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The point of my example is to show that the functions can exist even if their domain includes "indescribable" numbers. It shows that trivially the "indescribabl

The point of my example is to show that the functions can exist even if their domain includes "indescribable" numbers. It shows that trivially the "indescribable-ness" of numbers in the domain of the function is not relevant to whether the function exists. The author's argument relies on "indescribable-ness" somehow blocking the ability to "choose" an element. But that trades on a vague, intuitive understanding of "choosing", which doesn't capture what the AC is about. The AC is about the existence of certain functions. Since the "indescribable-ness" of the domain of a function is not relevant to its existence, the author's argument doesn't successfully attack the AC.

lisper 4y ago
> The point of my example is to show that the functions can exist even if their domain includes "indescribable" numbers. But that is not at issue. What is at issue is whether or not a function exists whose domain is an (infinite) set of indescribable numbers and whose range is a member of the set. The existence of such a function is far less clear. Of course you can postulate the existence of such a function. But you can also postulate the existence of an oracle for the halting problem. My gut tells me that these two ideas have a lot in common.
- boole1854 4y ago
  Remember that I am not here trying to argue for the truth of the AC; I'm arguing against the author's argument. The author's argument, as I read it, relies on an intuition that "choosing" an element that is not-finitely-describable is problematic. However, the mathematical meaning of "choosing", in the context of the AC, is a statement about the mathematical existence of a certain function. To respond to the author's intuition requires only showing that not-finitely-describable domains are not a "problem" for the existence of functions. Maybe it is difficult to "choose", in some natural language sense, an input to f(x) = x + 1 such that the output x + 1 is a real number (e.g., you can't write down such an input one digit at a time on a piece of paper, even an infinitely long piece of paper). But this does not mean it is problematic for the function to exist in the mathematical sense.
  - lisper 4y ago
    FYI, I am the author. > Remember that I am not here trying to argue for the truth of the AC; I'm arguing against the author's argument. Yep, I get that. > The author's argument, as I read it, relies on an intuition that "choosing" an element that is not-finitely-describable is problematic. Not quite. Choosing an element from a set where all of the members are not-finitely-describable is problematic. > However, the mathematical meaning of "choosing", in the context of the AC, is a statement about the mathematical existence of a certain function. Yep, I get that. > To respond to the author's intuition requires only showing that not-finitely-describable domains are not a "problem" for the existence of functions. No, you would have to show that defining a function on this particular set is not a problem. Note that this is not the same thing as defining a function on elements of this set. That is obviously not a problem. But defining a (non-trivial) function on the set itself is.
    - riversflow 4y ago
      I just don’t see what describability, or as jiggawatts calls it, naming, has to do with AC. AC is built upon an indexed set[1]. That way you can just reference the index. It seems extremely contrived to me to argue that being finitely nameable has any bearing on underlying mathematical principle, nothing about any form of infinity is relatable to the physical confines of this universe, of which language is a member, but the concept of infinite is very useful. [1]https://en.wikipedia.org/wiki/Axiom_of_choice https://en.wikipedia.org/wiki/Axiom_of_choice
      - lisper 4y ago
        > I just don’t see what describability, or as jiggawatts calls it, naming, has to do with AC. It doesn't have anything to do with AoC per se, it has to do (I claim) with people's intuitions about AoC. Specifically, an infinite set of arbitrary subsets of the set of indescribable reals is something for which a reasonable person might be a tad suspicious about postulating the existence of a choice function.
- syzarian 4y ago
  There are models of ZFC in which every real number is definable. You may be interested in the top answer at this mathoverflow.net question: https://mathoverflow.net/questions/44102/is-the-analysis-as-taught-in-universities-in-fact-the-analysis-of-definable-numb https://mathoverflow.net/questions/44102/is-the-analysis-as-...
  - lisper 4y ago
    That is very interesting (have an upvote!) but I don't see how it invalidates my argument. If every real is definable then the set of non-definable reals is empty and AoC does not apply at all.
    - syzarian 4y ago
