Question about the proof of V=L implies GCH

(Stefan Geschke's article: Models of Set Theory.)

Theorem.( V=L implies GCH)

Proof.

Assume $V=L$ and let $\kappa$ be an infinite cardinal. We show that $P(\kappa)\subseteq L_{\kappa+}$ and therefore $2^\kappa\leq \kappa^+$.

Let $a\subseteq\kappa$. By $V=L$ there is a regular cardinal $\lambda>\aleph_0$ such that $a\in L_\lambda$. By the Lowenheim-Skolem Theorem, there is an elementary submodel $M$ of $L_\lambda$ such that $|M|=\kappa$ and $\kappa\cup\{a\}\subseteq M$. Let $N$ be the transitive collpase of $M$. Since $a\subseteq\kappa$ and $\kappa$ is transitive, $\kappa\cup\{a\}$ is transitive. Since Mostolwski collapsing functions are the indentity on transitive classes, $a \in N$.

Since $N$ is isomorphic to an elementary submodel of $L_\lambda$, $N\models V=L$. Since $N$ is a transitive model of ZF without the Power Set Axiom, the absoluteness properties of $L$ give that $N=\bigcup_{\alpha\in N\cap Ord}L_\alpha=L_\beta$ where $\beta=\sup(N\cap Ord)$. Since $M$ and therefore $N$ are of size $\kappa$, $|\beta|=|L_\beta|=\kappa$. In particular, $\beta<\kappa^+$.

I know that $L_\kappa$ is a transitive model of ZF without Power Set Axiom if $\kappa>\aleph_0$ is regular,

but I don't understand why $N=\bigcup_{\alpha\in N\cap Ord}L_\alpha$.

By the way, I haven't proved the Godel's Condensation Lemma.

• Whoever downvoted this- I don't see what's wrong with it. – Colm Bhandal Jan 10 '18 at 13:04
• Maybe since I post a screenshot at first. My bad. – FAX Jan 10 '18 at 13:05
• In the 2nd paragraph of the proof: What is $M$ supposed to be? (Very likely, $M \prec L_{\lambda}$.) – Stefan Mesken Jan 10 '18 at 13:05
• @StefanMesken Sorry I lost it. I've edited it. – FAX Jan 10 '18 at 13:11

Let me fill in some details for you:

Assume $V=L$ and let $\kappa$ be an infinite cardinal. We show that $P(\kappa)\subseteq L_{\kappa+}$ and therefore $2^\kappa\leq \kappa^+$.

Let $a\subseteq\kappa$. By $V=L$ there is a regular cardinal $\lambda>\aleph_0$ such that $a\in L_\lambda$. By the Lowenheim-Skolem Theorem, there is an elementary submodel $M$ of $L_\lambda$ such that $|M|=\kappa$ and $\kappa\cup\{a\}\subseteq M$. Let $N$ be the transitive collpase of $M$. Since $a\subseteq\kappa$ and $\kappa$ is transitive, $\kappa\cup\{a\}$ is transitive. Since Mostolwski collapsing functions are the indentity on transitive classes, $a \in N$.

Let $m \colon M \to N$ be the Mostowski collapse. Since $\kappa \subseteq M$ and $\kappa$ is transitive, we have that $m \restriction \kappa = \mathrm{id}$. It follows that $a = m(a) \in N$. (Proving this is a nice (simple) exercise. Use the elementarity of $m$.)

Since $N$ is isomorphic to an elementary submodel of $L_\lambda$, $N\models V=L$. Since $N$ is a transitive model of ZF without the Power Set Axiom, the absoluteness properties of $L$ give that $N=\bigcup_{\alpha\in N\cap Ord}L_\alpha=L_\beta$ where $\beta=\sup(N\cap Ord)$.

"$V = L$" is the statement: for every set $x$ there is some ordinal $\alpha$ such that $x \in L_{\alpha}$. Here "$x \in L_{\alpha}$" is a shorthand for the formula that defines the sequence $(L_\beta \mid \beta \in \mathrm{Ord})$ and says that $x$ appears in the $\alpha$-th member of that sequence.

The construction of $(L_\beta \mid \beta \in \mathrm{Ord})$ is very 'robust'. Every transitive model of $\mathrm{ZF}^-$ correctly computes this sequence up to its own ordinal height. (Much less than $\mathrm{ZF}^-$ is needed here.) Hence $N$ correctly computes $(L_{\alpha} \mid \alpha \in N \cap \mathrm{Ord})$ (via the same formula that defines this sequence in $V$) and since $N \cap \mathrm{Ord} =: \beta$ is a limit ordinal (which is easy to prove) and $N \models V = L$, we have that $L_\beta = N$.

In simpler words: $N$ believes that every of its elements appears in some member of the $L_{\alpha}$-hierarchy. Since $N$ and $V$ agree on this hierarchy (up to the ordinal height of $N$), it is true (in $V$) that every element of $N$ appears in some $L_{\alpha}$ for $\alpha < N \cap \mathrm{Ord}$. Hence $N = \bigcup_{\alpha < \beta} L_{\alpha} = L_{\beta}$.

The Condensation Lemma, which isn't needed for this proof, is a more refined version of the result above.

Since $M$ and therefore $N$ are of size $\kappa$, $|\beta|=|L_\beta|=\kappa$. In particular, $\beta<\kappa^+$.

This is kind of self-explanitory. The only thing that may need some justification is that $| L_\beta | = | \beta |$. This can be proved (easily) by induction on all infinite ordinals $\beta$. For $\beta = \omega$ note that $L_\omega = V_\omega$ is countable (there is a well-known isomorphism $(\omega, <) \to (V_\omega; \in)$ due to Ackermann, see here). The step $\beta \mapsto \beta + 1$ is pretty much trivial since we only add $\omega$ many new sets. If $\beta$ is a limit ordinal, just note that $| L_{\beta} | = | \bigcup_{\alpha < \beta} L_{\alpha} |$ and apply the induction hypothesis.

• There are many ways to verify the 'robustness' of the $L_{\alpha}$-hierarchy that I claim above. See for example Jech's "Set Theory" (using Gödel operations), Kunen's "Set Theory" (using absoluteness of definability and unions / or alternatively -- if you are only interested in models of $\mathrm{ZF}^-$ -- absolutness of recursion of low complexity), or Schindler's "Set Theory" (using the $J_{\alpha}$-hierarchy instead). – Stefan Mesken Jan 10 '18 at 13:33
• Incredible efficiency and help. Thanks a lot. – FAX Jan 10 '18 at 13:59
• @FAX You're very welcome (: – Stefan Mesken Jan 10 '18 at 14:00