MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This is going to take some build up to completely describe what is a very strange question I seem to have walked into by accident:

For every partial order $\mathbb{P}$ and regular cardinal $\lambda > \omega$ we can define the following two statements

$$ \mathcal{C}(\mathbb{P}, \lambda) \iff 1 \Vdash_{\mathbb{P}} \forall \alpha \in \check{\lambda}\ \forall f: \alpha \to \check{\lambda}\ \exists \gamma \in \check{\lambda}\ \forall \xi \in \alpha\ (f(\xi) \neq \gamma)$$

(this is the formalized version of the statement "$\mathbb{P}$ preserves $\lambda$ is a cardinal" in the forcing language, this statement is normally certified by reasoning which does not involve the forcing relation and depends on the structure of $\mathbb{P}$-names)


$$ Cof(\mathbb{P}, \lambda) \iff 1 \Vdash_{\mathbb{P}} \forall \alpha \in \check{\lambda}\ \forall f:\alpha \to \check{\lambda}\ \exists \gamma \in \check{\lambda} \ (\sup(ran(f)) \le \gamma) $$

(Again a forcing language version of the statement $\mathbb{P}$ preserves $\forall \alpha \in \lambda \ (cf(\alpha) < cf(\lambda))$: we had to be careful here because we need to be able to distinguish between the two (If this is not the correct way to formalize this please let me know.))

Now, here comes the question: Does the following conjunction:

$\exists \lambda > \omega\ \exists\ \mathbb{P}$ such that

  • $\lambda$ is a Regular cardinal.
  • $\vert \mathbb{P} \vert = \lambda^{+}$
  • $\forall \mu \ (\mu$ is a cardinal $\implies \mathcal{C}(\mu,\mathbb{P}))$
  • $\neg Cof(\lambda, \mathbb{P})$

Imply there is an inner model with a measurable cardinal? (changed based on the answers.)

(Namba for $\omega_2$ and threading a generic square collapse cardinals; moreover if $ 0^\sharp $ exists then $\aleph_\omega^{V}$ is regular in $L$ producing a model in some sense)


(It was not my intention to scare a lot of nice mice)

(also, mice need to be more damn direct and stop subtly hinting things.... didn't realize what was going on until just now....)


locked by Scott Morrison Nov 19 '12 at 16:43

closed as no longer relevant by Michael Blackmon, Andrés E. Caicedo, David Corwin, Igor Rivin, Todd Trimble Nov 19 '12 at 7:08

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

Being forced by a dense set and being forced by $1$ are equivalent, as long as your poset has a $1$. Your formalization of $Cof(\mathbb{P},\lambda)$ says that $\mathbb{P}$ forces (and hence preserves) that $\lambda$ is regular. That's not the same as $\forall \alpha \in \lambda (cf(\alpha) < \lambda)$, which is just always true: $cf(\alpha) \leq \alpha < \lambda$. – Amit Kumar Gupta Nov 15 '12 at 10:25
PLEASE do not deface your own questions. People have put work in answering it, and removing the text makes their effort go to waste. – Mariano Suárez-Alvarez Nov 19 '12 at 4:54
Recent editing activity and comments make me feel that the question should probably be closed now as "no longer relevant". – Todd Trimble Nov 19 '12 at 7:08
I don't know what exactly has happened here, but if Joel or Andreas answered one or more of your questions satisfactorily, it might be good to mark one of the answers as accepted, and leave the question in its most usable form. If you have other things to ask, it might be best to open a new question. – S. Carnahan Nov 19 '12 at 8:27
I think it is a very natural and interesting question. – Joel David Hamkins Nov 19 '12 at 10:26

Let me answer the question that I believe you are trying to ask. Namely, if we can make a regular cardinal $\kappa$ into a singular cardinal $\kappa$ by forcing of size at most $\kappa^+$, without collapsing any cardinals, must $\kappa$ be measurable?

The question is very natural, since Prikry forcing is the main way to do something like that, but it requires a measurable cardinal. Nevertheless, the answer is no.

The reason is that we can have a non-measurable cardinal that becomes measurable, and so the combined forcing of first making it measurable and then using Prikry forcing can exhibit your features. Specifically, it is consistent with ZFC (relative to the existence of a measurable cardinal) that there is a non-measurable cardinal $\kappa$ that becomes measurable in a forcing extension, by forcing to add a Cohen subset to $\kappa$. This is explained in my answer to Trevor Wilson's question Can measures be added by forcing? One can arrange in that argument that the GCH holds and that there are no other measurable cardinals.

