What Makes the Axiom of Choice So Controversial?

[RandomAllTime]

Hi guys. So I've been wondering, what's so controversial about the axiom of choice? I heard it allows the Banach-Tarski Paradox to work. A little insight would be much appreciated, thanks.

 

[mathman]

Hi guys. So I've been wondering, what's so controversial about the axiom of choice? I heard it allows the Banach-Tarski Paradox to work. A little insight would be much appreciated, thanks.

I don't know you could call it controversial. The issue is that it is an independent axiom.

https://en.wikipedia.org/wiki/Axiom_of_choice

 

[RandomAllTime]

I don't know you could call it controversial. The issue is that it is an independent axiom.

https://en.wikipedia.org/wiki/Axiom_of_choice

I see. I guess it's because I heard that it sort of let's the Banach Tarski Paradox hold true. Thanks for the link.

 

[gill1109]

If you like Banach Tarski then you like Axiom of Choice. If you don't like Banach Tarski then you are free to trash the axiom of choice, and now no Banach Tarksi.

Axiom of Choice is usually thought to be a useful thing in mathematics since (a) it seems intuitive (b) it makes it easier to prove theorems claiming that certain things exists. Well that is fine as long as those are things you kind of like to exist, but at some point it also starts allowing things to exist which see counter-intuitive, and maybe you don't like that.

For most of practical mathematics, the axiom of countable choice is quite enough to do everything you want to do. https://en.wikipedia.org/wiki/Axiom_of_countable_choice
And you even need it to make sure that the characterisation of epsilon-delta defined convergence in terms of sequences is indeed a true theorem.

 

[RandomAllTime]

If you like Banach Tarski then you like Axiom of Choice. If you don't like Banach Tarski then you are free to trash the axiom of choice, and now no Banach Tarksi.

Axiom of Choice is usually thought to be a useful thing in mathematics since (a) it seems intuitive (b) it makes it easier to prove theorems claiming that certain things exists. Well that is fine as long as those are things you kind of like to exist, but at some point it also starts allowing things to exist which see counter-intuitive, and maybe you don't like that.

For most of practical mathematics, the axiom of countable choice is quite enough to do everything you want to do. https://en.wikipedia.org/wiki/Axiom_of_countable_choice
And you even need it to make sure that the characterisation of epsilon-delta defined convergence in terms of sequences is indeed a true theorem.

I see. Thanks

 

[WWGD]

Hi guys. So I've been wondering, what's so controversial about the axiom of choice? I heard it allows the Banach-Tarski Paradox to work. A little insight would be much appreciated, thanks.

Banach Tarski requires the existence of uncountably-many atoms, which does not hold " in this universe" . And, AFAIK, it requires infinitely-many operations.

 

[mathman]

Banach-Tarski uses a mathematical fact that the number of points in a sphere is uncountable, and with the axiom of choice it can be divided into a finite number of unmeasurable sets.

 

[WWGD]

Banach-Tarski uses a mathematical fact that the number of points in a sphere is uncountable, and with the axiom of choice it can be divided into a finite number of unmeasurable sets.

Isnt this equivalent to the existence of infinitely-many ( at least countably -) atoms? And isn't the cardinality of the operations resulting in the partition infinite?

 

[pbuk]

Isnt this equivalent to the existence of infinitely-many ( at least countably -) atoms?

No, atoms do not form a continuum. In any case you cannot "prove" anything about the real world using maths.

 

[WWGD]

No, atoms do not form a continuum. In any case you cannot "prove" anything about the real world using maths.

I mean one can argue reasonably -well that a ball containing uncountably-many points will contain infinitely-many atoms. But , yes, this would have to be laid out carefully.

 

[micromass]

Isnt this equivalent to the existence of infinitely-many ( at least countably -) atoms?

Yes, the Banach-Tarski paradox assumes the existence of uncountably many atoms. Although the word atom is confusing, since it has nothing to do with the real world atoms. Here, atom is just an indivisible point with zero volume.

And isn't the cardinality of the operations resulting in the partition infinite?

I don't really know what you mean with this.

 

[WWGD]

Yes, the Banach-Tarski paradox assumes the existence of uncountably many atoms. Although the word atom is confusing, since it has nothing to do with the real world atoms. Here, atom is just an indivisible point with zero volume.
I don't really know what you mean with this.

I mean the number of steps needed to do the partition.

 

[micromass]

What would be a step?

 

[WWGD]

A transformation on the Ball , to decompose it into the nonmeasurable parts. Let me see how to define it more clearly.

 

[micromass]

Why would you need to transform the ball to partition it??

 

[WWGD]

Why would you need to transform the ball to partition it??

Because we are assumming it is a physical , "real world" ball. How else would we go from a standard ball into the collection of non-measurable pieces?

 

[micromass]

I don't know, but Banach-Tarski isn't about how you would do it in practice. It involves the axiom of choice and thus an infinite amount of choices which is impossible in the real world anyway.

