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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) Th
by 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 agoThanks. 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 agoI 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.