So suppose that $V$ satisfies ZFC+GCH and there are no measurable cardinals in $V$, but $\kappa$ becomes measurable in $V[g]$, where $g$ was $V$-generic for the forcing to add a Cohen set $g\subset\kappa$. This does not collapse cardinals. Since $\kappa$ is measurable in $V[g]$, we may now perform Prikry forcing over $V[g]$ to add a Prikry sequence $s$, which changes the cofinality of $\kappa$ to $\omega$, while preserving all cardinals.

So in $V$, there were no measurable cardinals and $\kappa$ was regular, but the combined forcing to add $g\ast s$, forcing which has size $\kappa^+$ under the GCH, made $\kappa$ into a singular cardinal without collapsing any cardinals. Thus, this is a counterexample to the requested implication.

Meanwhile, although $\kappa$ is not measurable in $V$, it was measurable in an inner model of $V$, and this leads naturally to a closely related version of your question:

Question. If we can force a regular cardinal $\kappa$ to be singular with forcing of size at most $\kappa^+$ and without collapsing any cardinals, must there be an inner model with a measurable cardinal?

I don't know without further thought (although I recall having had conversations about this question). It seems likely that one might get $0^\sharp$ and perhaps much more out the hypothesis by combining the forcing with a collapse of $\kappa^+$, which would violate Jensen's theorem. We may have to wait for the inner model theory experts.

Joel, Thank you very much for the counter-example; this question has been bothering me for a couple of weeks, and for the life of me I couldn't figure out how to construct a forcing with these properties without first assuming either Con(ZFC) or \neg Con(ZFC); but as you seem to hint at, this is not at a problem because the extension of the forcing language to include the constant $\check{\lambda}$ for a fixed regular cardinal $\lambda$ is necessarily a more expressive language than that without constants. In particular forcing cheats way harder than I'd ever expected. – Michael Blackmon Nov 15 '12 at 18:01
(and makes me finally feel like I can sanely use forcing to produce consistency results again.) – Michael Blackmon Nov 15 '12 at 18:07
Oh no, don't get me wrong that is exactly where my intuition about forcing comes from and is why I was a bit worried, since proofs always carry more weight than intuitions. – Michael Blackmon Nov 15 '12 at 19:38

Here's an argument for an affirmative answer to Joel's modified version of the question. Suppose we have a forcing that preserves cardinals but singularizes some cardinal $\lambda$ that was regular in the ground model. Note that $\lambda$ had to be a limit cardinal, since otherwise singularizing it would collapse it down to its immediate predecessor cardinal (if not even lower). Now let $C$ in the forcing extension be a cofinal subset of $\lambda$ of smaller cardinality $\kappa$. I claim that C is not included in any set $D$ in the ground model of cardinality $\leq\max\{\kappa,\aleph_1\}$; in other words, I claim that $C$ is a counterexample to the assertion that the forcing extension satisfies the covering lemma over the ground model. Indeed, suppose we had such a $D$. Intersecting it with $\lambda$, we'd have a cofinal subset of $\lambda$ strictly smaller than $\lambda$ in the ground model, contrary to the assumption that $\lambda$ is regular in the ground model. ("Strictly smaller" in the preceding sentence uses that $\lambda>\aleph_1$, which is why I pointed out earlier that $\lambda$ has to be a limit cardinal.) So the forcing extension doesn't satisfy the covering lemma over the ground model. That implies the existence of an inner model with a measurable cardinal, by an ancient result of mine --- "Small extensions of models of set theory" in "Axiomatic Set Theory" (Proc. of 1983 Boulder Conference, edited by Baumgartner, Martin, and Shelah) Contemporary Math. 31 (1984) pp. 35-39.

Note that the size of the forcing notion doesn't matter here, as long as it's a set. – Andreas Blass Nov 15 '12 at 18:52
Great ! – Joel David Hamkins Nov 15 '12 at 19:13
Andreas, Thank you so much, I owe you a ladder system on \omega_1. – Michael Blackmon Nov 15 '12 at 22:56

Not the answer you're looking for? Browse other questions tagged or ask your own question.