5 ms·

I am having trouble grasping the result. As far as I understand it, any proof of any result has to be with respect to some axiom system. The continuum hypothes

by abnry 9y ago

I am having trouble grasping the result. As far as I understand it, any proof of any result has to be with respect to some axiom system. The continuum hypothesis result is that you cannot prove a infinity exists between the naturals and the real numbers under the usual axiom system A. How that was proved, I don't know, and I also don't know what axiom system was used in that proof. If the axiom system was also A, then it seems to me that it follows we cannot have p<t, and hence p=t, because otherwise by staying in the same axiom system A you violate the unprovability of the continuum hypothesis in A. This supplies a proof in A that p=t. But this seems trivial. So the axiom system used to prove that "the continuum hypothesis is unsolvable in A" must be a larger system A+. I think in A+ we again have p=t. The question this paper answers is whether we can prove "you cannot prove p=t using A" or "you can prove p=t using A". The latter means there is a straight up proof. Any people who really know this stuff can you comment if I got the gist right?

v64 9y ago
Disclaimer: I have a math background, but this isn't my field. The continuum hypothesis was proved to be independent of ZFC. From glancing at the first few paragraphs of the actual paper [2], it's been established for decades that the following inequality holds: aleph_1 <= p <= t, and that aleph_1 = p can be shown to imply aleph_1 = t. Note that the inequalities aren't strict inequalities, and this proof is essentially saying that they were equivalent all along. [1] https://en.wikipedia.org/wiki/Zermelo%E2%80%93Fraenkel_set_theory https://en.wikipedia.org/wiki/Zermelo%E2%80%93Fraenkel_set_t... [2] https://arxiv.org/pdf/1208.5424.pdf https://arxiv.org/pdf/1208.5424.pdf
clintonc 9y ago
The article is misleading on this point. The Continuum Hypothesis says that it is undecidable in the standard axioms whether there is a cardinality between countable and the continuum (i.e., the cardinality of the real). This work shows that p and t are different, which also means that t is of greater cardinality than the reals.
- samoright 9y ago
  What is p and t? You say that this work shows that p and t are different. What two entities are proven to be equal then in this work?
  - clintonc 9y ago
    p and t are implicitly defined cardinalities (i.e., the smallest cardinalities with given properties). See Definition 1.1 of the source article (https://arxiv.org/pdf/1208.5424.pdf https://arxiv.org/pdf/1208.5424.pdf)
- abnry 9y ago
  Oh, thank you. So the article just threw in Continuum Hypothesis because different infinities 'n stuff. But what is confusing to me is that the article suggested p and t are defined as collections of subsets of the natural numbers. The power set of the natural numbers has the same cardinality as the reals, right? Since p and t are both subsets of this power set, they must have cardinalities below the cardinality of the reals.
- cerved 9y ago
  Wait I'm confused. When you say p and t are different, you mean they are cardibalities of different sets but still equal in "size", right? Because I've understood they've proved p to be of equal "size" of t, which I guess means they have a one to one relation? Fuck this is confusing
deleted 9y ago
[deleted]
spi 9y ago
Same disclaimer as v64: my background is also mathematics, but not in this field. As he says, what was proved so far is that: a_0 < a_1 <= p <= t <= 2^(a_0) The first inequality is by definition (a_1 is "the smallest cardinal greater than a_0), the second one is a theorem by Hausdorff, the third one is supposedly easy, the last one is trivial. Also, if the first one is an equality, then the second one is, too. The continuum hypothesis claims that they are all equal; that's been proved by Cohen to be independent of what you call "A" (and is generally called ZFC), by "Forcing" which is not easy, technical, and has won Cohen a Fields medal... And you're right in saying that what is proved in an axiom set is true in any larger axiom set. However, the main point is that "we cannot have p<t, and hence p=t" is false; that is the whole point of indecidability. A statement can be "true", "false" or "not decidable". Many people apparently expected the p=t question to fall into this last group (with some enlarged axiom set "A+" permitting to prove the strict inequality). And no, the statement "The continuum hypothesis is undecidable in ZFC" is proved within ZFC, otherwise it wouldn't have been proved at all... To give a trivial counter-example, as a corollary of Cohen's proof, the axiom choice "ZFC+CH" (ZFC + continuum hypothesis) is as valid an axiom set as ZFC is. Within it, you can certainly prove that "ZFC implies CH" since... CH is simply true.
- abnry 9y ago
  Thanks. I think the idea I had in my head was this: If p=t or p!=t is decidable in ZFC+AoC, then it must be that p=t because otherwise that contradicts that the continuum hypothesis is undecidable. So a better way to sum up this result is: "The question of whether p=t or p!=t is decidable in ZFC+AoC. And oh yeah, they happen to be equal." How does that sound?
  - deleted 9y ago
    [deleted]
  - cerved 9y ago
    I hate to be that guy but a friendly correction: ZF + AoC is what ZFC refers to. I'm not sure I follow your line of thinking. If it's decidable it could either be p=t or p!=t, no? Ie. If the contium hypothesis can be proved in ZFC, it could be either. Afaik, Cohen proved, by using Gödel, that in fact p=t or p!=t is not provable in ZFC. Ie contium hypothesis is not provable in ZFC. As I understand, Malliaras and Shelah have proved p=t ie contium hypothesis is true by using set and model theory. As I understand they are not relying solely on ZFC to prove this since that would disprove Cohen and my understanding is they do not. Further, I understand this result to mean there can only be two infinities which can be said to be different, the cardinal of the countable rational numbers and the cardinal of the uncountable real numbers. But I don't really understand any of this shit so it is likely I've got it wrong.

