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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.

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You might also want to look at: en.wikipedia.org/wiki/… –  Thomas Andrews Feb 26 '13 at 15:47
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1 Answer 1

I

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.

II

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.

III

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).

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I strongly agree with the second comment. –  Asaf Karagila Feb 26 '13 at 16:44
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