A countable limit ordinal $\kappa$ has cofinality $\omega$. One proves this in ZF, say, using the usual trick for representing $\kappa$ as a countable set of reals having closed convex span $[0,1]$ (with the usual order) and then comparing with any increasing sequence in $[0,1]$ converging to 1.

Nevertheless, I suspect independence from ZF for the following uniform version of this claim:

There exists a function $f:\omega \times \omega_1 \rightarrow \omega_1$ such that

1) $f(\alpha,\beta) < \beta$;

2) ${\rm sup}_\alpha f(\alpha,\beta) =\beta $ for $\beta$ a limit ordinal.

(For fixed $\beta$ assume that $f(\alpha,\beta)$ increases with $\alpha$, if you like.)

Briefly, such an $f$ would support, by induction and coding tricks, the construction of an injection from $\omega_1$ to ${\Bbb R}$; that in turn would mean that CH implies the existence of a well-ordering of the reals. (Details on demand.)


1) Does this independence come up in the literature?

2) Can someone point me to a model of ZF+CH where the reals have no well-ordering?

3) Is it easy to get directly a model of ZF having no such $f$ by forcing?

4) Is the existence of $f$ equivalent to any well-known consequences of AC?

5) Are there toposes where even the original cofinality statement (on some reasonable interpretation) fails (for lack, say, of a global bijection between $\kappa$ and the natural number object)?

  • $\begingroup$ You can't get such models "easily" via forcing, because forcing the negation of AC requires that AC failed in the ground model, and that means usually that forcing is not that easy to work with (a lot of the standard tools are not very useful anymore). $\endgroup$ – Asaf Karagila Apr 8 '16 at 11:56
  • $\begingroup$ This might be relevant to some of your questions: mathoverflow.net/questions/86621 $\endgroup$ – Asaf Karagila Apr 8 '16 at 12:00

Yes, this comes up. For example, Raisonnier showed that the existence of an injection $\omega_1 \to \mathbb{R}$ implies the existence of a subset of $\mathbb{R}$ which is not Lebesgue measurable over ZF + DC.

Jean Raisonnier, A mathematical proof of S. Shelah’s theorem on the measure problem and related results, Israel J. Math. 48 (1984), no. 1, 48--56.

Important examples of models where this property fails are:

  • The Feferman-Levy model where $\mathbb{R}$ is a countable union of countable sets. [See Theorem 10.6 in Thomas J. Jech, The Axiom of Choice, Studies in Logic and the Foundations of Mathematics, Vol. 75. North-Holland Publishing Co., Amsterdam-London; Amercan Elsevier Publishing Co., Inc., New York, 1973. xi+202 pp.]

  • The Solovay model where every subset of $\mathbb{R}$ is Lebesgue measurable. [Robert M. Solovay, A model of set-theory in which every set of reals is Lebesgue measurable, Ann. of Math. (2) 92 (1970), 1--56.]

Both of these model constructions use an inaccessible cardinal. This is necessary for if $\omega_1$ is not inaccessible in $L$ then there is a real $x$ such that $\omega_1^{L[x]} = \omega_1$ and then the required function can be constructed in $L[x]$ (which always satisfies AC).

Raisonnier's Theorem in a topos-theoretic context was discussed my answer this MathOverflow question:

Alex Simpson, How strong is “all sets are Lebesgue Measurable” in weaker contexts than ZF?

  • $\begingroup$ In the Feferman-Levy model the continuum hypothesis actually fails. And it might be worth mentioning Truss' models which are a variant of Solovay's construction where you start with an arbitrary limit cardinal but still get ZF+"Every uncountable set of reals has a perfect subset". See also mathoverflow.net/questions/86621 and the answer there. $\endgroup$ – Asaf Karagila Apr 8 '16 at 11:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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