# FOM: CH and RVM: Reply to Insall

Joe Shipman shipman at savera.com
Tue Feb 22 10:10:30 EST 2000

Insall:

>>My point was only to indicate a reason that analysts and applied
mathematicians, who may have no inherent interest in  set theoretic
questions, but only see them as tools for answering their own  questions
in their own domain, CH is of interest *because* of its  relationship to
RVM, which appears to me to be inherently an analysis question.<<

Yes, although to a non-analyst CH may appear more fundamental than RVM,
being a simpler statement.

>>Apparently, you already have chosen to prefer AC, so  that you
interpret the idea of maximization'' in terms of more
(constructible) choice functions''.  But since AC does not imply even
CH, it  is possible to have AC *without* so many 1-1 correspondnces.
Thus, although  I mentioned the RVM problem as a motivation (for
analysts) to study CH, I  do not advocate it.<<

AC seems a VERY clear application of the "MAXIMIZE"  principle, because
it is equivalent to "a product of  nonempty sets is nonempty", and
denying it would  clearly be minimizing the size of the cartesian
product  set.

>> I would like to have, if possible, instead, the axiom c = \aleph_c,
meaning (a) The continuum is well-orderable, and (b) There are c initial
ordinals below c.  I have not yet seen any reason to deny this
possibility, although I have admittedly not read all that has been done
on large cardinals''.  If this axiom is not realizable, I will gladly
abandon it. However, it is my understanding that it is consistent  to
assume that c =  \aleph_{\alpha}, for any cardinal \aleph_{\alpha} not
explicitly ruled out by Königs Theorem, which states that c cannot be
the sum of countably many smaller cardinals''.<<

Yes, but c=aleph_c is not as strong as RVM, because c could simply be
the aleph_oneth (aleph_first?) fixed
point of the aleph function, you don't need to go  beyond ZFC to show
this is consistent.

It is an interesting question how much stronger the assumption of a
measure is than the assumption that
there are weak inaccessibles <= c.   I'm pretty sure that it is
consistent for c to be weakly inaccessible
but for no measure to exist.  Can anyone confirm this?

(Added in proof:  Simpson pointed out privately that "c weakly
inaccessible" requires the consistency strength of one inaccessible,
while RVM requires the higher consistency strength of a measurable, but
this does not quite give what I am asking for, because the models of "c
weakly inaccessible" constructed MAY have a real-valued measure on c.
I want a model where c is weakly inaccessible AND there is no measure.
If it happens that measurables are inconsistent, this will automatically
happen, but you need to do more than simply force c to be weakly
inaccessible in order to guarantee no measure.

One additional question:  It is consistent with ZFC that there is a
definable well-ordering of the continuum (since if V=L there is a
definable global well-ordering).  Is it consistent with ZFC+measurable
that there is a definable real-valued measure?)

-- Joe Shipman