[FOM] Refuting CH?/2

[Harvey Friedman]

Continuing from Refuting CH? which is reproduced here:

On Sat, Apr 23, 2016 at 6:38 PM, Harvey Friedman <hmflogic at gmail.com> wrote:
> GENERAL PRINCIPLE. Any simple existential property that holds of all
> simple Borel data, in fact holds of all such simple data, regardless
> of whether it is Borel or not.
>
> "THEOREM". The General Principle is consistent and refutes CH.
>
> So here is an example. We use n below for integers.
>
> THEOREM 1. Let f:R into R be Borel. There exists x,y such that x is
> not any f(y+n) and y is not any f(x+n).
>
> PROPOSITION 2. Let f_1,f_2,... be functions from R into R. There
> exists x,y such that x is not any f_i(y) and y is not any f_j(x).
>
> THEOREM 3. Proposition 2 is equivalent to not CH.

We look at the structure (R,<,~), where x ~ y iff x-y is an integer. A
very nice equivalence relation.

(R,<,~) is a very nice tame structure.

We look at statements of the forms

1) (for all Borel f:R into R)((R,<,~,f) satisfies phi)
2) (for all f:R into R)((R,<,~,f) satisfies phi)

where phi is a first order sentence in the language of (r,<,~,f).

CONJECTURE 1. The set of true sentences of the form 1) is decidable.
In fact, there is an intelligible decision procedure which provably
works in ATR_0. Weaker: Pi11-CA_0, Z_2,ZC, ZFC.

CONJECTURE 2. The statement "for every true sentence of form 1), the
corresponding sentence of form 2) is true" is equivalent to not CH.
Weaker: is consistent with ZFC.

THEOREM. The statement "for every true sentence of form 1), the
corresponding sentence of form 2) is true" refutes CH. In fact, this
is the case even if we only allow phi that are EEA with no nesting of
f.

The Conjectures can also be weakened to allow only phi that are in
various fragments such as the one used in the Theorem. A natural
important one is E...EA...A. Note that the fragment used in the
Theorem is finite.

Harvey Friedman