MathOverflow is a question and answer site for professional mathematicians. 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

I am talking about the principle that says that every set is the image of a projective set. For every set $x$ there is a surjection $f:y \twoheadrightarrow x$, such that for any set $u$ and function $j:y \to u$ and surjection $g:v \twoheadrightarrow u$ there is a function $h:y \to v$ such that $j=gh$.

There is a simple model of CZF in which this holds and the axiom of choice is violated. This also violates classical logic so is not a model of ZF (= CZF + classical logic). I am curious about the following comment of Andreas Blass on ZF:

...all the usual independence results involving weak axioms of choice remain true in the presence of the negation of EPSets. In particular, none of the axioms in the appendix of [Jech, the Axiom of Choice], except AC itself, implies EPSets. The following result ... is essentially all that is known about the strength of EPSets...

Theorem 6.2 EPSets implies the axiom of dependent choice ...

In particular, we do not know whether it implies AC ... we conjecture that it does not.

(Injectivity, Projectivity, and the Axiom of Choice, 1979)

Consequence of the Axiom of Choice (1998) listed the question of whether it implies AC as still open, and in fact unless I missed something it listed no new non-trivial results.

So, in 2010, is there really still nothing to be said other than ZF+DC < ZF+EPSets $\le$ ZFC ?

share|cite|improve this question
Where do you get "ZF+DC ≠ ZF+EPSets"? – Ricky Demer Oct 23 '10 at 18:49
Sorry, it's also in Blass' article, I added a bit more context. – Daniel Mehkeri Oct 23 '10 at 19:13
I thought David McCarty's model of IZF satisfies the presentation axiom but not choice. Am I wrong? – Andrej Bauer Oct 23 '10 at 20:23
Right, but it's also a model of ECT_0 (violates classical logic), so not of ZF. – Daniel Mehkeri Oct 23 '10 at 22:16
For reference, it is form 192 at the linked website. – David Roberts Nov 2 '11 at 23:26
up vote 1 down vote accepted

I haven't heard any new information about "enough projective sets" in the classical ZF context. Your formulation, however, looks weaker. "Enough projective sets" should say that for every $x$ there is a surjection $f:y \to x$ such that every surjection from any $w$ to $y$ splits. That implies your formulation (by taking $w$ to be the pullback of $y$ and $z$ over $x$) but I don't see the converse.

The principle "enough projective sets" has been used in constructive mathematics by Peter Aczel (and probably others) under the name "presentation axiom".

share|cite|improve this answer
The presentation axiom (projective set = "base") is how I heard of it, and the weaker formulation was just carelessness on my part, I changed it. – Daniel Mehkeri Oct 23 '10 at 19:04

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.