|
|
|
|
|
by cryvate1284
1598 days ago
|
|
You aren't very explicit what you mean, but I disagree I think. The main thing is willing things into existence (like ZFC, the Axiom of Choice) is equally consistent with not having it (ZF + not AC) or not deciding it (ZF) as shown by forcing. Russel's Paradox in particular was worked around by not allowing to "build sets" quantifying over all sets (see axiom schema of specification). If you are talking about avoiding "unprovable statements" (i.e. Godel's incompleteness theorem), you have to strip things way back further: it applies to systems that only have e.g. Peano arithmetic. An actual practical statement that is unprovable in Peano arithmetic is Goodstein's theorem:
- write out a number in its base 2, including its exponents (i.e. recursive)
- replace all the 2s with 3s
- subtract 1
- write out a number in base 3, including its exponents
- replace all the 3s with 4s
- subtract 1
- ...
The question is whether for all starting numbers this sequence eventually ends at 1. The answer is yes, trivially if you use ordinals (and replace all your base numbers with omega: the +1 will basically do nothing and -1 means it must be finite because ordinals well-ordered), but this cannot be used in Peano arithmetic. Fundamentally, like Paris-Harrington Theorem, I think the way to think about why this cannot be proved, is because these sequences get big before they end up at 1. |
|