What's so wrong with the Axiom of Choice ?

One of the Zermelo- Fraenkel axioms, called axiom of choice, is remarkably controversial. It links to linear algebra and several paradoxes- find out what is so strange about it !

(00:22) - Math objects as sets
(00:54) - What axioms we use ?
(01:30) - Understanding axiom of choice
(03:27) - Axiom leads to paradoxes

#Axiom
#ZFC
#Settheory

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[TheMaginor]

The symbol for the empty set is not a phi, it was introduced by the Bourbaki group inspired by the Danish-Norwegian Ø (which sounds a little like the i in bird). The axiom of choice is only necessary when dealing with infinite collections of sets. If you have a finite set of sets, you can just do exactly what you said - pick an element from each set in order. Similarly, finite-dimensional vector spaces have bases without the axiom of choice. Moreover, there is an axiom of countable choice that is strictly weaker than the axiom of choice, and it is in a way more intuitive. Separable Hilbert spaces have Hilbert space bases using just the axiom of countable choice. Bases not that often used for other types of infinite-dimensional vector spaces, so that is not the main issue. A bigger concern to functional analysts may be trying to develop functional analysis without the Hahn-Banach theorem, which relies on Zorn's Lemma. However this can be done e.g. in constructive theories like Homotopic Type Theory, or Nik Weaver's constructive theories.

[jgilferez]

Thanks for your video! I have several objections to its content, but I'll just point out what seems to me the most important, because it can easily lead to a misunderstanding. As you present it, it could be interpreted as saying that the Principle of Well-Ordering (PWO) is clearly contradictory with our intuitions, because "open intervals don't have smallest elements." But this has nothing to do with the PWO. What the PWO says about the real numbers is that there is an order, a way of putting the reals one after the other, such that, according to this order, every nonempty subset of real numbers has a least element. This order will then be very different from the standard order. This cannot contradict our perception of the reals, because we don't have any idea how this order would look like and no analysis of the standard order will help us to understand this alternative way of ordering the reals.

[ccdavis]

A of C is trivial for finite sets. Infinite sets give independence.

[jdzsf]

2:10 not exactly. If the family of sets is finite, the choice can be proven by induction. It cannot be proven for infinite families of sets. The axiom of choice is about infinity.

[tricanico]

I believe this missed the point.
I consider that the axiom of choice can be better described intuitively as "generalize to infinity whatever intuitive facts we know about the finite". (This is expressed by the "for every" in the formula.)

Just as a note, the axiom of choice is also surprisingly equivalent to the fact that any surjective function has a right inverse, but only in classical logic. Intuitionistically the latter is weaker.

[pecfexfextus]

i'd like to first point out that when u said functions are tuples, u wrote a set of sets instead of a set of tuples

[sufronausea]

{a, b} is very different from (a, b). {a, b}={b, a} is unordered and (a, b)≠(b, a) in general, unless a=b.

[Friek]

What you're saying at 4:14 is basically just "the usual ordering of R is not a well-ordering". The well-ordering principle states that there is _some_ ordering of R (which will look nothing like the usual one) that is a well-ordering.

[Testgeraeusch]

Consider an infinity shelf with containing pairs of shoes. If you want one of each kind, you simply choose the set of all leftys, and you're done even for an uncountable set. That's equivalent to choosing the set of all posivie reals out of the reals (and that get's messy already when go into the complex plane as there are no "positive" imaginary numbers...). But consider a shelf filled with pairs of socks. There is not way of universally distinguing between one sock and the other in each pair and at this point you need the AoC to declare wether the task is doable for an infinite abount of socks or not.

[swozzlesticks]

I don't think it's fair to say that "The well ordering principle is obviously false." It's just "an ordering", I see no reason at all why it should coincide with the standard ordering of the real numbers in any way.