      The notion of countable and not countable does have some nuance. In the models that every real is definable that model, from within, believes it is uncountable since there is no bijection with a countable set from within that model. However, from the outside that model is countable. The point is there is a lot of nuance and no matter what one believes regarding choice there will be unintuitive results. For instance, when the axiom of choice fails you can have an infinite set with not countably infinite subset. The reals can be a countable union of countable sets. There is a set that can be partitioned into strictly more equivalence classes than the original set has elements, and a function whose domain is strictly smaller than its range. In fact, this is the case in all known models.
  - Kranar 4y ago
    We need to be careful with the language here. It's not true that there are models of ZFC where the real numbers are actually definable. What's true is there are models of ZFC where what is considered a "real number" from within that model is always definable (but will not be definable within that model). However, those sets do not form the actual real numbers, they are simply sets that from within the model satisfy a given definition of the reals without actually being the reals. A similar result is that there are models of the reals in ZFC that are countable, even though the reals are not countable. How can this be? Well it's similar to the above argument, those models do not represent the actual real numbers. All this means is that it's not possible to categorically define real numbers in first order ZFC. Any axiomatic definition of the reals in first order ZFC will have some models that satisfy those axioms but are not actually real numbers.
- cgio 4y ago
  Intuitively I would argue the definition of indescribability is loaded with semantics that can sprinkle paradoxes at will. A set from my perspective has at least two interfaces, one to populate it and one to query it. By making the population interface “indescribable” you get an “indescribable” method to query. Worst case you start from a describable superset of the indescribable one and test the “indescribable “ set population method to get you an indescribable set member. Which makes the indescribable actually describable in terms of this process. “?” on this whole statement, late night intuition from an indescribably bad at math sleepy guy.
  - recursive 4y ago
    > one to populate it and one to query it I'm not a math expert, but "populating" a set strikes me as an imperative programming mindset. In math, I understand a set more like a function returning a boolean that indicates membership. If the contents of the set change, then it's just a new set.
    - lisper 4y ago
      Yeah, this. A set does not have "two interfaces", it only has one: a membership function that takes an object and returns a boolean indicating if the object is a member of the set or not. Note that this function need not be computable, only well-defined.
      - cgio 4y ago
        Is the membership function not equivalent to what I describe above on the query side? My difference is that I am making explicit that the set needs to be populated. The uttering of a set of “indescribable “ objects does not a set give you. That membership function works both inwards and outwards.
        recursive 4y ago
        This concept of "population" doesn't really seem to be consistent with my understanding of a set. Set A is the set of all widgets that have a frozz. The existence of Set A doesn't imply that there exists a widget that has a frozz. Perhaps the existence of such a widget is an open question. But the set is perfectly well defined.
- loicd 4y ago
  The nature of the elements of the set does not matter, since the existence of a choice function on the set A guarantees the existence of a choice function on the set B as soon as there is a bijection between A and B. Thus, no matters how counter-intuitive or unnatural one finds the elements of a set, or whether they model physical reality, what matters for the axiom of choice is whether one can construct a bijection with a set that has a known choice function. By the way, it is worth keeping in mind how Gödel proved the consistency of the axiom of choice. Roughly speaking, the steps are: start with a model of ZF, build from it an inner model where all sets are definable (in a sense) in terms of ordinals, that model (called the "constructible universe") satisfies the axiom of choice. In other words, the axiom of choice holds as soon as you assume that all sets are constructible.
  - lisper 4y ago
    > as soon as there is a bijection between A and B OK, but how are you going to construct a bijection from an arbitrary infinite set of undefinable numbers onto a set with a choice function without the AoC?
6gvONxR4sf7o 4y ago
It seems like the validity of what you're saying is intertwined with the axiom itself. You're going from a comment about the domain (the domain includes indescribable elements) to a comment about specific arguments (arguments may be indescribable). Ultimately, you're saying that if we have an indescribable x, then x+1 is conditionally describable (to abuse terminology). But I think that just sidesteps the question. Even then, if we're taking a computability lens, then indescribable(x) implies indescribable(x+1), which would seem to imply that applying your function to indescribable x's is a moot point. The fact that your function is finitely describable even though we can't describe each of its applications is admittedly a mindfuck.

