4 ms·

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 th

by 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.