Exploring the Controversy Around the Axiom of Choice

[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