• (disco) in reply to PWolff
    PWolff:
    Maybe if we take/assume the axiom of choice as granted (and therefore that the set of indescribable numbers can be well-ordered), we can run into a contradiction:

    I'm not sure that that's an entirely sound thing. We're talking about a set of irreducibly infinite objects (since if they had a reduction, they wouldn't be in the set in the first place) so stuff starts to get weird. For example, the amount of time to decide whether one object is less than another one in the well-order is not necessarily bounded.

  • (disco) in reply to RaceProUK
    RaceProUK:
    Seems my trolling was a little too subtle...

    I explained my reasoning. I even checked raw! Swearsies!

  • (disco) in reply to ScholRLEA
    ScholRLEA:
    IIUC, there is a good chance that the question of whether indescribable numbers exist as a sub-set of numbers is undecidable.

    I admit I didn't think of that while writing that stuff down, but I fully agree.

    ScholRLEA:
    What fun!

    I agree here, too.

  • (disco) in reply to dkf
    dkf:
    the amount of time to decide whether one object is less than another one in the well-order is not necessarily bounded.

    The point (and the definition) of a well-order is that every subset has one and only one smallest element. And when talking about a well-ordered set, mathematicians always postulate they can get at that element one way or the other.

    dkf:
    the amount of time

    Time isn't something mathematicians (in their role as mathematicians) are concerned with.

  • (disco) in reply to PWolff
    PWolff:
    Time isn't something mathematicians (in their role as mathematicians) are concerned with.

    As long as there's a finite decision procedure, I'd agree. What about when there isn't? Remember, these numbers are by definition irreducible, so any comparison for either equality or ordering may need to examine an unbounded number of digits/terms in order to reach a decision. It's this which sends things wonky.

    Truly infinite objects make many mathematicians' heads hurt so much that for a long time, there was a movement to deny that such things could even exist in a platonic sense.

  • (disco) in reply to dkf
    dkf:
    Truly infinite objects make many mathematicians' heads hurt so much that for a long time, there was a movement to deny that such things could even exist in a platonic sense.

    I don't believe that movement is extinct: https://en.wikipedia.org/wiki/Constructivism_%28mathematics%29

    And maybe Platon himself didn't think so: https://en.wikipedia.org/wiki/Constructivist_epistemology#History

    For to close the circle: The first of the above linked articlse has a link to the article about computable numbers, which are the only ones some flavors of constructivism accept.

  • (disco) in reply to PWolff
    PWolff:
    I don't believe that movement is extinct

    But as with a very large number of things in mathematics, it all depends on what axioms you really have. An assumption that you can decide whether two values are equivalent in finite time (i.e., that you've got a terminating decision procedure for numeric equivalence) is fine, but it is most certainly an assumption, an axiom.

  • (disco) in reply to dkf
    dkf:
    urkerab:
    Not only can't I describe any indescribable numbers, but now I find I can't describe my list of describable numbers, which surely means that some of those numbers aren't actually describable after all?
    Write an algorithm to create the list of describable numbers. Job done!
    That's the easy bit; you just enumerate all possible descriptions and find out what numbers they describe. The tricky bit is when Cantor comes along and describes a number that your algorithm will never output; if I've understood correctly (always a risky assumption) @PWolff says that this is because there's no such thing as the set of describable numbers.
  • (disco) in reply to urkerab

    Constructivists...I hate constructivists.

  • (disco) in reply to urkerab
    urkerab:
    @PWolff says that this is because there's no such thing as the set of describable numbers.

    And I say that this is because there are numbers that have “infinite descriptions” (i.e., that have infinitely many digits and an infinite Kolmogorov representation). Once you're dealing with infinite representations, Cantor proves that you can't enumerate them all.

  • (disco) in reply to boomzilla

    Why did it have to be constructivists?

    Filed under: be careful with that nuke, Indy.

  • (disco) in reply to RaceProUK
    RaceProUK:
    Seems my trolling was a little too subtle...
    [image]
  • (disco)

    Another question about representable numbers:

    Is the class of non-describable real numbers a set? (closely related: Is the class of real numbers a set?)

    If it is a set, what about the well-ordering theorem? Wouldn't "the first number of this well-ordering of the set of the non-describable numbers" be a description of a non-describable number?

    Or does this imply that there exist well-orderings of the set of the non-describable numbers, but there is no way to describe any of them?

    Can we conclude that there is no way to describe a well-ordering of the set of the real numbers, even if there exists at least one?

  • (disco) in reply to PWolff

    The sets of describable and undescribable numbers are dense in each other. They are well-ordered sets, but there is no smallest item. For comparison, consider the set { 1/n | n \in N }

  • (disco) in reply to PleegWat

    Of course. But in what way does this contribute?

  • (disco) in reply to PWolff
    PWolff:
    Is the class of non-describable real numbers a set?

    That's approximately equivalent to asking if the complement of a set is also a set. (You'd also need to establish the class of describable numbers is a set, but I suspect that isn't as deep a philosophical problem. The class of all real numbers is usually accepted as being a set, so we have a meaningful universe.)

  • (disco) in reply to PWolff

    The real numbers are well-ordered. Hence all subsets of the real numbers are well-ordered. The class of describable numbers is a well-defined (assumption) set, and hence so is it's complement in the real numbers, the set of undescribable real numbers.

    There may be a way to pick a specific number out of that set, but something as simple as 'the smallest' or 'the smallest in X range' will not work, as it is possible for any infinite subset of the reals to not have a smallest member. The set 1/n, with n a natural number, is an example of this - it's a subset of the real numbers, but it does not have a smallest element, even though all elements are larger than 0.

  • (disco) in reply to PleegWat
    PleegWat:
    There may be a way to pick a specific number out of that set, but something as simple as 'the smallest' or 'the smallest in X range' will not work

    Therefore I said "first" instead of "smallest" in

    PWolff:
    Wouldn't "the first number of this well-ordering

    I conjecture that there is no way to describe a well-ordering of the real numbers. That was my point.

    For that consideration I needed a well-ordering of the set of non-describable numbers.

  • (disco) in reply to PWolff

    It's one of those uncertain things that depends on what axioms you like. Quoth Wikipedia:

    From the ZFC axioms of set theory (including the axiom of choice) one can show that there is a well-order of the reals. Also Wacław Sierpiński proved that ZF + GCH (the generalized continuum hypothesis) imply the axiom of choice and hence a well-order of the reals. Nonetheless, it is possible to show that the ZFC+GCH axioms alone are not sufficient to prove the existence of a definable (by a formula) well-order of the reals. However it is consistent with ZFC that a definable well-ordering of the reals exists—for example, it is consistent with ZFC that V=L, and it follows from ZFC+V=L that a particular formula well-orders the reals, or indeed any set.
    So ZFC or ZF+GCH gives you a well-ordering. ZFC + GCH is not enough to prove a definable well-ordering. ZFC does not disprove a definable well-ordering. ZFC + V=L proves a definable well-ordering.

    Of course if you dislike AC and GCH then there might not be a well-ordering at all.

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