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>So we can use explicit examples to demonstrate that certain proofs are impossible Nonsense. The independence of an axiom shows in general nothing about what y

>So we can use explicit examples to demonstrate that certain proofs are impossible Nonsense. The independence of an axiom shows in general nothing about what you can and can not prove. A statement can just be false and therefore unprovable. If you want an actual example of an impossible proof you want to look at the continuums hypothesis. With is probably unprovable and probably unable to be disproven. >Banach-Tarski Theorem is a result that proves that a "Measure"[0] cannot have all four obviously desirable characteristics. The important point is that it can't have these properties on all sets. If you restrict yourself to specific sets, these properties can be achieved. E.g. it is false that you can separate a sphere into countable Lebesgue measurable sets which can be rearranged to form a set with larger Lebesgue measure.

ColinWright 2y ago
>> So we can use explicit examples to demonstrate that certain proofs are impossible > Nonsense. The independence of an axiom shows in general nothing about what you can and can not prove. A statement can just be false and therefore unprovable. You clearly know about these things, but I think you have not seen my intended meaning. It's easiest to use the example of Euclidean Geometry to clarify. We have axioms A1, A2, A3, and A4. We want to prove A5 from these. But we can show that it is impossible to do so by creating a system (for example, spherical geometry) wherein A1, A2, A3, and A4 are all true, but A5 is false. The existence of this system then shows that A5 cannot be proven from A1, A2, A3, and A4. Thus the explicit example of spherical geometry shows that it is impossible to prove A5 from A1, A2, A3, and A4. I wonder if your deeper knowledge of the subject is leading you to expect more of my statements than what I'm actually saying. >> Banach-Tarski Theorem is a result that proves that a "Measure"[0] cannot have all four obviously desirable characteristics. > The important point is that it can't have these properties on all sets. Yes, that is one of the four desirable properties. * Measures 1 on the unit object; * Countably additive; * Isometry invariant; * Defined on all sets. We can't have all four of these, even in one dimension. But if we relax "Countable Additivity" to "Finite Additivity" then we can have all four properties in one dimension, and in two dimensions. What the Banach-Tarksi Theorem shows is that even with the relaxation in property two of Countable to Finite, in three dimensions we still can't have a measure with all four desirable properties. That's why we don't relax that property, we relax property four, and no longer require the measure to be defined on all possible (sub)sets.
- lostmsu 2y ago
  > The existence of this system then shows that A5 cannot be proven from A1, A2, A3, and A4. It really doesn't. In general case given a system it is impossible to prove it is not self-contradictory.
  - constantcrying 2y ago
    >It really doesn't. It really does. The existence of a system where A1-4 are true, but A5 is false proves that you can't conclude A5 from A1-4. >In general case given a system it is impossible to prove it is not self-contradictory. Blatantly false, but also irrelevant.
    - lostmsu 2y ago
      > It really does. The existence of a system where A1-4 are true, but A5 is false proves that you can't conclude A5 from A1-4. No it does not. First you'd need to show that your A1-4 == true, A5 == false system is not self-contradictory. Cause if it happens to be you basically proved A5 by contradiction (assuming A1-4). > Blatantly false https://en.wikipedia.org/wiki/G%C3%B6del%27s_incompleteness_theorems https://en.wikipedia.org/wiki/G%C3%B6del%27s_incompleteness_... > but also irrelevant. See above.
      - wizzwizz4 2y ago
        It's impossible to prove that a (sufficiently-strong) system is consistent in that system. It's perfectly possible to prove that many systems are consistent with respect to the consistency of (say) ZF, and ZF's consistency isn't controversial. All proofs are, ultimately, proofs to one's satisfaction, so bringing Gödel's incompleteness into this isn't insightful, imo. It's like bringing up the Problem of Induction in a discussion about science: yeah, you're right, but now we're not talking about the interesting thing any more.
        lostmsu 2y ago
        Was it ever proven for A1-4 in ZF?

ColinWright 2y ago

>> So we can use explicit examples to demonstrate that certain proofs are impossible > Nonsense. The independence of an axiom shows in general nothing about what you can and can not prove. A statement can just be false and therefore unprovable. You clearly know about these things, but I think you have not seen my intended meaning. It's easiest to use the example of Euclidean Geometry to clarify. We have axioms A1, A2, A3, and A4. We want to prove A5 from these. But we can show that it is impossible to do so by creating a system (for example, spherical geometry) wherein A1, A2, A3, and A4 are all true, but A5 is false. The existence of this system then shows that A5 cannot be proven from A1, A2, A3, and A4. Thus the explicit example of spherical geometry shows that it is impossible to prove A5 from A1, A2, A3, and A4. I wonder if your deeper knowledge of the subject is leading you to expect more of my statements than what I'm actually saying. >> Banach-Tarski Theorem is a result that proves that a "Measure"[0] cannot have all four obviously desirable characteristics. > The important point is that it can't have these properties on all sets. Yes, that is one of the four desirable properties. * Measures 1 on the unit object; * Countably additive; * Isometry invariant; * Defined on all sets. We can't have all four of these, even in one dimension. But if we relax "Countable Additivity" to "Finite Additivity" then we can have all four properties in one dimension, and in two dimensions. What the Banach-Tarksi Theorem shows is that even with the relaxation in property two of Countable to Finite, in three dimensions we still can't have a measure with all four desirable properties. That's why we don't relax that property, we relax property four, and no longer require the measure to be defined on all possible (sub)sets.

lostmsu 2y ago
> The existence of this system then shows that A5 cannot be proven from A1, A2, A3, and A4. It really doesn't. In general case given a system it is impossible to prove it is not self-contradictory.
- constantcrying 2y ago
  >It really doesn't. It really does. The existence of a system where A1-4 are true, but A5 is false proves that you can't conclude A5 from A1-4. >In general case given a system it is impossible to prove it is not self-contradictory. Blatantly false, but also irrelevant.
  - lostmsu 2y ago
    > It really does. The existence of a system where A1-4 are true, but A5 is false proves that you can't conclude A5 from A1-4. No it does not. First you'd need to show that your A1-4 == true, A5 == false system is not self-contradictory. Cause if it happens to be you basically proved A5 by contradiction (assuming A1-4). > Blatantly false https://en.wikipedia.org/wiki/G%C3%B6del%27s_incompleteness_theorems https://en.wikipedia.org/wiki/G%C3%B6del%27s_incompleteness_... > but also irrelevant. See above.
    - wizzwizz4 2y ago
      It's impossible to prove that a (sufficiently-strong) system is consistent in that system. It's perfectly possible to prove that many systems are consistent with respect to the consistency of (say) ZF, and ZF's consistency isn't controversial. All proofs are, ultimately, proofs to one's satisfaction, so bringing Gödel's incompleteness into this isn't insightful, imo. It's like bringing up the Problem of Induction in a discussion about science: yeah, you're right, but now we're not talking about the interesting thing any more.
      - lostmsu 2y ago
        Was it ever proven for A1-4 in ZF?