Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

I know how constructible universe is created, but I also separatedly read that the universe is definable from ordinals - so I am wondering what it really means.

share|cite|improve this question
You might also want to look at:… – Thomas Andrews Feb 26 '13 at 15:47


This usually means that $L\subset\mathsf{HOD}$, the class of hereditarily ordinal definable sets, so that for every $x\in L$ there is a formula $\phi(y,t_1,\dots,t_n)$ and ordinals $\alpha_1,\dots,\alpha_n$ such that $$ x=\{y\mid \phi(y,\alpha_1,\dots,\alpha_n)\}. $$ (A decent reference for this is Lemmas 2.7 and 3.1 in Chapter II of Devlin's "Constructibility".) Note that the above is not characterizing $L$, in that in general there will be sets not in $L$ that are (hereditarily) definable by ordinals.


I suppose one could also use the phrase to claim that any two transitive models of enough set theory that have the same height coincide on what they think $L$ is.


Here one can actually say something stronger: As a consequence of his work on countable models of set theory, Harvey Friedman proved in

Harvey Friedman. Categoricity with respect to ordinals. In Higher Set Theory: Proceedings, Oberwolfach, Germany, April 13-23, 1977, Gerth Müller and Dana S. Scott, eds., Lecture Notes in Mathematics, vol. 669, pp. 17–20. Springer-Verlag, Berlin. MR0520185 (80m:03089),

that, in a technical sense, a theory $T$ extending $\mathsf{ZF}$ implies $V=L$ precisely when any model of $T$ is determined by its ordinals. More carefully: Friedman associates to a theory $T$ in the language of set theory a new theory $T^*$ in a three-sorted language, two sorts being interpreted as models of $T$, while the third is interpreted as the ordinals (which are the same for the two models). One of his requirements is that transfinite induction holds in $T^*$ (for all formulas).

Friedman proves that an extension $T$ of $\mathsf{ZF}$ proves $V=L$ if and only if, in any model of $T^∗$ there is a rank-preserving isomorphism between the first two sorts.

Note that the requirement on $T^*$ is stronger than the naive formulation "any model of $T$ is determined by its ordinals": Rosenthal proved that there are two models of $\mathsf{ZF}+V=L$ that are not elementarily equivalent, and yet their ordinals are order-isomorphic, see

John Rosenthal. Relations not determining the structure of $L$. Pacific J. Math. 37, (1971), 497–514. MR0304160 (46 #3295).

share|cite|improve this answer
I strongly agree with the second comment. – Asaf Karagila Feb 26 '13 at 16:44

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.