[FOM] Refuting CH?/4

Harvey Friedman hmflogic at gmail.com
Mon Apr 25 00:10:45 EDT 2016


Continuing from http://www.cs.nyu.edu/pipermail/fom/2016-April/019770.html

Recall

THEOREM 1. Let f:R into R be Borel. There exist x,y such that x is
not any f(y+n) and y is not any f(x+n).

PROPOSITION 2. Let f:R into R. There exist x,y such that x is not any
f(y+n) and y is not any f(x+n).

THEOREM 3. Proposition 2 is equivalent to not CH.

*****************

There is a more manageable language to do the Borel transfer. I expect
to be successful with this easier language to work with. It is still a
baby step towards the kind of strong Borel transfer that we would like
to have.

The first baby step is to look at all sentences of the form

(for all Borel f:R into R)(there exists x,y in R)(for all n in Z)(phi)

where phi is a conjunction of equations and inequations between
x,y,x+n,y+n.  This includes the situation with Theorem 1, Proposition
2, Theorem 3.

EXPECTED. The following is equivalent to not CH. Every true statement
of the above form is true if we remove "Borel".

The second baby step is to look at all sentences of the form

(for all Borel f:R into R)(there exists x,y in R)(for all n_1,...,n_k in Z)(phi)

where phi is a conjunction of equations and in equations between
n_1,...,n_k,x,y,x+n_1,...,x+n_k,y+n_1,...,y+n_k.

EXPECTED. The following is equivalent to not CH. Every true statement
of the above form is true if we remove "Borel".

The third baby step is to look at all sentences of the form

(for all Borel f:R into R)(there exists x_1,...,x_t in R)(for all
n_1,...,n_k in Z)(phi)

where phi is a conjunction of equations and in equations between
n_1,...,n_k,x_1,...,x_t, and the x_i+n_j.

EXPECTED. The following is equivalent to not CH. Every true statement
of the above form is true if we remove "Borel".

After this, there are many additional steps to take to strengthen the
Borel Transfer. To be continued...

Harvey Friedman


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