Can the Continuum Problem be Solved? - Menachem Magidor

Menachem Magidor
Hebrew University
December 6, 2011
This is a survey talk about different attempts to deal with the very troubling phenomena of independence in Set Theory.
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[@mylinhnguyenthi4805]

how did you do it can you share with me , thank you

[@Lucas-bf4pw]

What about only one axiom:
Axiom: A proposition is true if and only if it does not imply a contradiction
Formally: A proposition P is True if and only if, there is at least one proposition Q s.t. P and not Q.
The Axiom itself respects this property, because of Godel's theorem: P = Axiom and Q = Every True proposition can be derived from P

[@urigeorgepeterzil130]

You cannot quantify over propositions.

[@Lucas-bf4pw]

@@urigeorgepeterzil130 Logically, every proposition that can be derived from P is already "contained" in P, so we can redefine the statement avoiding the quantifier: P is True if and only if not P implies P. No quantifiers.

[@urigeorgepeterzil130]

This seems circular, but I am no longer in my comfort zone

[@tdbcclubmanager3419]

@@Lucas-bf4pw The statement (for proposition P): [ (not-P implies P) implies P ] is a tautology in classical logic (true for any proposition P) and even more: [ P if and only if (not-P implies P)]. The reason: Either P or not-P; if we assume P, of course P follows; if we assume not-P, P also follows (this is what [not-P implies P] means); so either way we must have P. Thus, [ (not-P implies P) implies P ] . For the other direction: Assume P. You can assume whatever you like (not-P, Q, R or S, etc.), P is still assumed true. So, P. Thus, [P implies (not-P implies P) ]. In intuitionistic logic [ (not-P implies P) implies P ] is rejected (since [P or not-P] does not always hold) and relevance logic rejects [P implies (Q implies P)], since Q is an irrelevant side assumption in the derivation of P.

[@annaclarafenyo8185]

This "axiom" doesn't work at all, as it is not decidable what is an axiom. You can add a statement "reals are aleph-1" and add "reals are aleph-11" and each are separately consistent, but not both.

[@janezskubic781]

Same here. Submitted article based on this idea to Journal of Symbolic Logic 1977. Refused.

[@aligator7181]

What idea, can you elaborate ?

[@annaclarafenyo8185]

That's because anyone after Cohen with any sense can see that the continuum problem is not meaningful.

[@elizabethharper9081]

Why not meaningful? There are consequences of CH in topology or analysis.

[@annaclarafenyo8185]

@@elizabethharper9081 These problems come out different in different models of the real numbers. After Cohen, you stop being a Platonist regarding the real numbers, they become an enormous thing, much larger than any set. This is because they are intrinsically uncountable, while models of set theory are entirely countable, by the Skolem theorem.
This is the huge breakthrough of forcing, it shows you how to modify models of set theory by changing the model of the real numbers, or any other uncountable set of infinite powerset size.

[@elizabethharper9081]

@@annaclarafenyo8185 Meaningfulness of a statement comes from it's consequences when it is combined with an assumed theory, not it's independence from it.