# Another question on saturated models of ZFC

Let $M$ be a saturated model of ZFC and let $\kappa$ be any cardinal in $M$. Now let \begin{align*} p_\kappa(f) &= \{ “f \text{ 1-1 function”} \wedge \operatorname{dom}(f) \subseteq \omega \wedge \operatorname{ran}(f) \subseteq \kappa \} \\ &\qquad\cup \{n \in \operatorname{dom}(f) : n \in \omega \} \cup \{ \alpha \in \operatorname{ran}(f) : \alpha \in \kappa \}. \end{align*} $p_\kappa(f)$ is fin. realized in $M$ and therefore it is realized in $M$ (since $M$ is saturated and $\kappa < \operatorname{card}(M)$). So there exist $f_\kappa : \omega \to \kappa$ that is $1-1$ and onto, which of course is a contradiction. What am I doing wrong?

Note : This is question is related to another question that I asked earlier, "A question on saturated models of ZFC". I don't know if I should merge the two questions.

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Does the condition "$f$ is 1-1 function" state that $f$ is a bijection? –  user17090 Oct 31 '11 at 11:05
@Ali: $f$ is a bijection between its domain and range (= image), but those aren’t specified. –  Brian M. Scott Oct 31 '11 at 11:08

The problem with the argument you propose is that your type has too many parameters. When you say that $M$ is saturated, then you mean that it has some cardinality $\delta=|M|$, and any finitely consistent type using fewer than $\delta$ many parameters is realized in the model. Your type uses as parameters all the nonstandard natural numbers $n$ of $\omega$, meaning the $\omega$ of $M$ and so this includes all the nonstandard natural numbers of $M$, as well as the ordinals $\alpha$ that $M$ thinks are less than $\kappa$, which you fixed.
For your argument to work, you need that this amounts to fewer than $\delta$ many parameters. But it is not. In fact, if $M$ is saturated, then there is an easier type showing that $M$ must have $|M|$ many nonstandard natural numbers. The reason is that otherwise we can write down the type asserting that $x$ is a natural number not among them, and this type will be finitely realized by not realized in $M$.
Meanwhile, you type is indeed consistent, and it is realized in another model. In fact, it is realized in the forcing extension of $M$ collapsing $\kappa$ to become countable.