – Mohammad Golshani Jun 27 '14 at 13:37

In my paper "Partial strong compactness and squares", I investigate those questions. The main results of this paper are:

• If every $$\kappa$$-complete filter on $$\kappa$$ can be extended to a $$\kappa$$-complete ultrafilter then $$\square(\kappa)$$ and $$\square(\kappa^+)$$ fail. The consistency strength of this assertion is not fully known (at least for $$\kappa > 2^{\aleph_0}$$), and the current lower bound is the existence of non-domestic mouse which implies the consistency of $$\mathrm{AD}_{\mathbb{R}}$$. In the paper "Equiconsistencies at subcompact cardinals", Neeman and Steel prove that under some inner model theoretical assumptions the failure of square on two consecutive cardinals, where the lower one is large (Woodin) implies that there is an inner model with ($$\Pi^2_1$$-)subcompact cardinal.
• On the other hand, less than $$2^\kappa$$-supercompactness of $$\kappa$$ is sufficient in order to get that every $$\kappa$$-complete filter on $$\kappa$$ can be extended to a $$\kappa$$-complete ultrafilter (indeed, Neeman-Steel's $$\Pi^2_1$$-subcompact is enough). Thus in a model of $$\mathrm{GCH}$$ and level-by-level equivalence between strong compactness and supercompactness, the least $$\kappa$$ which is $$\kappa$$-compact is strictly smaller than the least $$\kappa$$ which is $$\kappa^{+}$$-strongly compact.

The proof of the failure of squares from the filter extension property on $$\kappa$$ relies on showing that $$\kappa$$ being $$\kappa$$-compact is equivalent to the $$\kappa$$-compactness of $$\mathcal{L}_{\kappa,\kappa}$$ for theories of size $$\leq 2^\kappa$$ (namely, every $$\mathcal{L}_{\kappa,\kappa}$$-theory $$T$$ of size $$\leq 2^\kappa$$ such that every sub-theory $$T' \subseteq T$$, $$|T'| < \kappa$$, has a model, also has a model). This shows that the filter extension property lies in the hierarchy of partial strong compactness, strictly below $$2^\kappa$$-strong compactness of $$\kappa$$.

Similarly, for $$\theta$$ such that $$\theta^{<\kappa} = \theta$$, the filter extension property of $$\kappa$$-complete filters on $$\theta$$ is equivalent to $$\kappa$$-compactness of $$\mathcal{L}_{\kappa,\kappa}$$ for languages of size $$\leq 2^\theta$$.