I am trying to solve Exercise IV.3.31 from Kunen's Foundations of Mathematics. I think I have a solution but I am confused by one of the hints.

By request, here is the text of the exercise.

Exercise IV.3.31 Prove the following version of the Mostowski Collapsing Lemma: Let $A$ be any set and assume that the $\in$-model built on $A$ (as in Section II.17) satisfies the Axiom of Extensionality. Then there is a transitive set $T$ such that $(A;\in) \cong (T;\in)$. Furthermore, the isomorphism $F : A \overset{\text{1-1}}{\underset{\text{onto}}{\to}} T$ has the property that $F(x)=x$ for all $x\in A$ such that $\operatorname{trcl}(x) \subseteq A$.

Conclude from this that there is a countable transitive $T \vDash ZFC-P$ such that $\approx$ is not absolute for $T$ (see the discussion on page 165 after Notation II.17.12).

Hint. First define $F(x) = \{F(v) : v \in A \cap x\}$; this is a recursion on $\in$; see Exercise I.14.24. Then, prove that $F \restriction A$ is an isomorphism from $A$ onto $T := \operatorname{ran}(F \restriction A)$. You use the Axiom of Extensionality in $A$ to prove that $F \restriction A$ is 1-1. As an example, if $A = \{0,1,2,\omega,\{\omega\}\}$, then $T = \{0,1,2,3,\{3\}\}$ with $F(\omega)=3$ and $F(2)=2$.

For the "Conclude from", use the Downward Löwenheim-Skolem-Tarski Theorem II.16.6 to get a countable $A \preccurlyeq H(\aleph_3)$, and then get $T \cong A$.

Notation: $\operatorname{trcl}(x)$ is the transitive closure of $x$ (with respect to $\in$); $P$ is the Power Set axiom; $x \approx y$ means "there exists a bijection from $x$ to $y$"; $H(\kappa)$ is the set $\{ x : |\operatorname{trcl}(x)| < \kappa\}$, where $\kappa$ is a cardinal.

I am fine with the first part of the question (the Collapsing Lemma) using the hint.

Here's my solution to the second part.

Downward Löwenheim-Skolem-Tarski provides a countable elementary submodel $A \preccurlyeq H(\aleph_3)$. We have previously shown (Exercise I.14.12) that for any regular cardinal $\kappa > \omega$, $H(\kappa) \vDash ZFC-P$, so the same is true for $A$. In particular, $A$ satisfies Extensionality, so by the first part it is isomorphic to a transitive model $T$. Of course $T$ is also countable and $T \vDash ZFC-P$.

Now let us show $\approx$ is not absolute in $T$. Let $|x| \ge \omega$ be the formula "there exists an injection from $\omega$ to $x$". Now $H(\aleph_3)$ contains uncountable sets, so we have $H(\aleph_3) \vDash \exists b (|b| \ge \omega \land \lnot b \approx \omega)$. $(\omega$ is definable so it is kosher to use it in this formula.) Therefore, we also have $T \vDash \exists b (|b| \ge \omega \land \lnot b \approx \omega)$. Let $b \in T$ be one of the sets whose existence is thus asserted; then $T$ contains an injection from $\omega$ to $b$, so indeed $b$ is infinite in "real life". To say it another way, the formula $|b| \ge \omega$ is $\Sigma_1$ (right?) so it remains true in any transitive superset of our transitive set $T$ (this is Kunen's Lemma II.17.24); in particular $V \vDash |b| \ge \omega$. But on the other hand, since $T$ is transitive we have $b \subset T$, and $T$ is countable, so in "real life" $b$ is countable, i.e. $V \vDash b \approx \omega$. However $T \vDash \lnot b \approx \omega$, so we have shown $\approx$ is not absolute in $T$.

My questions are:

  1. Does this solution look correct?

  2. Why did we have to start with $H(\aleph_3)$? As far as I can tell, all we used was that $H(\aleph_3)$ contains an uncountable set $b$ and an injection from $\omega$ to $b$, so we could just as easily have used $H(\aleph_2)$ (taking for instance $b = \omega_1$). This makes me wonder if I have missed something.

  • $\begingroup$ For those of us which don't have the book available, could you please add the exact exercise? $\endgroup$ – Asaf Karagila Jun 10 '14 at 7:00
  • $\begingroup$ @AsafKaragila: Done! $\endgroup$ – Nate Eldredge Jun 10 '14 at 12:31
  • 1
    $\begingroup$ Your argument is fine. I have no idea why Kunen wanted to work with $H_{\aleph_3}$. The only thing that comes to mind is that you could take your $A$ to contain $\aleph_1$ and $\aleph_2$, so $T$ would think their collapses have those cardinalities but they are really countable. But this is a really flimsy reason. $\endgroup$ – Miha Habič Jun 11 '14 at 2:17

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