The axiom of choice is required for the Banach-Tarski paradox to be true.

The axiom of choice is required for the Banach-Tarski paradox to be true. But because the Banach-Tarski paradox is absurd, does that not prove that the axiom of choice is not valid?

Because if the AoC is valid, then you can apparently decompose a ball into pieces and assemble them again into two identical balls. Obviously that's impossible.

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  1. 3 weeks ago
    Anonymous

    >The axiom of choice is required for the Banach-Tarski paradox to be true. But because the Banach-Tarski paradox is absurd, does that not prove that the axiom of choice is not valid?
    No, [math]B Rightarrow A[/math] does not entail [math]neg B Rightarrow neg A[/math].

    • 3 weeks ago
      Anonymous

      Well something must be wrong with the Banach-Tarski paradox and if it's not the axiom of choice then it's some other axiom. But the AoC is the most controversial one so it's probably that.

      • 3 weeks ago
        Anonymous

        Why must it be wrong? You know they're not real balls, right?

        • 3 weeks ago
          Anonymous

          We ca prove that something is wrong with the axioms of set theory by using the Sherlock Holmes logic (picrelated).

          Once we eliminate the impossible (and getting two new balls from one ball of same size is impossible mathematically), then whatever remains must be true.

    • 3 weeks ago
      Anonymous

      ((B IMPLY A) AND B) = A NOR B

      • 3 weeks ago
        Anonymous

        AND NOT B I meant.

    • 3 weeks ago
      Anonymous

      is the sideways L symbol an inversion operator?

      • 3 weeks ago
        Anonymous

        it's the "not" operand

  2. 3 weeks ago
    Anonymous

    >BRO LOOK AT THIS MEASURABLE OBJECT... WELL NO YOU CAN'T MEASURE THAT ONE THOUGH BUT LIKE LOOK AT IT NOW I MEASURE TWO!!
    Set theorists deserve to be crucified

  3. 3 weeks ago
    Anonymous

    >oh no, I deliberately disassembled this object into unphysical non-measurable sets and now I get an unphysical result
    >how could this ever have happened??
    >math isn't mathing

    • 3 weeks ago
      Anonymous

      YOU DON'T GET IT CHUD SET THEORY SAYS 1 = 2 AND SET THEORY IS MY RELIGION AND YOU'RE A RACIST BIGOT IF YOU DISAGREE
      Did you know 100% of set theorists are black

      • 3 weeks ago
        Anonymous

        Can confirm. Thought set theory prof was a asiatic until he peeled his skin off to reveal a big sweaty African man underneath

  4. 3 weeks ago
    Anonymous

    Or there is something wrong with proof derivation.
    IMO just because something theoretically exists doesn't mean you can use it in a proof without constructing it first - construction itself my put constraints on the object that existence alone does not.

  5. 3 weeks ago
    Anonymous

    Meanwhile, it would be a form of confirmation bias to discuss only counterintuitive consequences of the axiom of choice, without also discussing the counterintuitive situations that can occur when the axiom of choice fails. Although mathematicians often point to what are perceived as strange consequences of the axiom of choice, a fuller picture is revealed by also mentioning that many of the situations that can arise when one drops the axiom of choice are perhaps even more bizarre.
    For example, it is relatively consistent with the axioms of set theory without the axiom of choice that there can be a nonempty tree T, with no leaves, but which has no infinite path. That is, every finite path in the tree can be extended to further steps, but there is no path that goes forever. This situation can arise even when countable choice holds (so countable families of nonempty sets have choice functions), and this highlights the difference between the countable choice principle and the principle of dependent choice, where one makes countably many choices in succession. Finding a branch through a tree is an instance of dependent choice, since the later choices depend on which choices were made earlier.
    Without the axiom of choice, a real number can be in the closure of a set of real numbers X ⊂ R, but not the limit of any sequence from X. Without the axiom of choice, a function f : R R can be continuous in the sense that every convergent sequence xₙ x has a convergent image f(xₙ) f(x), but not continuous in the ε, δ sense. Without the axiom of choice, a set can be infinite, but have no countably infinite subset. Indeed, without the axiom of choice, there can be an infinite set, with all subsets either finite or the complement of a finite set. Thus, it can be incorrect to say that ℵ0 is the smallest infinite cardinality, since these sets would have an infinite size that is incomparable with ℵ0.
    1/2

