What Makes the Axiom of Choice So Controversial?

[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