# $\mathbf{V} = \mathbf{L} \rightarrow \text{AC}+\text{GCH}$.

In Kunens "Set Theory An Introduction To Independence Proofs", thm 4.6 in VI, we are given the following:

Theorem. If $$\mathbf{V} = \mathbf{L}$$, then for all infinite ordinals $$\alpha$$, $$\mathscr{P}(L(\alpha)) \subset L(\alpha^+)$$.

The proof goes like this:

Let $$\chi$$ be a finite conjuction of axioms of $$\sf{ZF}+\mathbf{V} = \mathbf{L}$$ such that $$\forall M(M \ \text{transitive} \land \chi^M \rightarrow M = L(o(M))).$$ This is possible by theorem 3.9.

Assume $$\mathbf{V} = \mathbf{L}$$, and fix $$A \in \mathscr{P}(L(\alpha))$$. Let $$X = L(\alpha) \cup \{A\}$$. Then $$|X| = |\alpha|$$ by lemma 1.14 (1.14 uses AC, but we have just seen that $$\mathbf{V} = \mathbf{L} \rightarrow \text{AC}$$). By a Löwenheim-Skolem argument followed by the Mostowski Collapsing Theorem there is a transitive $$M$$ such that $$|M| = |\alpha|$$, $$X \subset M$$ and $$\chi^M \leftrightarrow \chi^{\mathbf{V}}$$ (we have applied IV 7.10 with $$\mathbf{Z} = \mathbf{V}$$). But $$\chi^{\mathbf{V}}$$ is true by $$\mathbf{V} = \mathbf{L}$$, so $$\chi^{M}$$ holds, whence $$M = L(o(M))$$. Since $$|M| = |\alpha|$$, $$|o(M)| < \alpha^+$$. Thus $$A \in L(o(M)) \subset L(\alpha^+)$$.

I think I understand the structure of the argument, but there is one step I don't understand: "But $$\chi^{\mathbf{V}}$$ is true by $$\mathbf{V} = \mathbf{L}$$". Why is $$\chi^{\mathbf{V}}$$ true just because $$\mathbf{V} = \mathbf{L}$$?

Some other reflections: The Löwenheim-Skolem (or perhaps Downwards Löwenheim-Skolem; let's abbreviate this as DLS) $$\qquad (*)$$ gives us

$$\forall B \ \forall X \ \exists A[X \subset A \subset B \land A \prec B \land |A| \leq \max(|X|,|\mathcal{L}|)] \qquad \qquad (1)$$ where in our case, $$|\mathcal{L}| = \omega$$ since the language is countable/finitary (if that is the correct usage of finitary).

Now, we note that $$A \subset L(\alpha)$$. Since $$\mathbf{V} = \mathbf{L}$$ and $$A \in \mathbf{V}$$, it follows that $$A \in \mathbf{L}$$, so it follows that $$A \in L(\beta)$$ for some $$\beta$$, presumably $$\beta > \alpha$$ (if no such $$\beta$$ existed, then we would already be done, I claim). Then by reflection, we can find some $$\zeta > \beta$$ so $$\chi^{L(\zeta)}$$ holds.

By DLS applied to $$X,(L(\zeta),\in)$$ (taking $$B = L(\zeta)$$ and $$X = X$$ in $$(1)$$) where we use that $$A \in L(\beta) \subset L(\zeta)$$ so that $$X = L(\alpha) \cup \{A\} \subset L(\zeta)$$ we can find $$\mu$$ such that $$X \subset \mu \subset L(\zeta)$$, $$\mu \prec L(\zeta)$$ and $$|\mu| \leq \max(\omega,|X|) = \max(\omega,|\alpha|)$$. Now, I suppose $$\mu$$ inherits the order $$\in$$ from $$L(\beta)$$, so that we can talk about $$(\mu,\in)$$. But we still don't know whether $$\mu$$ is transitive.

If we now apply the Mostowski Collapse to $$\mu$$, we find that there is some $$(N,\in)$$ so that $$(\mu,\in) \cong (N,\in)$$, and furthermore $$N$$ is transitive.

Now, since $$\mu \prec L(\beta)$$, we have (by lemma 1.10 in Kunen, V) that for every formula $$\phi(x_0,\ldots,x_{n-1})$$, $$\phi^{\mu}(x_0,\ldots,x_{n-1}) \leftrightarrow \phi^{L(\beta)}$$. But by our earlier use of reflection, this means that $$\chi^{\mu}$$ holds, and by extension, it follows that $$\chi^N$$ holds. It follows from this ($$N$$ being transitive and $$\chi^N$$ holding) together with thm 3.9 (VI) that $$N = L(o(N))$$, where $$o(N) = N \cap \mathbf{ON}$$. Now, we know that $$X \subset \mu$$ and that we had an isomorphism $$f:\mu \overset{\simeq}{\to} N$$ for the $$\in$$-relation to $$N$$. My teacher claimed that it will follow that $$f$$ will act as the identity on $$X$$, so that $$X \subset N$$ (if someone has a more elaborate argument for this part I'd be happy) $$\qquad$$ (**).

Furthermore, we of course have \begin{align*} |\mu| &= |N|\\ &= |\alpha|.\end{align*}

Then we see that since $$L(\alpha) \subset N$$, $$\alpha \leq o(N)$$, and from $$|N| = \alpha$$, we see that $$|o(N)| \leq |N| = |\alpha|$$, so that $$o(N) < \alpha^+$$, since if $$o(N) \geq \alpha^+$$ then \begin{align*} |o(N)| &\geq |\alpha^+|\\ &= \alpha^+ \qquad \qquad (\text{since \alpha^+ is a cardinal})\\ &> \alpha\\ & \geq |\alpha|, \end{align*} contradicting $$|o(N)| \leq |\alpha|$$.

Then $$A \in L(o(M)) \subset L(\alpha^+)$$ (which follows from $$o(N) < \alpha^+$$).

We have shown that $$\mathscr{P}(L(\alpha)) \subset L(\alpha^+)$$.

(*) I should comment and say that perhaps I should have called the Downward Löwenheim-Skolem theorem the Downward Löwenheim-Skolem-Tarski theorem.

(**) What I would most like to know a more detailed argument about, is the step where we say that $$f$$ acts like the identity on $$L(\alpha) \cup \{A\}$$.

(***) Please point out if there are any general errors in the proof.

• $\chi$ is a finite conjunction of axioms of ZF + V=L. Commented Aug 1 at 15:00
• Hm, not sure I follow. I mean, in the more elaborate part I wrote down, it followed from the fact that we first by DLS found an elementary submodel (I believe it is so called), and then by mostowski collapse (again, I believe so called) found an isomorphism to some set $N$ with $\in$-relation respected by said isomorphism. Commented Aug 2 at 0:18
• I’m just answering your first question why $\chi$ holds in $V$. $V$ satisfies ZF+V=L. $\chi$ is a conjunction of axioms of ZF+V=L. That’s all there is to it. Commented Aug 2 at 1:58
• For the question about why the Mostowski collapse is the identity on $L(\alpha)\cup{A},$ note this set is transitive. (cf Kunen's reference to IV 7.10). Commented Aug 2 at 2:30
• Not sure I follow all that but yes, $\sf ZF^-\vdash Con(ZF)$ would be violate the incompleteness theorem. And yes $\bf V=WF$ is equivalent to foundation over $\sf ZF^-,$ but bringing $\bf WF$ into things is missing the point... the point is that we're working in $\sf ZF$, so we're assuming $\bf V$ satisfies $\sf ZF.$ That's what it means to be working in $\sf ZF.$ Commented Aug 2 at 3:42