    • 3 weeks ago
      Anonymous

      Without the axiom of choice, there can be an equivalence relation on R, such that the number of equivalence classes is strictly greater than the size of R. That is, you can partition R into disjoint sets, such that the number of these sets is greater than the number of real numbers. Bizarre! This situation is a consequence of the axiom of determinacy and is relatively consistent with the principle of dependent choice and the countable axiom of choice.
      Without the axiom of choice, there can be a field with no algebraic closure. Without the axiom of choice, the rational field Q can have different nonisomorphic algebraic closures. Indeed, Q can have an uncountable algebraic closure as well as a countable one. Without the axiom of choice, there can be a vector space with no basis, and there can be a vector space with bases of different cardinalities. Without the axiom of choice, the real numbers can be a countable union of countable sets, yet still uncountable. In such a case, the theory of Lebesgue measure is a complete failure.
      To my way of thinking, these examples support a call for balance in the usual conversation about the axiom of choice regarding counterintuitive or surprising mathematical facts. Namely, the typical way of having this conversation is to point out the Banach-Tarski result and other counterintuitive consequences of the axiom of choice, heaping doubt on the axiom of choice; but a more satisfactory conversation would also mention that the axiom of choice rules out some downright bizarre phenomena — in many cases, more bizarre than the Banach-Tarski-type results.
      — Joel David Hamkins, Lectures on the Philosophy of Mathematics

    • 3 weeks ago
      Anonymous

      Without the axiom of choice, there can be an equivalence relation on R, such that the number of equivalence classes is strictly greater than the size of R. That is, you can partition R into disjoint sets, such that the number of these sets is greater than the number of real numbers. Bizarre! This situation is a consequence of the axiom of determinacy and is relatively consistent with the principle of dependent choice and the countable axiom of choice.
      Without the axiom of choice, there can be a field with no algebraic closure. Without the axiom of choice, the rational field Q can have different nonisomorphic algebraic closures. Indeed, Q can have an uncountable algebraic closure as well as a countable one. Without the axiom of choice, there can be a vector space with no basis, and there can be a vector space with bases of different cardinalities. Without the axiom of choice, the real numbers can be a countable union of countable sets, yet still uncountable. In such a case, the theory of Lebesgue measure is a complete failure.
      To my way of thinking, these examples support a call for balance in the usual conversation about the axiom of choice regarding counterintuitive or surprising mathematical facts. Namely, the typical way of having this conversation is to point out the Banach-Tarski result and other counterintuitive consequences of the axiom of choice, heaping doubt on the axiom of choice; but a more satisfactory conversation would also mention that the axiom of choice rules out some downright bizarre phenomena — in many cases, more bizarre than the Banach-Tarski-type results.
      — Joel David Hamkins, Lectures on the Philosophy of Mathematics

      That's cool and all, but he's defending AC against people advocating for a simple ZF rather than ZFC, which doesn't include people critiquing the whole thing in general. OP didn't specify though so w/e.

      • 3 weeks ago
        Anonymous

        >OP didn't specify though so w/e.
        Are there any other systems besides ZF that have "Choice -> Banach-Tarski" as a theorem? Maybe one of those that avoid classical logic, but that seems unlikely by Diaconescu's theorem.

  6. 3 weeks ago
    Anonymous

    infinite sets are obviously the issue that led to this whole thing.
    >the axiom of choice is obviously true, the well-ordering principle is obviously false, and who can tell about Zorn's lemma?
    >t. Jerry Bona

    • 3 weeks ago
      Anonymous

      > Infinite sets are the issue

      Sure. Infinite sets and limits cause things to get complicated. Do you have an alternative that doesn't require something absurd like "there exists some N which is the largest integer?"

      • 3 weeks ago
        Anonymous

        >Do you have an alternative that doesn't require something absurd like "there exists some N which is the largest integer?"
        I think claiming there is no largest integer is more absurd than claiming that there is a largest useful integer, which can evolve with time and need, but still remains finite.
        In fact, for every application, we could fix an N that is large enough and meet all our needs with such an N.

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