4 ms·

Fun fact about considering R as a vector space over Q: the Hamel basis A mentioned in the post is not only infinite, but even uncountable.

by nmc 13y ago

scotty79 13y ago
I can see now why some mathematicians would like to do without axiom of choice.
- ColinWright 13y ago
  Consider the product of two sets, A and B: A x B = { (a,b): a in A, b in B } Pretty obviously if A and B are both non-empty, the product is non-empty. The axiom of choice says that this is still true even when you take the product of infinitely many sets. Without the axiom of choice, the product of infinitely many non-empty sets could be empty. Is that a better choice?
  - scotty79 13y ago
    Infinty is unreal. Any formal intuition we build around it is free to choose. If you can imagine base of R then surely you can imagine infinite number of nonempty sets that have empty cartesian product. They might all be not empty but be sort of asymptotically empty, like probably not empty but with that probabiliy going to zero as number of sets goes to infinity. The question is, does any good physics come from the axiom of choice?
    - ColinWright 13y ago
      > If you can imagine base of R then surely you can > imagine infinite number of nonempty sets that have > empty cartesian product. So you think it's OK to have infinitely may sets and not be able to choose one from each of them, even though they're all non-empty? That's a choice you're free to make, of course, but to me it seems more perverse than the alternatives. > The question is, does any good physics come from > the axiom of choice? Does it matter? Once you're dealing with uncountable infinities you're already outside physics and pursuing things because they're interesting, not because they have immediate practical implications. Of course, bits of pure math has a tendency to crop up and be useful decades later, and you never know which bits that will be.
      - scotty79 13y ago
        Weird things happen when you go from "every" to "all" for uncountable infinities. Loosing ability to pick elements wouldn't be the weirdest. Physics matters. But since math is purely intelectual thing of course you are welcomed to pursue both paths. After all physics is not the only game we play. If someting is not useful in the real world it can be useful in virtual worlds we construct. All math is eventually applied math is a self fullfilling prophecy. Most mathematitians choose to accept axiom of choice because it makes lots of things easier and lots even possible. It isn't better in any other sense than the oposite.
      - Tloewald 13y ago
        How do you feel about well ordering the reals? Because the Axiom of Choice gives you that too and so much more. I'm not a zealot one way or another, but don't argue for it based on intuition, it plays both ways.
        ColinWright 13y ago
        As the old joke goes: We all know that the Axiom of Choice is true, Well-Ordering of the Reals is false, and who knows about Zorn's Lemma. I use AC when I need it, avoid it when I can do without it, and sometimes look at what it allows versus what it implies.
        thaumasiotes 13y ago
        I've never really understood why the well-ordering theorem is the version of the axiom of choice that's supposed to be "obviously false". It's pretty trivial to see how you get it from the AC, and since the AC is "obviously true" I've always felt the well-ordering theorem is intuitive as well. (Actually, my mental model of the process generally leads me to want to think "all sets are countable"... c'est la vie :/ ) The Banach-Tarski theorem, on the other hand, is much more robustly "obviously false".
      - jerf 13y ago
        "So you think it's OK to have infinitely may sets and not be able to choose one from each of them, even though they're all non-empty?" There's a lot hidden behind the word "choose". Of course I have no objection to you just grabbing whatever out of each set, but when you "choose" by performing an arbitrarily unbounded computation, it sounds less un-perverse. I am aware I've implicitly used a constructivist formulation by reference any sort of "computation". Personally I'm ambivalent about the axiom itself, I just think it should always be clear whether or not you've chose it, which is generally itself not a problem, so, generally I'm happy either way. It is the case that when you use the Axiom of Choice, you've just irretrievably left the physical universe, which is fine, but worth being aware of.
    - thaumasiotes 13y ago
      > The question is, does any good physics come from the axiom of choice? In the first place, this isn't the question. In the second place, obviously, no good physics comes from the axiom of choice, just as no good physics comes from denying the axiom of choice. If there were any known practical applications, they would be mentioned as soon as you heard of the thing.
    - pfortuny 13y ago
      Infinity is unreal... Well, it allows quite a few arguments concerning asymptotics of algorithms, for example. The "reality" of a concept is something on which many litres of ink and blood have been shee.
      - scotty79 13y ago
        Yes. We reason about infinities. But when it comes to result we are always interested in with what's well before it.
  - adsche 13y ago
    > Without the axiom of choice, the product of infinitely many non-empty sets could be empty. This sounds like a stupid question, but I can't really figure it out: How could it be empty, given your above formula? (Math level: Engineer) Edit: Wait, is it because every set contains the empty set? And without choosing you could end up with the empty set?
    - ColinWright 13y ago
      This is what the axiom of choice is about. It says: Given a collection of non-empty sets, there exists a "choice function" that chooses one element from each set. (AC) The problem is, AC can be used to prove all sorts of counter-intuitive results. The example that turns up here on HN all the time is Banach-Tarski[0][1], but this submission about x^2 being the sum of three periodic functions is another. AC has been proven to be independent of the usual axioms of set theory, so in some sense you can choose whether to believe it or not. If it's true then Banach-Tarski is true. If it's false then you can have infinitely many non-empty sets with an empty Cartesian product. (corrected to read "an empty" instead of the error "a non-empty" - thanks thaumasiotes) Pure math be crazy. And cool. [0] http://en.wikipedia.org/wiki/Banach%E2%80%93Tarski_paradox http://en.wikipedia.org/wiki/Banach%E2%80%93Tarski_paradox [1] https://www.hnsearch.com/search#request/all&q=banach+tarski https://www.hnsearch.com/search#request/all&q=banach+tarski ---- Edit: No, it's not because of your edit: > ... is it because every set contains the empty set? > And without choosing you could end up with the empty set? Not every set contains the empty set as an element, but every set has the empty set as a subset. We are talking about choosing elements, and the it that says "non-empty set" means that for every individual set we can choose one of its elements. The problem comes in that when there are infinitely many, including possibly uncountably infinitely many, we need to do all these choices "at once" (in some sense).
      - thaumasiotes 13y ago
        > If it's false then you can have infinitely many non-empty sets with a non-empty Cartesian product. with an empty Cartesian product, I think?
      - 13y ago
  - CJefferson 13y ago
    As a (beginner) constructionist, the problem is not that the product "could be empty", but just that the product can't be constructed. I view this in a similar way to Russell's Paradox (while I'm not claiming the axiom of choice leads to a contradiction) -- we can define how we can construct new objects, and we can say you can't invoke the axiom of choice, but have to give an explicit choice function.
Grue3 13y ago
This easily follows from the fact that R itself is uncountable. Otherwise it's trivial to enumerate all real numbers based on their decomposition.

