# Question about Jech's proof of V = L implies GCH

On pg 190 of Jech's Set Theory, he proves V = L implies GCH. I understand it all except the following:

Thus let X ⊂ $$ω_α$$. There exists a limit ordinal δ>$$ω_α$$ such that X ∈ $$L_δ$$. Let M be an elementary submodel of $$L_δ$$ such that $$ω_α$$ ⊂ M and X ∈ M, and $$|M| = \aleph_{\alpha}$$.

I think this uses the Lowenheim Skolem theorem to get the model of size $$\aleph_{\alpha}$$, but why should $$\omega_{\alpha} \subset M$$ and why should $$X \in M$$, or how do we construct such an M?

I know there are other proofs of this theorem which uses a version of the reflection principle which gives you this directly, but the version of the reflection principle I know is the more basic one which doesn't deal with the cardinality of the sets for which the formulas are absolute (the beginning reflection theorems in Kunen).

Basically, I would prefer an explanation that uses the Lowenheim Skolem theorem instead. How is this done?

• He is starting from the assumption that $X$ is an element of $L$; since these are the only sets relevant to establishing $\mathsf{GCH}$ holds in $L$. – Not Mike Jun 3 '19 at 11:11
• Also, to construct $M$, you take the Skolem-Hull of $\omega_\alpha\cup \{X\}$. This yields the required properties. – Not Mike Jun 3 '19 at 11:14
• I understand that $X \in L$ beause $V = L$ but I was wondering why $X \in M$ the elementary submodel of $L_\delta$. Re: your second comment; My model theory is a bit rusty, so I may be wrong here, but that's not the typical way the Lowenheim Skolem theorem is applied is it? So the construction of M just doesn't use the theorem? If you DO take the Skolem Hull of $\omega_{\alpha} \cup \{X\}$, how do we know that gives us an elementary submodel of $L_\delta$? – lost_set_theory_student Jun 3 '19 at 11:21
• Initial segments of $L$ have very well-behaved, definable Skolem-functions, which allow you to quickly form Hulls and guarantee elementarity. – Not Mike Jun 3 '19 at 11:23
• I see. I must have skipped the part when Jech went through this. Do you know where he covers this? – lost_set_theory_student Jun 3 '19 at 11:29

## 2 Answers

The (downward) Löwenheim-Skolem theorem has several formulations. The one I use says that, if $$\mathfrak A$$ is a structure with universe $$A$$ and vocabulary (also called language and signature) $$L$$, and if $$S\subseteq A$$, then $$\mathfrak A$$ has an elementary substructure $$\mathfrak B$$ whose universe $$B$$ satisfies both $$S\subseteq B$$ and $$|B|\leq\max\{|S|,|L|,\aleph_0\}$$. Applying this theorem with $$\mathfrak A=(L_\delta,\in)$$ and $$S=\omega_\alpha\cup\{X\}$$ gives you the result that Jech needs.

Some people use other (often weaker) formulations of the Löwenheim-Skolem theorem. If you're one of those people, go back to the proof of that formulation and tweak it slightly to get the formulation I use.

• Yep I was unfamiliar with this application of the Lowenheim Skolem theorem. You're write though, the result seems easy enough from the original proof. Thanks! – lost_set_theory_student Jun 3 '19 at 12:56

You assume $$\omega_\alpha\subseteq M$$ and $$X\in M$$ so that $$X$$ belongs to the transitive collapse of $$M$$ (because if $$\pi$$ is the collapsing map, $$\pi(X)=\pi[X]=X$$. You assume $$|M|=\aleph_\alpha$$ so that the transitive collapse of $$M$$ has size $$\aleph_\alpha$$. Since you also have that this transitive collapse is of the form $$L_\beta$$ for some $$\beta$$, it follows that $$|\beta|=\aleph_\alpha$$.

Building $$M$$ is a direct application of the downward Lowenheim-Skolem theorem (so $$M\preceq L_\delta$$). If you do not find this apparent, I suggest you read the section(s) in Chang and Keisler's classic model theory book that cover this result, which they do in more detail than in Jech's or Kunen's texts.