Consequences of the Axiom of Choice in the Constructible Universe

Question

$L$ is a proper subclass of $V$ in every model other than $L$, so it's not the biggest model of $ZFC$, but by how much?

Specifically, I'd like to know what objects and properties allowed by $ZFC$ that are not in $L$. Additionally, are there any properties or objects allowed by $ZFC$, but not by $ZF$, that are not in $L$?

Answer 1

Let me treat your comment "$L$ is smaller (or weaker) than $ZFC$". You're comparing two different things: a model and a theory. I believe this is the source of your confusion.

The situation is the following: $ZFC+V=L$ is a stronger theory than $ZFC$ (= it proves more). However, $L$ is a small model: $ZFC$ proves that there is no inner model strictly contained in $L$, so it is consistent with $ZFC$ that $V$ (the model) is strictly larger than $L^V$ (its version of $L$).

The size of the model has nothing to do with the strength of a theory describing it: a stronger theory has fewer models (by virtue of saying more - that is, being more specific), not necessarily larger ones. In the same way, the theory of abelian groups is stronger than the theory of groups, even though the general theory of groups has "larger" models in the sense that no subgroup of an abelian group is nonabelian. (A more direct, but less interesting, example might be the theory of groups versus the theory of groups of cardinality $\le60$ - the latter theory is stronger, but has "smaller" models).

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A comment on the relationship between $L$ and $ZFC$: it is provable in $ZF$ (Choice isn't needed) that $L\models ZFC$. Choice, in particular, comes from the fact that $L$ has a definable well-ordering, so in fact $L$ satisfies global choice. (Incidentally, the fact that $L$ satisfies GCH (and other combinatorial principles) requires more work, and hinges on the condensation lemma; and more complicated combinatorial principles require more work.)

If you're interested in $L$ as a "small" model, then I think the right questions to ask are: what are the properties permitted by $ZFC$ which provably don't hold in $L$? There are a number of these; here are a few of my favorites:

• In $L$, the projective sets are badly behaved: if I recall correctly, coanalytic sets need not be measurable, or have the perfect set or Baire properties.

• The generalized continuum hypothesis holds in $L$, which means that various arithmetic phenomena can't occur: for instance, we can't have $2^\kappa=2^\lambda$ for $\kappa\not=\lambda$. Strong forcing axioms fail, and the theory of cardinal characteristics of the continuum trivializes.

• If $V=L$, there are no nontrivial countably complete ultrafilters. More generally, larger large cardinal axioms are incompatible with $L$.

Answer 2

The statement $V=L$ implies $\sf AC$, in fact it implies a stronger axiom: $\sf GCH$. In $\sf ZF$ we can show that $\sf GCH$ implies $\sf AC$.

So every implication of Choice holds in $L$, but on the other hand, we know that if $\sf ZFC$ is consistent at all, then it does not prove $V=L$. Therefore not every consequence of $V=L$ follow from the axiom of choice. These exist in abundance: for example the existence of Suslin trees, or $\lozenge_{\omega_1}$ and generally $\sf CH$-related statements.

Answer 3

The Erdős cardinals provide a nice example. Let $\kappa$ be an infinite cardinal. Then:

1. If $\kappa\rightarrow(\alpha)^{\lt\omega},$ where $\alpha$ is a countable ordinal of $L,$ then $L\models\kappa\rightarrow(\alpha)^{\lt\omega}.$ (So any large cardinal of this form is consistent with $V=L,$ if it's consistent at all.)

2. If the ordinal $\alpha$ is uncountable in $L,$ then $L\models\kappa\not\rightarrow(\alpha)^{\lt\omega}.$ (So large cardinals of this form are not consistent with $V=L,$ even though it's generally believed that they're consistent with ZFC.)

Here $\kappa\rightarrow(\alpha)^{\lt\omega}$ means that for every collection $S$ of finite subsets of $\kappa,$ there exists $Z\subseteq\kappa$ of order type $\alpha$ such that for every $n\lt\omega,$ the set of $n$-element subsets of $Z$ is either contained in $S$ or disjoint from $S.$

And of course $\kappa\not\rightarrow(\alpha)^{\lt\omega}$ means that $\kappa\rightarrow(\alpha)^{\lt\omega}$ is false.

Answer 4

If ZF is consistent then so is ZF+V=L. And ZF+V=L inplies AC (Choice), so ZF+V=L is equivalent to ZFC+V=L. So unless ZF is inconsistent we cannot prove from ZF (or from ZFC) that there are any larger class-models for ZF than L.

Since Paul Cohen's discovery of the method called Forcing, there is now a huge array of properties that are known to be equiconsistent with ZF (or ZFC) that each imply V$\ne$ L. For example:

(i). (ZF). There exists a set that cannot be well-ordered.

(ii).(ZFC). $\neg CH.$

(iii).(ZFC). There does not exist an $\omega_1$-Souslin line. (ZFC+V=L implies there does exist an $\omega_1$-Souslin line.)

(iv).(ZFC). $|(\omega_1)^L|=|(2^{\omega_1})^L|=\omega.$