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This is a question from Jech's Set Theory (Ex. 17.12) which I'm reading at the moment and pretty much stuck on.

If $D$ is a normal measure on $\kappa$ and $\{ \aleph_\alpha \colon > 2^{\aleph_\alpha} \le > \aleph_{\alpha+\beta}\} \in D$ (for some constant $\beta < \kappa$), then $2^\kappa > \le \aleph_{\kappa + \beta}$

He gives the following hint: If $f$ is such that $f(\aleph_\alpha) = \aleph_{\alpha+\beta}$ for all $\alpha < \kappa$, then $[f]_D = (\aleph _{ \kappa+j(\beta)})^M$

I think that I am just confused about the whole representation in $M$ and how to use it to solve this problem. Hints, partial or complete solutions are most welcomed.

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I have to run and teach so don't have time to give a full solution or check what I came up with, but here's a start I guess: Let $g(\aleph _{\alpha}) = 2^{\aleph _{\alpha}}$ and let $f$ be as in the hint. The set where $f$ and $g$ agree is in $D$, thus $\kappa$ belongs to the set where $j(f)$ and $j(g)$ agree. So $j(f)(\kappa)= j(g)(\kappa)$, and so $[f] = [g]$. Since $\kappa$ is measurable it's inaccessible, so $\beta < \kappa$, so $j(\beta) = \beta$, so $[f] = \aleph _{\kappa + \beta}$. –  Amit Kumar Gupta Oct 28 '10 at 17:52
What do you mean by "the whole representation in $M$"? What is missing beyond Amit's sketch? (There may be several potentially confusing points: How Los's lemma is used, or the ultrapower construction in itself, or Mostowski collapsing lemma...) –  Andrés Caicedo Oct 28 '10 at 20:34
I think that I didn't fully understand the idea of some function $f$ such that $[f] = \alpha$ for some ordinal $\alpha$. The so-called confusing points which you mention are actually fairly clear to me. –  Asaf Karagila Oct 28 '10 at 20:55
Asaf, is what you mention about $f$ clear now? –  Andrés Caicedo Oct 28 '10 at 21:13
Andres, I think so. It means that $f(\gamma) = \alpha$ for almost all $\alpha < \kappa$, right? –  Asaf Karagila Oct 28 '10 at 21:44

3 Answers 3

up vote 5 down vote accepted

The question you've stated isn't the question in Jech, you've made a minor typo. Here's the actual problem:

If $\beta < \kappa$ and {$\aleph _{\alpha} : 2^{\aleph _{\alpha}} \leq \aleph _{\alpha + \beta}$} $\in D$ and $D$ is a normal measure on $\kappa$, then $2^{\aleph _{\kappa}} \leq \aleph _{\kappa + \beta}$

Note that since $\kappa$ is measurable, $\aleph _{\kappa} = \kappa$.

Okay, now we know that a normal measure extends the club filter, and the set of cardinals below $\kappa$ is club in $\kappa$, hence it makes sense in the hint to define $f(\aleph _{\alpha}) = \aleph _{\alpha + \beta}$ without specifying how $f$ acts on non-cardinals. Following my comment, let $g(\aleph _{\alpha}) = 2^{\aleph _{\alpha}}$. Then $g \leq f$ almost everywhere, and so:

$M \vDash [g] \leq [f]$


$M \vDash j(g)(\kappa) \leq j(f)(\kappa)$


$M \vDash 2^{\kappa} \leq \aleph _{\kappa + j(\beta)}$

Since $\beta < \kappa$, $j(\beta) = \beta$. Thus there is an injection from $(2^{\kappa})^M$ to $\aleph _{\kappa + \beta} ^M$. Since $P(\kappa) = P^M(\kappa)$, it means there's an injection from $2^{\kappa}$ to $\aleph _{\kappa + \beta}^M$. Finally, $\aleph _{\kappa + \beta} ^M \leq \aleph _{\kappa + \beta}$ since $M \subseteq V$.

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I assumed that if someone knows what a measurable cardinal is, he'll know that it's a fixed-point of the Aleph function. :) Many thanks. –  Asaf Karagila Oct 29 '10 at 6:00
I think you do need to say explicitly that $\beta\lt\kappa$, because otherwise the hypothesis is free when $\beta=\kappa$, but the conclusion may fail when $\beta=\kappa$. That is, if $\beta=\kappa$, then $2^{\aleph_\alpha}\leq \kappa\leq\aleph_{\alpha+\kappa}$, but it could happen that $2^\kappa=2^{\aleph_\kappa}\gt \aleph_{\kappa+\kappa}$. For example, one can force this over a model with a supercompact cardinal (but it has weaker strength). –  Joel David Hamkins Nov 4 '10 at 16:26
You're absolutely right, I'll edit my response. I think I was thinking the condition was that $2^{\aleph _{\alpha}} = \aleph _{\alpha + \beta}$ (that might've been how the question was originally (erroneously) stated. –  Amit Kumar Gupta Nov 4 '10 at 20:58

I just wanted to fix my answer, which I couldn't do yesterday as it was already midnight and I was too tired (nevertheless the answer already given by Amit is elegant and true)

As $D$ is normal $\kappa$ is represented in $M \cong Ult_{D} (V)$ by the diagonal function $ d: \kappa \to \kappa$, and as $\kappa$ is measurable, each element of $M$ is already determined by a function defined only on the cardinals below kappa.

Now if $x \in P(\kappa)^{M}$ then there exists a function $h: \kappa \to V$ such that $ x = h_{D}$, and as $M \models h_{D} \subset \kappa$ it follows that {$\aleph_{\alpha} < \kappa : h (\aleph_{\alpha}) \subset \aleph_{\alpha}$} $\in D$. Thus $M \models P(\kappa) \subset g_{D}$ where $g_D$ denotes the equivalence class of the function $g: \aleph_{\alpha} \to P(\aleph_{\alpha})$. This leads us to $M \models |P(\kappa)| \le |g_{D}|$. But the cardinal $|g_{D}|$ is represented by the function $f: \aleph_{\alpha} \to 2^{\aleph_{\alpha}}$.

Invoking the hint we may conclude $$M\models 2^{\kappa} \le f_{D} \le \aleph_{\kappa + \beta}$$ and as $P(\kappa)^{M} = P(\kappa)$ we finally have $2^{\kappa} \le (2^{\kappa})^{M} \le (\aleph_{\kappa + \beta})^{M} \le \aleph_{\kappa + \beta}$

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Thanks for the answer, firstly I'd like to stress that in my re-reading the question on Jech I found several mistakes in my question here and corrected them, one of those was weakening the equality to a weak inequality. Secondly, my biggest problem with this question is that the intended result is not to have some transitive model in which $2^\kappa = \aleph_{\kappa + \beta}$ but rather the original model. –  Asaf Karagila Oct 28 '10 at 21:49

Please compare with Lemma 17.11. It has all the details for the case $\beta = 1$. From there it should be easy to infer the general case.

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