FOM: Re: CH and 2nd-order validity
Harvey Friedman
friedman at math.ohio-state.edu
Sun Oct 15 15:55:08 EDT 2000
Reply to Jones 5:28PM 10/14/00:
First of all, I want to say that this whole thread concerning CH and second
order validity consists of observations and comments that have been well
known for38 (Cohen) to 62 (Godel) years. I will go over the basic facts
here.
1. There is a semantic system called second order ZFC. According to our
usual way of understanding this semantics, we have the following results:
a. Either the continuum hypothesis (CH) is a semantic consequence of second
order Zermelo, or the negation of CH is a semantic consequence of second
order Zermelo.
b. The existence of an inaccessible cardinal is not a semantic consequence
of second order ZFC.
c. It is widely believed the the nonexistence of an inaccessible cardinal
is not a semantic consequence of second order ZFC.
Results b,c imply the weaker results
b'. The existence of an inaccessible cardinal is not provable in ordinary ZFC.
c'. The nonexistence of an inaccessible caridnal is not provable in
ordinary ZFC.
However, result a in no way shape or form shed any light on the following
questions:
CH is provable in ordinary ZC? CH is provable in ordinary ZFC?
notCH is provable in ordinary ZC? NotCH is provable in ordinary ZFC?
In fact, these questions were answered in the negative about 38 and 62
years ago:
i) CH is not provable or refutable in ordinary ZC, or ordinary ZFC.
2. We now come to deductive forms of second order logic. In every proposed
form of deductive second order logic - i.e., where axioms and rules are
given - we have
i') CH is not provable or refutable in deductive second order ZFC.
The reason i') holds is that ZFC based on all proposed forms of deductive
second order logic becomes a trivial variant of either ordinary ZFC or some
trivial variant of ordinary ZFC.
3. In fact, any reasonably standard mathematical problem is either a
consequence of 2nd order ZFC or its negation is a consequence of 2nd order
ZFC. However, this result - known for perhaps 80-100 years -gives
absolutely no clue as to which of these two possibilities holds. So for one
main line of investigation in f.o.m., semantic 2nd order logic and 2nd
order logic formulations of ZFC are of no apparent value.
>If the language ZF is construed as being about all the models of ZF,
>then CH does not, indeed, have a determinate truth value.
ZF is not about models of ZF. ZF is about sets.
>However, at least two respondents to my question asserted or implied
>that CH does have a definite truth value, and reading between their
>lines I guess that they were admitting only standard interpretations of ZF
>in which power sets are complete.
That is irrelevant. People who think that CH has a definite truth value
think that it has a definite truth value as a statement of set theory. Some
people think - or at least assert - that any sentence of set theory has a
definite truth value.
>1. First it is desirable to make the semantics of first order set theory
>sufficiently precise that CH has a definite truth value under that
>semantics.
The first order semantics of first order set theory are no different than
the first order semantics of predicate calculus generally. But this does
not have anything to do with definite truth values of statements such as CH.
>2. Then there is the problem of deciding what that truth value is.
>3. Finally one could consider candidate axioms to add to
>ZF to enable CH or its negation to be derived (as appropriate).
>(CH or its negation would do!)
Item 3, together with the justification of the axiom candidates, is the
standard method of attack on 2.
>Reflecting a little longer on (1), it seems that settling on a notion of
>standard model for ZF similar in spirit to the idea of standard model
>in second order logic suffices to make the truth value of CH determinate
>(though not to make clear what it is).
My own view is that the notion
the truth value of phi is determinate
is itself in serious need of an explanation. I don't know of a clear enough
explanation of this notion that would be sufficient for me to make
determinations as to which sentences do or do not have determinate truth
values with any real confidence - except perhaps in cases where I have
(some sort of) a proof or refutation of the statement in question.
>This I take to be equivalent to adding the second order validities as
>axioms.
What does it mean to take a set of statements as axioms where there is no
decision procedure for determining membership in that set? Sure, one can
make sense of this as a mathematical construction. But what does it mean in
terms of mathematical practice? Probable answer: nothing.
>Since 2nd-order validity is regarded in some quarters with suspicion
>it might be worth asking whether there are any weaker and less
>controversial semantic notions which suffice to settle CH.
An example of why people think that 2nd order logic is unsuitable for
various kinds of important foundational investigations is just what we are
talking about: the trivial result that
either CH is a consequence of 2nd order ZFC or notCH is a consequence of
2nd order ZFC
yet with no clue as to which.
>For example:
> Is CH independent of true arithmetic?
The answer is no, and this has also been known for 38 and 62 years.
>(and if this were not the case, would it suffice to interpret
>set theory in well-founded model's of ZF for CH to be determinate?)
The answer is no, and this has also been known for 38 and 62 years.
4. The informal question goes like this. Prove that any deductive
formulation of anything like 2nd order ZFC will not decide the continuum
hypothesis (CH). In fact, prove that any reasonable weakening of 2nd order
semantics results in a formulation of ZFC which does not decide CH.
CONJECTURE. Any reasonable weakening of 2nd order semantics results in a
formulation of ZFC which does not decide CH.
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