lisper 4y ago

> The point of my example is to show that the functions can exist even if their domain includes "indescribable" numbers. But that is not at issue. What is at issue is whether or not a function exists whose domain is an (infinite) set of indescribable numbers and whose range is a member of the set. The existence of such a function is far less clear. Of course you can postulate the existence of such a function. But you can also postulate the existence of an oracle for the halting problem. My gut tells me that these two ideas have a lot in common.

boole1854 4y ago
Remember that I am not here trying to argue for the truth of the AC; I'm arguing against the author's argument. The author's argument, as I read it, relies on an intuition that "choosing" an element that is not-finitely-describable is problematic. However, the mathematical meaning of "choosing", in the context of the AC, is a statement about the mathematical existence of a certain function. To respond to the author's intuition requires only showing that not-finitely-describable domains are not a "problem" for the existence of functions. Maybe it is difficult to "choose", in some natural language sense, an input to f(x) = x + 1 such that the output x + 1 is a real number (e.g., you can't write down such an input one digit at a time on a piece of paper, even an infinitely long piece of paper). But this does not mean it is problematic for the function to exist in the mathematical sense.
- lisper 4y ago
  FYI, I am the author. > Remember that I am not here trying to argue for the truth of the AC; I'm arguing against the author's argument. Yep, I get that. > The author's argument, as I read it, relies on an intuition that "choosing" an element that is not-finitely-describable is problematic. Not quite. Choosing an element from a set where all of the members are not-finitely-describable is problematic. > However, the mathematical meaning of "choosing", in the context of the AC, is a statement about the mathematical existence of a certain function. Yep, I get that. > To respond to the author's intuition requires only showing that not-finitely-describable domains are not a "problem" for the existence of functions. No, you would have to show that defining a function on this particular set is not a problem. Note that this is not the same thing as defining a function on elements of this set. That is obviously not a problem. But defining a (non-trivial) function on the set itself is.
  - riversflow 4y ago
    I just don’t see what describability, or as jiggawatts calls it, naming, has to do with AC. AC is built upon an indexed set[1]. That way you can just reference the index. It seems extremely contrived to me to argue that being finitely nameable has any bearing on underlying mathematical principle, nothing about any form of infinity is relatable to the physical confines of this universe, of which language is a member, but the concept of infinite is very useful. [1]https://en.wikipedia.org/wiki/Axiom_of_choice https://en.wikipedia.org/wiki/Axiom_of_choice
    - lisper 4y ago
      > I just don’t see what describability, or as jiggawatts calls it, naming, has to do with AC. It doesn't have anything to do with AoC per se, it has to do (I claim) with people's intuitions about AoC. Specifically, an infinite set of arbitrary subsets of the set of indescribable reals is something for which a reasonable person might be a tad suspicious about postulating the existence of a choice function.
syzarian 4y ago
There are models of ZFC in which every real number is definable. You may be interested in the top answer at this mathoverflow.net question: https://mathoverflow.net/questions/44102/is-the-analysis-as-taught-in-universities-in-fact-the-analysis-of-definable-numb https://mathoverflow.net/questions/44102/is-the-analysis-as-...
- lisper 4y ago
  That is very interesting (have an upvote!) but I don't see how it invalidates my argument. If every real is definable then the set of non-definable reals is empty and AoC does not apply at all.
  - syzarian 4y ago
    The notion of countable and not countable does have some nuance. In the models that every real is definable that model, from within, believes it is uncountable since there is no bijection with a countable set from within that model. However, from the outside that model is countable. The point is there is a lot of nuance and no matter what one believes regarding choice there will be unintuitive results. For instance, when the axiom of choice fails you can have an infinite set with not countably infinite subset. The reals can be a countable union of countable sets. There is a set that can be partitioned into strictly more equivalence classes than the original set has elements, and a function whose domain is strictly smaller than its range. In fact, this is the case in all known models.
