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!

  • $\begingroup$ 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.) $\endgroup$ – Andrés E. Caicedo Apr 28 '12 at 3:55
  • $\begingroup$ Cool, good to know it works! Any idea if you could push down the large cardinal assumptions even further? $\endgroup$ – lesnikow Apr 28 '12 at 4:37
  • $\begingroup$ Equivalently, what's the strongest (consistency-wise) large cardinal notion that an L(A) is able to support? $\endgroup$ – lesnikow Apr 28 '12 at 4:42
  • $\begingroup$ Not "equivalently". Large cardinal assumptions way past supercompactness in consistency strength are consistent with V being L(A) for some set A. $\endgroup$ – Andrés E. Caicedo Apr 28 '12 at 6:15
  • $\begingroup$ Thanks, point taken. I was reasoning that taking the contraposition of the Theorem: (ZFC + exists some set such A such that V=L(A) ) "there exist no supercompact or strong cardinals" gives us that L(A), assuming it satisfies choice, cannot have supercompacts or strongs in it. Is it the models you mention with stronger cardinals are choiceless? Or that consistency-wise stronger cardinals exist, but none that directly imply supercompacts or strongs? I'm curious to think out the implications of the theorem for models of large cardinals... $\endgroup$ – lesnikow Apr 28 '12 at 10:07

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.

  • $\begingroup$ 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. $\endgroup$ – lesnikow Apr 29 '12 at 2:54
  • $\begingroup$ Well, then, allow me to be the first to welcome you to mathoverflow! $\endgroup$ – Joel David Hamkins Apr 29 '12 at 4:50
  • $\begingroup$ Very appreciated :-) $\endgroup$ – lesnikow Apr 29 '12 at 5:45

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