# Finding $A \subseteq \omega_1$ such that $\omega_1^{L[A]} = \omega_1$ and $\omega_2^{L[A]} = \omega_2$

Exercise 13.31 of Jech's Set Theory says:

If $$\omega_2$$ is not inaccessible in $$L$$, then there exists $$A \subseteq \omega_1$$ such that $$\omega_1^{L[A]} = \omega_1$$ and $$\omega_2^{L[A]} = \omega_2$$.

Jech also specifically noted that $$\mathsf{AC}$$ is needed for this exercise.

If we weaken the statement to allowing $$A \subseteq \omega_2$$, then we may encode the order-type of all ordinals $$\alpha < \omega_2$$ into subsets of $$\omega_1$$, say $$A_\alpha$$, and by $$\mathsf{AC}$$ we obtain a set $$A$$ such that $$A_\alpha = \{\xi : (\alpha,\xi) \in A\}$$. Then one may prove that $$\omega_1^{L[A]} = \omega_1$$ and $$\omega_2^{L[A]} = \omega_2$$ in $$L[A]$$, and convert $$A$$ to a subset of $$\omega_2$$ via the canonical well-ordering. Here we did not use the fact that $$\omega_2$$ is not inaccessible in $$L$$.

The following is my guess on the approach to obtain an $$A \subseteq \omega_1$$ instead: Since $$\omega_2$$ is not inaccessible in $$L$$, there exists a cardinal $$\kappa$$ such that $$(\omega_2 \leq 2^{\kappa^L})^L$$. Since $$\kappa^L$$ is an ordinal of cardinality $$\leq \aleph_1$$, we may perhaps be able to "collect" all the $$A_\alpha$$'s above using subsets of $$\kappa^L \subseteq \omega_1$$, so that $$\omega_1^{L[A]} = \omega_1$$ and $$\omega_2^{L[A]} = \omega_2$$. But I'm not sure if this is the right direction to go.

• First note that, since GCH holds in $L$, $\omega_2$ (of $V$) must be a successor cardinal of $L$, say the successor of $\kappa$. So, in $V$, $\kappa$ has cardinality $\aleph_1$ and can therefore be coded by a subset of $\omega_1$. Include that in your $A$. Apr 30, 2022 at 0:31

With the help of Andreas Blass' comment I was able to solve the exercise. I will post the full solution here.

Lemma (Exercise 13.28). Let $$\alpha \geq \omega$$ be a countable ordinal. There exists $$A \subseteq \omega$$ such that $$\alpha$$ is countable in $$L[A]$$.

Proof. Let $$f : \omega \to \alpha$$ be one-to-one and onto. Define $$W \subseteq \omega \times \omega$$ by stipulating that $$n \, W \, m$$ iff $$f(n) \in f(m)$$. Then $$W$$ is a well-ordering of order-type $$\alpha$$. Since "$$W$$ is a well-ordering of order-type $$\alpha$$" is $$\Delta_0$$ relative to $$\alpha$$, we have that $$L[W] \models$$ "$$W$$ is a well-ordering of order-type $$\alpha$$", so $$L[W] \models \alpha$$ is countable.

Let $$A = \Gamma(W) \subseteq \omega$$, where $$\Gamma$$ is the canonical well-ordering of $$\mathbf{ORD}^2$$ (recall that $$\Gamma(\omega_\alpha \times \omega_\alpha) = \omega_\alpha$$ for all ordinals $$\alpha$$). Since $$\Gamma$$ is definable without parameters, we have that $$A \in L[W]$$ and $$W \in L[A]$$. Thus $$L[A] = L[W]$$. Note that $$\mathsf{AC}$$ is not used in this proof. $$\square$$

Lemma (Exercise 13.30). There exists $$A \subseteq \omega_1$$ such that $$\omega_1 = \omega_1^{L[A]}$$.

Proof. By the above lemma, for each $$\alpha < \omega_1$$ there exists $$A_\alpha \subseteq \omega$$ such that $$\alpha$$ is countable in $$L[A_\alpha]$$. Let $$F$$ be the function $$\alpha \mapsto A_\alpha$$ (this is a choice function, so $$\mathsf{AC}$$ is required), and let $$A = \bigcup_{\alpha < \omega_1} \{\alpha\} \times F(\alpha) \subseteq \omega_1 \times \omega_1$$. Then $$L[A_\alpha] \subset L[A]$$, and since "$$X$$ is countable'' is $$\Sigma_1$$, hence upward absolute, $$\alpha$$ is countable in $$L[A]$$. Thus $$\omega_1^{L[A]} = \omega_1$$.

To ensure that $$A \subseteq \omega_1$$, we simply consider the canonical well-ordering $$\Gamma$$ on $$\omega_1 \times \omega_1$$. Then $$L[\Gamma(A)] = L[A]$$, as we argued in the previous lemma. $$\square$$

Theorem (Exercise 13.31). If $$\omega_2$$ is not inaccessible in $$L$$, then there exists $$A \subseteq \omega_1$$ such that $$\omega_1^{L[A]} = \omega_1$$ and $$\omega_2^{L[A]} = \omega_2$$.

Proof. Since $$L \models \mathsf{GCH}$$, we have that in $$L$$, $$\omega_2$$ is inaccessible iff it is a regular limit cardinal (weakly inaccessible). As "$$\alpha$$ is a regular cardinal'' is $$\Pi_1$$we have that $$\omega_2$$ is a regular cardinal in $$L$$. Therefore, if $$\omega_2$$ is not inaccessible in $$L$$, then $$\omega_2$$ is a successor cardinal in $$L$$.

Let $$\alpha < \omega_2$$ be an ordinal such that $$\omega_2 = \alpha^+$$ in $$L$$. By repeating the proof in the first lemma but for $$\alpha \geq \omega_1$$ and $$|\alpha| = \omega_1$$, one may obtain an $$A \subseteq \omega_1$$ such that $$|\alpha| = \omega_1^{L[A]}$$ in $$L[A]$$. By the second lemma, one obtain another $$B \subseteq \omega_1$$ such that $$\omega_1 = \omega_1^{L[B]}$$ ($$\mathsf{AC}$$ is used here). Thus we have that $$\omega_1^{L[A \times B]} = \omega_1$$ and $$\omega_2^{L[A \times B]} = \omega_2$$. Now convert $$A \times B \subseteq \omega_1 \times \omega_1$$ to a subset of $$\omega_1$$ using the canonical well-ordering $$\Gamma$$. $$\square$$