[FOM] Refuting CH?/7
Harvey Friedman
hmflogic at gmail.com
Sun May 1 01:11:26 EDT 2016
I now want to focus on some entirely precise mathematical issues
surrounding this apparent asymmetry between CH and not CH discussed in
Refuting CH/1-6..
1. There are very simple statements of the form (for all f:R into
R)(there exists x,y in R)(forall n in N)(phi) where phi is even in a
very small fragment of propositional combinations of linear
inequalities in x,y,n, which are
i. Provable in Z for all Borel f.
ii. Consistent with ZFC.
iii. Implies not CH over ZC.
iv. Moreover, is equivalent to not CH over ZC.
Note that iv implies ii,iii. see *) below.
2. There appears to be no comparatively simple statements of the form
(for all f:R into R)(phi) which are consistent with ZFC and imply CH.
So let's turn this into some precise results and conjectures.
It is well known that there is the following asymmetry between CH and
not CH. Not CH is equivalent to a statement of the form indicated, but
Ch is not equivalent even to a statement of the form (for all f:R into
R)(phi), where phi is an arbitrary first order formula over
(R,f,Z,+,dot,<).
However, we don't consider this kind of asymmetry nearly as reasonable
an argument in favor of accepting not CH as we would have if we had
this asymmetry:
1. There are very simple statements of the form (for all f:R into
R)(phi), with phi in very restrictive languages, which are consistent
with ZFC and imply not CH
2. No statements at all of the form (for all f:R into R)(phi), phi an
arbitrary first order formula over (R,f,Z,+,dot,<), are consistent
with ZFC and imply CH.
HOWEVER, 2 is well known to be FALSE.
We have already seen a tremendously clear example of 1 which we repeat there.
*) FOR ALL f:R into R THERE EXIST x,y in R SUCH THAT x IS NOT ANY
f(y+n) AND y IS NOT ANY f(x+n). Here n can be taken to range over
positive integers, nonnegative integers, all integers, or all
rationals. And we actually get provable equivalence with not CH over
ZC.
QUESTION: How simple, in various senses, can we find phi so that 2 is
invalidated: I..e. so that
3) (forall f:R into R)(phi) is consistent with ZFC and implies CH?
It seemst that no example of 3) is going to get anywhere near
whatsoever to *) in terms of the simplicity of phi, no matter how you
reasonably measure simplicity here.
I have the distinct impression that exploring just how simple phi can
be, from various viewpoints, for 3), is going to be a greatly
challenging and rich investigation.
So let's start the exploration.
HOW SIMPLE CAN WE MAKE PHI FOR 3)?
Here we make a crude start. We are going to express "every real is
constructible" in a relevant form. This is consistent with ZFC and
implies CH.
The idea is to use Baire space N^N and then convert to the real line.
We can express every real is constructible in the form
(for all x in N^N)(there exist y in N^n)(for all n in N)(phi(n,x,y))
where phi is a recursive predicate of n and the finite sequence
numbers for x,y up through n.
The idea is to convert this to the reals, obtaining
(for all x in R)(there exist y in R)(for all n in N)(phi(n,x,y))
where phi is quite complicated by the standards of our *). We can
obviously put this in the form
(for all f:R into R)(there exist y in R)(for all n in N)(phi(n,f(0),y))
which is very ugly. But we would like to take advantage of the front
third order quantifier in order to greatly simplify the phi. However,
it is still not clear how to avoid a pretty nasty mess.
Harvey Friedman
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