9
$\begingroup$

Let consider the Vitali set $V \subset \mathbb R$, which is constructed using the axiom of choice. (I could take any other mathematical "object" that can be constructed using the axiom of choice, but I chose the Vitali set just to make the purpose clearer).

On the one hand, the Vitali set exists thanks to the axiom of choice. On the other hand, the power set $\mathcal P (\mathbb R)$ exists in ZF without assuming AC. In some sense, the existence of any $A \in \mathcal P (\mathbb R)$ is independant on (i.e. does not require) AC.

But then, I had the following reflection : the existence of the Vitali set $V \in \mathcal P (\mathbb R)$, as an element of this power set, should not require the axiom of choice... There must be some fallacy here ; that's why I would like some clarifications.

To be concise : can we say that the axiom of choice "affects" the existence of elements in $\mathcal P (\mathbb R)$ (as it seems to be the case for the existence of the Vitali set), although the construction of $\mathcal P (\mathbb R)$ (in ZF) has nothing to do with AC?

Thank you!

$\endgroup$
  • 6
    $\begingroup$ I can give you instructions on how to build a house from a pile of bricks that sits in front of you. If you ask me, "does the house contain a gold brick?" "I don't know, if there is a gold brick in that pile then the house will contain a gold brick." $\endgroup$ – CommonerG Dec 30 '15 at 22:03
  • $\begingroup$ @CommonerG A gold brick that can be disassembled, and reassembled into two gold bricks!!!! $\endgroup$ – Matemáticos Chibchas May 18 '17 at 0:20
6
$\begingroup$

$\mathcal P (\mathbb R)$ exists in any set theoretic universe with the Power Set axiom in which $\mathbb R$ is a set. AC vs $\neg$AC changes the universe of sets, and therefore what exactly $\mathcal P(\mathbb R)$ consists of.

$\endgroup$
  • $\begingroup$ AC is equivalent to : Every set can be well-ordered.In other words that every set is the bijective image of an ordinal.The graph of a function, to a set-theorist, is the function itself. AC asserts the existence of a certain class of sets: these bijections. (In set theory a function is its graph, which is a certain kind of set of ordered pairs .And the ordered pair (a,b) is {a, {a,b} }.) $\endgroup$ – DanielWainfleet Dec 30 '15 at 22:50
  • 1
    $\begingroup$ $\lnot \text{AC}$ does not necessarily change the powerset of the reals; AC can hold for sets of small cardinality and not for some sets of large cardinality. $\endgroup$ – Carl Mummert Jan 1 '16 at 13:44
8
$\begingroup$

Most axioms of set theory assert the existence of certain things. Why should $\mathcal P(\Bbb R)$ exist in the first place? Or $\Bbb R$? We use axioms that assert their existence.

There are set theories, like Pocket set theory, or very weak variants of Zermelo-Fraenkel (most notably Kripke-Platek set theory) which cannot prove the existence of $\Bbb R$. You can still do quite a bit of mathematics in these theories, but the real numbers do not form a set there.

So it shouldn't be an actual surprise that the axiom of choice which asserts the existence of various objects, can be used to prove the existence of various objects.

Equally, the axioms of the field to not assert the existence of $\sqrt2$. We know that it is consistent that $\sqrt2$ exists, but also that it does not. Why should that be any different from Vitali sets, or any other intangible object?

$\endgroup$
2
$\begingroup$

It is important to distinguish between actual objects (in our case 'sets') and definitions thereof. While we can prove in $\operatorname{ZF}$ (with or without choice) that every model of $\operatorname{ZF}$ has to contain an object that satisfies (inside this model) the definition of '$\mathcal P(\mathbb R)$', this doesn't tell us the whole story of what this particular object actually is.

(In the following, let's say that we defined $\mathbb R$ to be the set of all functions $f \colon \omega \to \omega$. It doesn't really matter, but we have to be a bit careful as how to define the reals as a set for some absoluteness reasons... Let's also take $\operatorname{ZFC}$ as our background theory.)

For example: Let $(M; \in)$ be a countable transitive model of (a sufficiently large fragement of) $\operatorname{ZF}$. Then there will be some $x \in M$ such that $$ (M; \in) \models 'x = \mathcal P (\mathbb R)' $$ (by which I mean that $M$ thinks that $x$ is the powerset of the reals). As $M$ is countable and transitive, we have that $x \subseteq M$ and thus $x$ has to be countable as well. But (this is the point where it matters how we defined the reals as a set) from the point of view of our background universe, $x$ really consists of subsets of reals, so $x$ is a subset of the 'true' powerset of the reals. It's just the case that it misses most subsets. In particular, $x$ may or may not contain some $y \in x$ such that $$ (M; \in ) \models 'y \text{ is a Vitali set}' $$

If $(M; \in)$ satisfies choice, then there will be indeed be such a $y \in x$.

$\endgroup$

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.