Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Jensen claimed that for any finite increasing sequence countable admissible ordinals $\omega= \alpha_0<\alpha_1\cdots <\alpha_n$, there is a real $x$ so that, for each $m\leq n$, $\alpha_m$ is the $m+1$-th admissible ordinal relative to $x$.

Anybody knows the proof? Or where to find it?

share|cite|improve this question
2 (The paper titled "Admissible sets") –  Andrés Caicedo Apr 5 '12 at 3:05
Thanks. It's really painful to read the handwritten manuscripts... –  喻 良 Apr 5 '12 at 3:32
(I just emailed you a typeset version of the notes.) –  Andrés Caicedo Apr 5 '12 at 6:27
Hello Andres. Can you email me typset version of the notes too? –  user16974 Aug 25 '12 at 13:46
Andres me too? –  Peter Gerdes Jan 12 '13 at 11:40

1 Answer 1

up vote 7 down vote accepted

There is a model theoretic proof of the generalization to countable sequences which appears in a paper by Simpson and Weitkamp.

High and low Kleene degrees of coanalytic sets Stephen G. Simpson & Galen Weitkamp Journal of Symbolic Logic 48 (2):356-368 (1983)

I believe that this proof is due to Harrington, based on Jensen's and Friedman's model theoretic proof of Sacks's theorem that every countable admissible ordinal is the least admissible relative to some real.

share|cite|improve this answer
Hi Ted, nice to see you here! –  Andrés Caicedo Feb 3 '13 at 20:31
Yes, Ted, welcome to MathOverflow. –  Joel David Hamkins Feb 3 '13 at 21:11
Ted, thanks. In Simpon-Weitkamp's paper, it is claimed that Harrington has a model theoretical proof. But where to find it? –  喻 良 Feb 4 '13 at 4:14
I asked Prof. Jensen, when he was in NUS, whether he has a model theoretical proof, or by applying Barwise compactness, of his result. He said no. So I guess Harrington's proof must be highly nontrivial. –  喻 良 Feb 4 '13 at 7:14
Yu, the result is Theorem 4.1 of the Simpson and Weitkamp paper and they include a proof. The proof comes before the attribution to Harrington. –  Theodore Slaman Feb 5 '13 at 2:48

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.