[FOM] ACA0, PA and natural language

Dean Buckner Dean.Buckner at btopenworld.com
Sat May 10 04:39:00 EDT 2003


Harvey,

Thanks for your posting.  The technique you use (in your parable about the
two friends) is well-known to philosophers.  Let's re-use it as follows.  A
set theorist, Seth, and a nominalist, Norm, are arguing whether "sets"
exist.  Norm says that just Alice and Bob exist.  Seth claims that some
entity {Alice, Bob} exists "as well", and it's getting pretty heated until
Harvey turns up.  "Don't you see you're not disagreeing at all?  We can
translate anything Seth says into something Norm says, and conversely.  When
Seth says  "{Alice, Bob} exists", we can translate this into "Alice and Bob
exist", which Norm agrees with.  Seth forms plurals with curly braces, Norm
with the word "and".  Stop arguing".

A caveat.  Seth believes in the truth of an axiom, stating that if Alice and
Bob exist, then {Alice, Bob} exists.  Imagine the world changing so that the
axiom becomes false.  For example

     Alice exists and Bob exists & ~ {Alice, Bob} exists

stated in Seth's language.  Stated in Norm's language this is

    Alice exists and Bob exists & ~ Alice and Bob exist.

which is false in any world, because "Alice exists and Bob exists" simply
means "Alice and Bob exist" in Norm's language.  In Norm's language, what
Seth holds as an axiom, and hence as contigently true, is a necessary truth.
So can't quite translate set-theoretical language exactly into nominalist
language.  What are axioms, and therefore contingent truths in the one
language, are necessary truths in the other.  It follows that Seth is
asserting the existence of an object {Alice, Bob}  which Norm genuinely
cannot countenance.  (As also becomes obvious if we consider singleton sets.
Both Seth and Norm have a name for "Alice".  But only Seth has a name for
{Alice}.) There ARE objects that Seth is talking about, or trying to talk
about, that CANNOT be talked about in Norm's language.

Is this important? I suspect it is.  Systems that do not use plural
quantification require certain axioms to make quantification possible at
all.  These axioms "invent" singular objects (sets) which we can substitute
for variables, and which make possible the general, universal statements
that are the essence of mathematics.  These objects may not be necessary.
They allow quantifiers and variables, which are necessary.  But the objects
themselves are only necessary, if there is no other way of quantifying.

My point is that there are other ways of quantifying.  If I am right, there
is a formal system underlying natural language (call this NL) which uses
plural reference and quantification.  This can do without certain objects
implied to exist by ACA0 and PA*.  Thus ACA0 is not necessarily conservative
over NL.  I understand this means ACA0 does not prove any theorems that NL
does not prove (given a suitable translation or "interpretation" of the
theorems of course).  And it's clear that ACA0 (for example) must have
theorems that NL does not have, since it allows us to proves the existence
of certain sets (e.g. the set {0}) that NL does not.  (Forgive me if I have
not understood that point).

In particular, there will be some analogue of Cantor's Theorem in ACA0, that
establishes the existence of an infinite set

 {x in N: x is not in f(x)}

This, as I have pointed out very mnay times on FOM, is an existence proof.
We really do need to prove that an object referred to by the curly bracket
expression exists.  Otherwise we could suppose it does not, and Cantor's
Theorem would be as worthless as proving "Pegasus lies outside the range of
f".  But there need be no such analogue in NL.  There need be no term in NL
that corresponds to "{x in N: x is not in f(x)}".  Which leads to the
following

CONJECTURE: is there a system whose axioms support real analysis, but which
do not support the existence of the objects required for Cantor's Theorem
(i.e. objects such as {x in N: x is not in f(x)}, or any analogue thereof)?

If the conjecture is true, every object, including every real number, that
is generated by NL will lie WITHIN the range of some counting function f.
I.e. every object would be "countable".  This would effectively be a
solution of CH.

In summary, are we absolutely certain that there is not some more limited
set of assumptions that would suffice for "ordinary mathematics", but not
suffice for the theorem that the reals are uncountable?  It's absolutely
certain that using plural quantification we can do without certain axioms.
Are we certain that these very axioms are not also the ones that generate
Cantor's Theorem?  Has this ever been proved or examined?  By whom?




Dean


Dean Buckner
London
ENGLAND

Work 020 7676 1750
Home 020 8788 4273




More information about the FOM mailing list