Fun fact about considering R as a vector space over Q: the Hamel basis A mentioned in the post is not only infinite, but even uncountable.

by nmc 13y ago

Fun fact about considering R as a vector space over Q: the Hamel basis A mentioned in the post is not only infinite, but even uncountable.

scotty79 13y ago

I can see now why some mathematicians would like to do without axiom of choice.

ColinWright 13y ago
Consider the product of two sets, A and B: A x B = { (a,b): a in A, b in B } Pretty obviously if A and B are both non-empty, the product is non-empty. The axiom of choice says that this is still true even when you take the product of infinitely many sets. Without the axiom of choice, the product of infinitely many non-empty sets could be empty. Is that a better choice?
- scotty79 13y ago
  Infinty is unreal. Any formal intuition we build around it is free to choose. If you can imagine base of R then surely you can imagine infinite number of nonempty sets that have empty cartesian product. They might all be not empty but be sort of asymptotically empty, like probably not empty but with that probabiliy going to zero as number of sets goes to infinity. The question is, does any good physics come from the axiom of choice?
  - ColinWright 13y ago
    > If you can imagine base of R then surely you can > imagine infinite number of nonempty sets that have > empty cartesian product. So you think it's OK to have infinitely may sets and not be able to choose one from each of them, even though they're all non-empty? That's a choice you're free to make, of course, but to me it seems more perverse than the alternatives. > The question is, does any good physics come from > the axiom of choice? Does it matter? Once you're dealing with uncountable infinities you're already outside physics and pursuing things because they're interesting, not because they have immediate practical implications. Of course, bits of pure math has a tendency to crop up and be useful decades later, and you never know which bits that will be.
    - scotty79 13y ago
      Weird things happen when you go from "every" to "all" for uncountable infinities. Loosing ability to pick elements wouldn't be the weirdest. Physics matters. But since math is purely intelectual thing of course you are welcomed to pursue both paths. After all physics is not the only game we play. If someting is not useful in the real world it can be useful in virtual worlds we construct. All math is eventually applied math is a self fullfilling prophecy. Most mathematitians choose to accept axiom of choice because it makes lots of things easier and lots even possible. It isn't better in any other sense than the oposite.
    - Tloewald 13y ago
      How do you feel about well ordering the reals? Because the Axiom of Choice gives you that too and so much more. I'm not a zealot one way or another, but don't argue for it based on intuition, it plays both ways.
      - ColinWright 13y ago
        As the old joke goes: We all know that the Axiom of Choice is true, Well-Ordering of the Reals is false, and who knows about Zorn's Lemma. I use AC when I need it, avoid it when I can do without it, and sometimes look at what it allows versus what it implies.
      - thaumasiotes 13y ago
        I've never really understood why the well-ordering theorem is the version of the axiom of choice that's supposed to be "obviously false". It's pretty trivial to see how you get it from the AC, and since the AC is "obviously true" I've always felt the well-ordering theorem is intuitive as well. (Actually, my mental model of the process generally leads me to want to think "all sets are countable"... c'est la vie :/ ) The Banach-Tarski theorem, on the other hand, is much more robustly "obviously false".
    - jerf 13y ago
