1
$\begingroup$

Suppose that using Zorn's lemma, we have proven that an object with some properties exists and then we've proven that such object is unique. Can we always conclude that we can prove the existence (and uniqueness) of the object without using Zorn's lemma (or any other equivalent form of axiom of choice)?

I was studying differentiable manifolds and I arrived to a point where it was proven that any differentiable atlas is contained in some unique maximal differentiable atlas. Then I found out that it can be proven without using axiom of choice. But isn't that because the maximal atlas is unique?

Everywhere I remember that it was necessary to use axiom of choice for proving the existence of some object, it was later proven that such object is not unique and in fact, there are uncountably many of such objects. So I asked the above question, but couldn't find any good answer.

$\endgroup$
5
$\begingroup$

The part about atlases is, as Asaf noted, a duplicate, but let me answer the general question whether uniqueness lets you avoid the axiom of choice. The first counterexample that comes to mind is the smallest ordinal number whose cardinality (i.e., the number of smaller ordinals) equals the cardinality of the real line. The existence of this ordinal is equivalent to the statement that the real line can be well-ordered, so it's provable with the axiom of choice but not without the axiom of choice. Yet the ordinal in question is unique because of "smallest" in its definition.

$\endgroup$
2
$\begingroup$

Not necessarily. You can prove in $\sf ZFC$ that the algebraic closure of $\Bbb Q$ exists and it is unique up to isomorphism. And while you can always prove that $\Bbb Q$ has an algebraic closure, you cannot prove it is unique without appealing to the axiom of choice.

Another example in the same vein is the well-defined notion of a dimension of a vector space. You can prove in $\sf ZFC$ that every vector space has a basis and every two bases have the same cardinality. Proving that every space has a basis implies the axiom of choice (as proven by Andreas Blass in 1984); but even if you know that there is a basis to your space, proving that every two bases have the same cardinality requires you to appeal to the axiom of choice (although it certainly does not imply the axiom of choice in general).

$\endgroup$
  • $\begingroup$ There can be algebraic closures of $\Bbb Q$ not isomorphic to the regular one in ZF? What do these look like? $\endgroup$ – Akiva Weinberger Nov 8 '15 at 5:42
  • $\begingroup$ Yes, there can be. What does it look like? It looks like madness. $\endgroup$ – Asaf Karagila Nov 8 '15 at 5:43
  • 1
    $\begingroup$ I'm not sure it looks like madness; the madness might not be something you can see. For example, if you force a generic well-ordering of a mad algebraic closure of $\mathbb Q$, then it becomes isomorphic to the "regular" one. From this point of view, one can imagine that the madness is not a defect of the mad algebraic closure but rather of the ambient universe, which fails to provide the isomorphism that "ought" to be there. $\endgroup$ – Andreas Blass Nov 8 '15 at 10:50
  • $\begingroup$ @Andreas: Sure, that is a valid point. But you can argue that everything is countable, since you can always force a bijection between a set and $\omega$. And it isn't hard to show a lot of isomorphism theorems (e.g. every two blahs are isomorphic) are automatically true if both objects in question are well-orderable. So you can argue that we're really working in some second-order arithmetic, and the model is just insane since it omits a lot of bijections! :-) $\endgroup$ – Asaf Karagila Nov 8 '15 at 10:52
  • $\begingroup$ @Akiva: Here I sketch a construction of a quadratic closure of $\mathbb F_p$ that is not well-orderable and so not isomorphic to the usual one. I think similar techniques could be used to make a "mad" algebraic closure of $\mathbb Q$, and at least give an idea what one may look like. $\endgroup$ – Henning Makholm Nov 8 '15 at 14:31
0
$\begingroup$

If unique can mean unique up to isomorphism, then no: In ZFC, every field has an algebraic closure, unique up to isomorphism; but there is a model of ZF in which some field has no algebraic closure. Proof of existence of algebraic closures doesn't require full AC — the ultrafilter theorem (aka, dually, the prime ideal theorem) suffices. See for example Can one construct an algebraic closure of fields like $\mathbb{F}_p(T)$ without Zorn's lemma?.

$\endgroup$

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.