 

[WWGD]

I don't know, but Banach-Tarski isn't about how you would do it in practice. It involves the axiom of choice and thus an infinite amount of choices which is impossible in the real world anyway.

If it were possible, I would be rich by now, buying .1 oz of gold and doubling its volume many times. I don't know if there are physical models of non-measurable sets.

 

[micromass]

If it were possible, I would be rich by now, buying .1 oz of gold and doubling its volume many times.

Not necessarily. Just because it is possible doesn't mean it's practically feasible. Physics still doesn't know whether there are nonmeasurable sets out there. So they might still exist.

 

[WWGD]

Not necessarily. Just because it is possible doesn't mean it's practically feasible. Physics still doesn't know whether there are nonmeasurable sets out there. So they might still exist.

Well, maybe contrived, but if I can come up with a way and convince someone of it, pretty sure I can borrow enough to have it done. But this may be far OT. And this practically feasible aspect has to see with the fact that this cannot be done in a finite number of steps, if at all. EDIT: maybe tautological, but if it could be done in a number of steps, it would be feasible.

 

[glaucousNoise]

so, if we are an ultrafinitist, we needn't bother with the axiom of choice?

 

[micromass]

so, if we are an ultrafinitist, we needn't bother with the axiom of choice?

Indeed, the axiom of choice is something that arises from the notion of infinity.

 

[glaucousNoise]

hmm, what's the use of infinity, as a pure notion, rather than a practical one?

for a physicist, infinity is "a scale >> the characteristic scale of the system", but it's always actually a finite number.

when I say use, I mean for a mathematician asking a pure mathematical question, since for an applied mathematician it's clearly only useful as an approximation (except in rare philosophical circumstances where one ponders whether or not space is actually continuous etc).

 

[micromass]

What's the use of anything in pure mathematics? Sure, pure math has nice applications. But when you talk about "usefulness" in pure mathematics, you must elaborate what that would be to you.

 

[glaucousNoise]

What motivates mathematicians to retain this philosophically difficult definition of infinity?

 

[micromass]

Because mathematicians don't find it difficult at all. Mathematically, infinity is very well understood.

 

[glaucousNoise]

hmm, so why did it stop being controversial?

 

[micromass]

Because it is now very well understood and it shows up in all of pure mathematics.

 

[glaucousNoise]

sorry, I was referring to the axiom of choice and was unclear

 

[micromass]

Same answer.

 

[WWGD]

What motivates mathematicians to retain this philosophically difficult definition of infinity?

I think it is useful in e.g., probabilities, where sample spaces are often infiite, i.e., there are infnitely-many possible outcomes.

 

[gill1109]

The Banach-Tarski theorem needs the axiom of choice. The axiom of choice is independent of the other axioms of set theory. If you don't like the axiom of choice you can say that you don't want to have it. Instead, you can have a different axiom making all sets measurable and making Banach-Tarski theorem false. So whether or not the Banach-Tarski theorem is true has got nothing whatsoever to do with the real world but only to do with reasoning about infinite sets. What rules do we accept and what rules do we not accept?

 

[glaucousNoise]

I think it is useful in e.g., probabilities, where sample spaces are often infiite, i.e., there are infnitely-many possible outcomes.

but are there? suppose you take a sample of heights, and then fit a continuous distribution to those heights. When you calculate the probability of sampling a height between two data values, you're really making a guess based upon the data. Heights among humans may actually vary in discrete increments, but these increments may be so tiny (maybe on the order of nanometers) that it's reasonable to just come up with a rule (a distribution) that produces a value for any interval of rational numbers you feed it, even if it's actually invalid for small intervals (on the order of a nanometer, perhaps).

Is there a need for a notion of real numbers here? Your rule assigns a value to any rational number you give it, and there are as many of those as you'd like, but there's a discrete scale at which the distribution is effective anyway. In real statistical applications you normally need to numerically integrate anyway, which requires discretizing your continuous distribution.

 

[micromass]

Is there a need for a notion of real numbers here?

No, there absolutely isn't. Everything you can do with real numbers is something you can do with rational numbers. But working with real numbers makes things vastly more simple, and that's why we work with them.

 

[WWGD]

but are there? suppose you take a sample of heights, and then fit a continuous distribution to those heights. When you calculate the probability of sampling a height between two data values, you're really making a guess based upon the data. Heights among humans may actually vary in discrete increments, but these increments may be so tiny (maybe on the order of nanometers) that it's reasonable to just come up with a rule (a distribution) that produces a value for any interval of rational numbers you feed it, even if it's actually invalid for small intervals (on the order of a nanometer, perhaps).

Is there a need for a notion of real numbers here? Your rule assigns a value to any rational number you give it, and there are as many of those as you'd like, but there's a discrete scale at which the distribution is effective anyway. In real statistical applications you normally need to numerically integrate anyway, which requires discretizing your continuous distribution.