- Kranar 4y ago
  We need to be careful with the language here. It's not true that there are models of ZFC where the real numbers are actually definable. What's true is there are models of ZFC where what is considered a "real number" from within that model is always definable (but will not be definable within that model). However, those sets do not form the actual real numbers, they are simply sets that from within the model satisfy a given definition of the reals without actually being the reals. A similar result is that there are models of the reals in ZFC that are countable, even though the reals are not countable. How can this be? Well it's similar to the above argument, those models do not represent the actual real numbers. All this means is that it's not possible to categorically define real numbers in first order ZFC. Any axiomatic definition of the reals in first order ZFC will have some models that satisfy those axioms but are not actually real numbers.
cgio 4y ago
Intuitively I would argue the definition of indescribability is loaded with semantics that can sprinkle paradoxes at will. A set from my perspective has at least two interfaces, one to populate it and one to query it. By making the population interface “indescribable” you get an “indescribable” method to query. Worst case you start from a describable superset of the indescribable one and test the “indescribable “ set population method to get you an indescribable set member. Which makes the indescribable actually describable in terms of this process. “?” on this whole statement, late night intuition from an indescribably bad at math sleepy guy.
- recursive 4y ago
  > one to populate it and one to query it I'm not a math expert, but "populating" a set strikes me as an imperative programming mindset. In math, I understand a set more like a function returning a boolean that indicates membership. If the contents of the set change, then it's just a new set.
  - lisper 4y ago
    Yeah, this. A set does not have "two interfaces", it only has one: a membership function that takes an object and returns a boolean indicating if the object is a member of the set or not. Note that this function need not be computable, only well-defined.
    - cgio 4y ago
      Is the membership function not equivalent to what I describe above on the query side? My difference is that I am making explicit that the set needs to be populated. The uttering of a set of “indescribable “ objects does not a set give you. That membership function works both inwards and outwards.
      - recursive 4y ago
        This concept of "population" doesn't really seem to be consistent with my understanding of a set. Set A is the set of all widgets that have a frozz. The existence of Set A doesn't imply that there exists a widget that has a frozz. Perhaps the existence of such a widget is an open question. But the set is perfectly well defined.
loicd 4y ago
The nature of the elements of the set does not matter, since the existence of a choice function on the set A guarantees the existence of a choice function on the set B as soon as there is a bijection between A and B. Thus, no matters how counter-intuitive or unnatural one finds the elements of a set, or whether they model physical reality, what matters for the axiom of choice is whether one can construct a bijection with a set that has a known choice function. By the way, it is worth keeping in mind how Gödel proved the consistency of the axiom of choice. Roughly speaking, the steps are: start with a model of ZF, build from it an inner model where all sets are definable (in a sense) in terms of ordinals, that model (called the "constructible universe") satisfies the axiom of choice. In other words, the axiom of choice holds as soon as you assume that all sets are constructible.
- lisper 4y ago
  > as soon as there is a bijection between A and B OK, but how are you going to construct a bijection from an arbitrary infinite set of undefinable numbers onto a set with a choice function without the AoC?

6gvONxR4sf7o 4y ago

It seems like the validity of what you're saying is intertwined with the axiom itself. You're going from a comment about the domain (the domain includes indescribable elements) to a comment about specific arguments (arguments may be indescribable). Ultimately, you're saying that if we have an indescribable x, then x+1 is conditionally describable (to abuse terminology). But I think that just sidesteps the question. Even then, if we're taking a computability lens, then indescribable(x) implies indescribable(x+1), which would seem to imply that applying your function to indescribable x's is a moot point. The fact that your function is finitely describable even though we can't describe each of its applications is admittedly a mindfuck.