[wyrmofvt]

In my experience, the Barnard-Tarsky Theorem and other weird things that happen with the AoC reveal that all the counterintuitive consequences of AoC depend on the existence of pathological functions that break something in our intuitive understanding of those systems, and reveal how limited our intuitions are. In B-T, the sets are not measurable — that is, there's no way to assign to each set a volume that makes sense. Because the sets don't have a sensible way to assign volume, there is an intermediate state where the sets don't have a total volume that is well-defined, so of course you can end up with a case where you start with some volume V and end up with volume 2V. If you disallow such sets, demanding that each component set to still be measurable, then the B-T theorem collapses. Similarly with the W-O theorem; the ordering function is not the usual ordering function, and would have to be pretty gonzo in order for the W-O theorem to work. Order, after all, is additional structure on the real numbers, and a different order will have different properties. Our usual ordering of the reals is out choice out of the infinity of ordering functions available, and the AoC reveals that some of them are strange beasts indeed.

[thestemgamer]

Not all mathematical objects can be regarded as sets. Case in point, categories which are distinctly quite a lot larger than sets and we can talk about the category of all categories, but not really the set of all sets.

[geraldthaler]

4:14 confuses the natural order with wellorder. It would also apply to an open interval of rationals wich is trivially wellorderable as it's countable. A wellorder of any uncountable set get's quite "long" in a sense and is thus hard to imagine.

[randomstrategy]

An ordering where every open set of reals has a smallest element can actually be constructed as follows without the axiom of choice: Given x and y:
1) If x and y are both irrational, x<y if and only if x<y in the classical sense.
2) if x is rational and y is irrational, then x<y (and vice-versa)
3) If x and y are both rational, add up their respective numerator and denominators in reduced form; smallest sum is smallest; if the sums are equal, smallest denominator is smallest. And then if those are equal, then the negative version is smaller than the positive version.

Then since every open interval contains a rational, you can find the smallest rational in our ordering, which will be the smallest element of the entire open interval. This ordering we constructed is of course not a well-ordering, but does still satisfy the property we wanted.

[abdulmasaiev]

Is the axiom of choice actually part of Zermelo-Fraenkel? To my knowledge it was actually an optional extra one which when included turns regular ZF into ZFC

[taxicabnumber]

I despise the Axiom of Choice. Although Gödel proved that it is consistent with the other ZF axioms of standard set theory, his proof relies on a significant loophole in the axiom of Powerset. The axiom of Powerset asserts that every set has a powerset, but it does not clearly define the powerset of an infinite set, such as N. Consequently, the ZF axioms permit the construction of a set theory model where the "powerset" of an infinite set like N includes only the subsets with finite descriptions. This model is known as the Constructible Universe.

The Constructible Universe has an extremely sparse notion of the powerset, where all sets are essentially countable. (Although the powerset of N isn't technically countable within the model, as the diagonal argument still applies.) Unsurprisingly, the Axiom of Choice holds true in the Constructible Universe, making it consistent with ZF. However, this only indicates that the ZF axioms allow for pathological models of set theory. Many people mistakenly believe that the Axiom of Choice has been proven consistent with the intuitive notion of sets and powersets, but this is not the case.

[ahoj]

The problem of finding a basis for any vector space is indeed puzzling, even in the enumerable case. Start with the field of rationals Q. The polynomials Q[X] are also enumerable and, as a Q-vector space, have the powers of X as a basis. No need of AC for that. Now consider the algebraic numbers over Q. They also form an enumerable Q-vector space. Yet we don’t have an ACTUAL privileged basis for them. We know it exists, we even know how to compute one, by enumerating algebraic numbers and checking if they are linearly independant from the previous ones. But we cannot describe such a basis in its entirety.

[hanna]

For 4:00, there seems to be an easy definition to make it well-ordered, just let x ∈ (a, b), and the less-or-equal-than is defined by comparing abs(x1 - (a+b)/2) <= abs(x2 - (a+b)/2). Then x0 = (a+b)/2 is the "smallest" element. Please point out where I am wrong...

[benjaminpedersen]

If you actually dig into the proof of Banach-Tarski, then the argument I find most paradoxical is not the application of choice; it is the fact that if you have the set of all sequences of cardinal steps (left, right, up, down) that end with a step up, you can step them down to get absolutely all sequences of cardinal steps.

[SteinGauslaaStrindhaug]

I thought axioms were unprovable "premises" we just chose to assume is true (or not) and mathematics is what happens when we explore the consequences of the axioms we've chosen to use...