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I know that sometime ago Vopenka proved this:

Theorem: Assume there is a strongly compact cardinal. Then for any set $A$, $V \neq L(A)$.

Can we get by with a consistency-wise strictly weaker assumption?

Namely, call an uncountable cardinal $\kappa$ strong if for any ordinal $\gamma$, there is an elementary embedding $j: V \rightarrow M$ into a transitive $M$ such that $\kappa = crit(j) > \gamma$ and $V_{\gamma} \subset M$. That is, a rank initial segment of $V$ is contained in $M$.

Could someone assess whether a strong cardinal is enough? The argument would be:

Assume for contradiction that $V = L(A)$, for some set $A$. Assume further there is a strong cardinal, let $\kappa$ be the least. Let $\lambda$ be bigger than the rank of $A$, and let $j: V \rightarrow M$ with critical point $\kappa$ and $V_\lambda \subset M$. Since $\lambda$ was big enough, then $A \in M$. Since $M$ is an inner model, the transitive closure of $A$ is in $M$ also. Now we prove by induction that $L(A) \subset M$: $L_0(A) = tc(A) \in M$. Successors and limits (I think) both work, for successors, since transitive models are correct about calculating definable powersets. But we assumed $V = L(A)$, so $M = L(A) = V$. So we have an elementary $j: V \rightarrow V$, which is impossible by Kunen's Theorem.

Any fine points to be careful about here? Thanks!

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Nice question. Thank you! –  François G. Dorais Apr 28 '12 at 1:42
    
Yes, it is a well known result that a strong cardinal suffices for this. Your argument is the expected one, and it is fine. (And too much detail, really.) –  Andres Caicedo Apr 28 '12 at 3:55
    
Cool, good to know it works! Any idea if you could push down the large cardinal assumptions even further? –  student Apr 28 '12 at 4:37
    
Equivalently, what's the strongest (consistency-wise) large cardinal notion that an L(A) is able to support? –  student Apr 28 '12 at 4:42
    
Not "equivalently". Large cardinal assumptions way past supercompactness in consistency strength are consistent with V being L(A) for some set A. –  Andres Caicedo Apr 28 '12 at 6:15

1 Answer 1

up vote 5 down vote accepted

The way to think about it is this.

Whenever a property in $V$ is witnessed in an absolute manner by the existence of a single set with certain properties, then we will be able to take that set, $A$, and form the universe $L(A)$, which will have the witness and thus verify that the property is true.

For example, it is a fun exercise to prove that a set-theoretic assertion $\varphi$ is equivalent to a $\Sigma_2$ assertion (in the Levy hierarchy) if and only if it is equivalent to an assertion of the form "$\exists\gamma H_\gamma\models\phi$," for any complexity $\phi$, or similarly to an assertion of the form "$\exists \alpha V_\alpha\models\phi$." Such assertions are absolute to the corresponding $L(H_\gamma)$ and thus compatible with $V=L(A)$. (If you want ZFC, then one should adjoint not just the set, but also a well-ordering of that set.)

Many large cardinal properties are witnessed by the existence of single sets in this way, and thus are $\Sigma_2$ definable. For example, you can tell if $\kappa$ is measurable by looking at $V_{\kappa+2}$. Similarly, you can tell if $\kappa$ is superstrong or almost huge or huge by looking at the appropriate $V_\gamma$, which is large enough to see the extender giving rise to the embedding. Basically, a large cardinal property that is witnessed by a single extender or ultrafilter will be of this type, and thus compatible with $V=L(A)$.

In contrast, large cardinal properties like strongness, strong compactness or supercompactness are not witnessed by a single embedding or extender, and have complexity $\Pi_3$. You have to say: for every $\theta$ there is a normal fine measure on $P_\kappa\theta$, and this is witnessed by a proper class sequence of measures rather than a single set. From this perspective, what your and Vopenka's arguments amount to is showing that yes, strongness and strong compactness are not $\Sigma_2$ properties, but do require complexity $\Pi_3$.

Meanwhile, to address the point raised in the comments, one thing is that it is easy to have large cardinal properties with very high strength that are compatible with V=L. For example, the large cardinal axiom "there is a transitive model of ZFC in which there is a proper class of supercompact cardinals" is true in $V$ if and only if it is true in $L$. This is because there is a such a model if and only if there is a countable model of that theory, and the assertion that there is a real coding a well-founded model of any particular theory is a $\Sigma^1_2$ assertion, which by the Shoenfield theorem is absolute between $V$ and $L$. So if one replaces the actual existence of the large cardinal properties with the assertion that they hold merely inside a transitive model, then one can retain $V=L$ and have the desired large cardinal strength.

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Nice answer, thanks Joel! I just joined this site, and it's nice to get such detailed comments on some of the questions I haven't been able to make progress on. –  student Apr 29 '12 at 2:54
    
Well, then, allow me to be the first to welcome you to mathoverflow! –  Joel David Hamkins Apr 29 '12 at 4:50
    
Very appreciated :-) –  student Apr 29 '12 at 5:45

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