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?

  • $\begingroup$ 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 $L$. $\endgroup$ – Not Mike Jun 3 at 11:11
  • $\begingroup$ Also, to construct $M$, you take the Skolem-Hull of $\omega_\alpha\cup \{X\}$. This yields the required properties. $\endgroup$ – Not Mike Jun 3 at 11:14
  • $\begingroup$ 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 $L_\delta$? $\endgroup$ – lost_set_theory_student Jun 3 at 11:21
  • $\begingroup$ Initial segments of $L$ have very well-behaved, definable Skolem-functions, which allow you to quickly form Hulls and guarantee elementarity. $\endgroup$ – Not Mike Jun 3 at 11:23
  • $\begingroup$ I see. I must have skipped the part when Jech went through this. Do you know where he covers this? $\endgroup$ – lost_set_theory_student Jun 3 at 11:29

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.

  • $\begingroup$ 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! $\endgroup$ – lost_set_theory_student Jun 3 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.


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