MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

It is known that $L\models 2^\kappa=\kappa^+$, and that for a set of ordinals $A$ we know that $L[A]\models \exists\lambda\forall\kappa>\lambda(2^\kappa=\kappa^+)$.

In this sense, there is some similarity between $L$ and $L[A]$. Both models have definable well orderings, and both models have a very nice sense of minimality deep within them.

Assuming $0^\sharp$ exists we have the class of Silver indiscernibles from which can define $L$ (using the definable Skolem functions, and the Skolem hull of $I$). Assuming that $A^\sharp$ exists we have a similar class for $L[A]$ as well.

Denote by $I$ the Silver indiscernibles and $I_A$ the corresponding class of $L[A]$.

Is there any intersection between them? Is there some $\alpha$ such that $I\setminus\alpha=I_A\setminus\alpha$?

If the answer to both questions is no in the general case, can we say anything on particular cases known?

share|cite|improve this question
up vote 5 down vote accepted

There are several issues.

  • First, of course, when $A\in L$ then $L[A]=L$ and consequently $I_A=I$.

  • In any case, there will be large overlap in $I_A$ and $I_B$, since they are both class clubs, and hence intersect in a closed unbounded class of ordinals.

  • If $0^\sharp\in L[A]$, then every cardinal of $L[A]$ is a cardinal in $L[0^\sharp]$, and hence is an $L$-indiscernible, but of course, not every cardinal of $L[A]$ is an $L[A]$-indiscernible. So the eventual agreement you requested does not generally hold.

  • In any case, $I_A\subset I$.

  • It can be that $I_A=I$ even when $A\notin L$. For example, I believe that this is the case when $A$ is an $L$-generic Cohen real added by forcing, since one can lift any $j:L\to L$ to $j:L[A]\to L[A]$.

share|cite|improve this answer
Thanks, Joel. I did not think about the club-ness of the class. I did have in mind the last point you had. Thanks for the quick answer! :-) – Asaf Karagila Aug 15 '11 at 0:18

It can also be the case that $I_A$ is a periodic (but club) subclass of $I$: by Jensen Coding one can define (necessarily by class forcing) reals $a\subset\omega$ with $0^\sharp \notin L[a]$ and that there are countable ordinals $\alpha,\beta$ so that for all $\tau\in On$ the $\tau $'th silver indiscernible for $a$ is the $\alpha + \beta\cdot\tau$'th indiscernible for $L$.

Also: this must be a class forcing, for any set forcing $P\in L$, and any $G$, $P$-generic over $L$, the $L[G]$-indiscernibles are exactly the $L$-indiscernibles from some point on.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.