by abnry 9y ago

v64 9y ago

Disclaimer: I have a math background, but this isn't my field. The continuum hypothesis was proved to be independent of ZFC. From glancing at the first few paragraphs of the actual paper [2], it's been established for decades that the following inequality holds: aleph_1 <= p <= t, and that aleph_1 = p can be shown to imply aleph_1 = t. Note that the inequalities aren't strict inequalities, and this proof is essentially saying that they were equivalent all along. [1] https://en.wikipedia.org/wiki/Zermelo%E2%80%93Fraenkel_set_theory https://en.wikipedia.org/wiki/Zermelo%E2%80%93Fraenkel_set_t... [2] https://arxiv.org/pdf/1208.5424.pdf https://arxiv.org/pdf/1208.5424.pdf

clintonc 9y ago

The article is misleading on this point. The Continuum Hypothesis says that it is undecidable in the standard axioms whether there is a cardinality between countable and the continuum (i.e., the cardinality of the real). This work shows that p and t are different, which also means that t is of greater cardinality than the reals.

samoright 9y ago
What is p and t? You say that this work shows that p and t are different. What two entities are proven to be equal then in this work?
- clintonc 9y ago
  p and t are implicitly defined cardinalities (i.e., the smallest cardinalities with given properties). See Definition 1.1 of the source article (https://arxiv.org/pdf/1208.5424.pdf https://arxiv.org/pdf/1208.5424.pdf)
abnry 9y ago
Oh, thank you. So the article just threw in Continuum Hypothesis because different infinities 'n stuff. But what is confusing to me is that the article suggested p and t are defined as collections of subsets of the natural numbers. The power set of the natural numbers has the same cardinality as the reals, right? Since p and t are both subsets of this power set, they must have cardinalities below the cardinality of the reals.
cerved 9y ago
Wait I'm confused. When you say p and t are different, you mean they are cardibalities of different sets but still equal in "size", right? Because I've understood they've proved p to be of equal "size" of t, which I guess means they have a one to one relation? Fuck this is confusing

deleted 9y ago

[deleted]

spi 9y ago

Same disclaimer as v64: my background is also mathematics, but not in this field. As he says, what was proved so far is that: a_0 < a_1 <= p <= t <= 2^(a_0) The first inequality is by definition (a_1 is "the smallest cardinal greater than a_0), the second one is a theorem by Hausdorff, the third one is supposedly easy, the last one is trivial. Also, if the first one is an equality, then the second one is, too. The continuum hypothesis claims that they are all equal; that's been proved by Cohen to be independent of what you call "A" (and is generally called ZFC), by "Forcing" which is not easy, technical, and has won Cohen a Fields medal... And you're right in saying that what is proved in an axiom set is true in any larger axiom set. However, the main point is that "we cannot have p<t, and hence p=t" is false; that is the whole point of indecidability. A statement can be "true", "false" or "not decidable". Many people apparently expected the p=t question to fall into this last group (with some enlarged axiom set "A+" permitting to prove the strict inequality). And no, the statement "The continuum hypothesis is undecidable in ZFC" is proved within ZFC, otherwise it wouldn't have been proved at all... To give a trivial counter-example, as a corollary of Cohen's proof, the axiom choice "ZFC+CH" (ZFC + continuum hypothesis) is as valid an axiom set as ZFC is. Within it, you can certainly prove that "ZFC implies CH" since... CH is simply true.

abnry 9y ago
Thanks. I think the idea I had in my head was this: If p=t or p!=t is decidable in ZFC+AoC, then it must be that p=t because otherwise that contradicts that the continuum hypothesis is undecidable. So a better way to sum up this result is: "The question of whether p=t or p!=t is decidable in ZFC+AoC. And oh yeah, they happen to be equal." How does that sound?
- deleted 9y ago
  [deleted]
- cerved 9y ago
  I hate to be that guy but a friendly correction: ZF + AoC is what ZFC refers to. I'm not sure I follow your line of thinking. If it's decidable it could either be p=t or p!=t, no? Ie. If the contium hypothesis can be proved in ZFC, it could be either. Afaik, Cohen proved, by using Gödel, that in fact p=t or p!=t is not provable in ZFC. Ie contium hypothesis is not provable in ZFC. As I understand, Malliaras and Shelah have proved p=t ie contium hypothesis is true by using set and model theory. As I understand they are not relying solely on ZFC to prove this since that would disprove Cohen and my understanding is they do not. Further, I understand this result to mean there can only be two infinities which can be said to be different, the cardinal of the countable rational numbers and the cardinal of the uncountable real numbers. But I don't really understand any of this shit so it is likely I've got it wrong.