Normal measures and Elementary Embeddings - MathOverflow

Normal measures and Elementary Embeddings

2010-10-28T17:33:35Z

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.

Answer by Stefan Hoffelner for Normal measures and Elementary Embeddings

2010-10-28T21:29:10Z

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 $2^{\kappa} \le (2^{\kappa})^{M} \le (\aleph_{\kappa + \beta})^{M} \le \aleph_{\kappa + \beta}$

Answer by Amit Kumar Gupta for Normal measures and Elementary Embeddings

2010-10-29T01:48:11Z

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:

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

i.e.

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

i.e.

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

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 $M \subseteq V$.

Answer by Eran for Normal measures and Elementary Embeddings

2010-11-04T08:58:28Z

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.