      "So you think it's OK to have infinitely may sets and not be able to choose one from each of them, even though they're all non-empty?" There's a lot hidden behind the word "choose". Of course I have no objection to you just grabbing whatever out of each set, but when you "choose" by performing an arbitrarily unbounded computation, it sounds less un-perverse. I am aware I've implicitly used a constructivist formulation by reference any sort of "computation". Personally I'm ambivalent about the axiom itself, I just think it should always be clear whether or not you've chose it, which is generally itself not a problem, so, generally I'm happy either way. It is the case that when you use the Axiom of Choice, you've just irretrievably left the physical universe, which is fine, but worth being aware of.
  - thaumasiotes 13y ago
    > The question is, does any good physics come from the axiom of choice? In the first place, this isn't the question. In the second place, obviously, no good physics comes from the axiom of choice, just as no good physics comes from denying the axiom of choice. If there were any known practical applications, they would be mentioned as soon as you heard of the thing.
  - pfortuny 13y ago
    Infinity is unreal... Well, it allows quite a few arguments concerning asymptotics of algorithms, for example. The "reality" of a concept is something on which many litres of ink and blood have been shee.
    - scotty79 13y ago
      Yes. We reason about infinities. But when it comes to result we are always interested in with what's well before it.
- adsche 13y ago
  > Without the axiom of choice, the product of infinitely many non-empty sets could be empty. This sounds like a stupid question, but I can't really figure it out: How could it be empty, given your above formula? (Math level: Engineer) Edit: Wait, is it because every set contains the empty set? And without choosing you could end up with the empty set?
  - ColinWright 13y ago
    This is what the axiom of choice is about. It says: Given a collection of non-empty sets, there exists a "choice function" that chooses one element from each set. (AC) The problem is, AC can be used to prove all sorts of counter-intuitive results. The example that turns up here on HN all the time is Banach-Tarski[0][1], but this submission about x^2 being the sum of three periodic functions is another. AC has been proven to be independent of the usual axioms of set theory, so in some sense you can choose whether to believe it or not. If it's true then Banach-Tarski is true. If it's false then you can have infinitely many non-empty sets with an empty Cartesian product. (corrected to read "an empty" instead of the error "a non-empty" - thanks thaumasiotes) Pure math be crazy. And cool. [0] http://en.wikipedia.org/wiki/Banach%E2%80%93Tarski_paradox http://en.wikipedia.org/wiki/Banach%E2%80%93Tarski_paradox [1] https://www.hnsearch.com/search#request/all&q=banach+tarski https://www.hnsearch.com/search#request/all&q=banach+tarski ---- Edit: No, it's not because of your edit: > ... is it because every set contains the empty set? > And without choosing you could end up with the empty set? Not every set contains the empty set as an element, but every set has the empty set as a subset. We are talking about choosing elements, and the it that says "non-empty set" means that for every individual set we can choose one of its elements. The problem comes in that when there are infinitely many, including possibly uncountably infinitely many, we need to do all these choices "at once" (in some sense).
    - thaumasiotes 13y ago
      > If it's false then you can have infinitely many non-empty sets with a non-empty Cartesian product. with an empty Cartesian product, I think?
    - 13y ago
- CJefferson 13y ago
  As a (beginner) constructionist, the problem is not that the product "could be empty", but just that the product can't be constructed. I view this in a similar way to Russell's Paradox (while I'm not claiming the axiom of choice leads to a contradiction) -- we can define how we can construct new objects, and we can say you can't invoke the axiom of choice, but have to give an explicit choice function.

Grue3 13y ago

This easily follows from the fact that R itself is uncountable. Otherwise it's trivial to enumerate all real numbers based on their decomposition.