Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. 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

We know that if $0^\#$ exists then it's not in $L$. For an infinite ordinal $\alpha$, denote by $I_\alpha$ the initial segment of length $\alpha$ of Silver indiscernibles.

Question: For which $\alpha$ (if any) is $I_\alpha$ in $L$?

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

No infinite set of Silver indiscernibles is in $L$. (Of course, every finite set of Silver indiscernibles is in $L$ since $L$ contains all finite sets of ordinals.)

To see this, assume $0^\#$ exists in $V$ and suppose $(\eta_i)_{i<\omega}$ is an infinite increasing sequence of Silver indiscernibles. I will show that $0^\#$ is definable from $(\eta_i)_{i<\omega}$ using parameters from $L$. So if $(\eta_i)_{i<\omega}$ were in $L$, then $0^\#$ would be in $L$ too. Since $0^\# \notin L$, it follows that $(\eta_i)_{i<\omega}$ is also not in $L$.

Pick an uncountable $V$-cardinal number $\kappa$ which is greater than all the $\eta_i$. Then $L_\kappa \prec L$. So if $\phi(v_0,\dots,v_{n-1})$ is any formula in the language of set theory, then $${}^\ulcorner\phi{}^\urcorner \in 0^\# \Leftrightarrow L \vDash \phi(\eta_0,\dots,\eta_{n-1}) \Leftrightarrow L_\kappa \vDash \phi(\eta_0,\dots,\eta_{n-1}).$$ Since $\kappa$, $L_\kappa$, and the satisfaction relation for $L_\kappa$ are all in $L$, it follows that $0^\#$ is definable from $(\eta_i)_{i<\omega}$ using parameters from $L$.

share|cite|improve this answer
Thank you, Francois. – ZeroDagger Jul 18 '11 at 22:44
No problem! It was a pleasure! – François G. Dorais Jul 19 '11 at 4:17

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.