How would you deal with numbers like $e, \pi$ , which are not "made in a lab " (i.e., they come about from "real world" scenarios/situations)? Would you approximate them by Rationals to the needed level of accuracy?

 

[pwsnafu]

How would you deal with numbers like $e, \pi$ , which are not "made in a lab " (i.e., they come about from "real world" scenarios/situations)? Would you approximate them by Rationals to the needed level of accuracy?

You can do that, or just work out which constants you care about ($e, \pi, \sqrt2, \log2$ etc) and consider the field extension over the rationals. This is what computer algebra systems do.

 

[Demystifier]

For most of practical mathematics, the axiom of countable choice is quite enough to do everything you want to do. https://en.wikipedia.org/wiki/Axiom_of_countable_choice

If I understood it correctly, with axiom of countable choice (ACC) there is no Banach-Taraski theorem/paradox. It looks like a good reason to use only ACC, and not the standard AC. Or is there a good example of some practical mathematics for which we still need the full standard AC?

 

[micromass]

If I understood it correctly, with axiom of countable choice (ACC) there is no Banach-Taraski theorem/paradox. It looks like a good reason to use only ACC, and not the standard AC. Or is there a good example of some practical mathematics for which we still need the full standard AC?

Define practical mathematics.

 

[Demystifier]

Define practical mathematics.

A tentative answer: Any branch of mathematics except logic, set theory, and category theory.

Or let me reformulate my question. Is there a theorem which (unlike Banach-Tarski) most mathematicians find intuitively appealing, and which can be proved by AC, but not by ACC?

 

[micromass]

The axiom of choice is equivalent to saying that every vector space has a basis. Does that answer your question?

 

[Demystifier]

The axiom of choice is equivalent to saying that every vector space has a basis. Does that answer your question?

Oh yes, that's great!
Is then ACC equivalent to saying that every separable vector space has a basis? Or something like that?

 

[micromass]

Uh well, what does "separable vector space mean"? I don't think a vector space has a canonical topology in infinite dimensions.

Other equivalenties:
1) Tychonoff theorem: the product of compact spaces is compact (the definition of compact matters here: here it is that every open cover has a finite subcover).
2) Every nontrivial unital ring has a maximal ideal.
3) Perhaps set theoretic but still: every product of sets is nonempty.

There are entire books written on these subject (including my thesis, I guess). So do ask if you want to know more.

 

[WWGD]

You can do that, or just work out which constants you care about ($e, \pi, \sqrt2, \log2$ etc) and consider the field extension over the rationals. This is what computer algebra systems do.

But I think GlaucousNoise was referring to the theoretical need for the existence of Irrationals.

 

[Demystifier]

Uh well, what does "separable vector space mean"?

https://en.wikipedia.org/wiki/Hilbert_space#Separable_spaces

 

[S.G. Janssens]

So this is a case where one should listen to what you mean, not what you say?

 

[Demystifier]

Other equivalenties:
1) Tychonoff theorem: the product of compact spaces is compact (the definition of compact matters here: here it is that every open cover has a finite subcover).
2) Every nontrivial unital ring has a maximal ideal.
3) Perhaps set theoretic but still: every product of sets is nonempty.

There are entire books written on these subject (including my thesis, I guess). So do ask if you want to know more.

That reminded me of a problem that I had in mathematical physics:
The integral can be thought of as precise way to define a continuous sum of uncountably many terms.
Is there a precise way to define a continuous product of an uncountable many terms? (I mean, without explicitly using logaritms which reduce products to sums.)

 

[Demystifier]

So this is a case where one should listen to what you mean, not what you say?

Maybe, but Hilbert space is a kind of vector space too, so not necessarily.

 

[gill1109]

That reminded me of a problem that I had in mathematical physics:
The integral can be thought of as precise way to define a continuous sum of uncountably many terms.
Is there a precise way to define a continuous product of an uncountable many terms? (I mean, without explicitly using logaritms which reduce products to sums.)

Yes, it is called the product-integral. https://en.wikipedia.org/wiki/Product_integral

 

[Demystifier]

Yes, it is called the product-integral. https://en.wikipedia.org/wiki/Product_integral

As I suspected, they use logarithms to reduce it to ordinary integrals. I was hoping that it can be done without the logarithms, but I guess my hope was groundless. Nevertheless, I am glad to see that at least there is a standard notation for such a thing.

 

[gill1109]

As I suspected, they use logarithms to reduce it to ordinary integrals. I was hoping that it can be done without the logarithms, but I guess my hope was groundless. Nevertheless, I am glad to see that at least there is a standard notation for such a thing.

Some people do it with logarithms. I do not. https://projecteuclid.org/euclid.aos/1176347865
Ann. Statist.
Volume 18, Number 4 (1990), 1501-1555.
A Survey of Product-Integration with a View Toward Application in Survival Analysis
Richard D. Gill and Soren Johansen