Freiling's axiom is wonderful, but is it the only reasonable
extension of Zermelo-Fraenkel's axioms? Is it powerful enough
to solve many problems known as undecidable in ZF or ZFC?
Two powerful extensions of ZF are well known:
- the axiom of constructibility (known as V=L, i.e.
all sets are constructible),
- the axiom of determinateness (AD).
Most writers reject V=L as a proper foundation for set theory
because of its "non evidence". I think, this is a somewhat
strange position. For me, V=L is some form of the Church's
thesis. Have You ever seen a non-constructible set?
If not, let us assume that V=L. Why not?
Is the axiom of choice "evident"? Let it be evident, if
some people like so. But You can demonstrate that AD
is "evident" also:
For any set A of sequences of natural numbers:
Ex[0]Ax[1]Ex[2]Ax[3]Ex[4]... (sequence x belongs to A)
or
Ax[0]Ex[1]Ax[2]Ex[3]Ax[4]... (sequence x does not belong to A)
Now convert the second member of disjunction as follows:
not(Ex[0]Ax[1]Ex[2]Ax[3]Ex[4]... (sequence x belongs to A).
I.e. AD is "equivalent" to the law of the excluded middle,
and therefore, "evident".
A wonderful situation for a determined syntacticist (like me):
the axiom of choice is "evident", AD is "evident" also, but
these two postulates contradict each other.
Hence, You can extend ZF axioms in many ways obtaining different
wonderful new set theories (call your own extension "evident",
if You like so).
I agree that "it is too narrow a view of mathematics to consider
it as only concerning the consequences of an initially chosen set
of axioms". This is only the first (and the main) component of
mathematics. The second one is the art of inventing interesting,
powerful, wonderful etc. sets of axioms.
Karl Podnieks, Dr.Math.,
determined syntacticist,
podnieks@mii.lu.lv
Institute of Mathematics and Computer Science
University of Latvia
Raina Blvd. 29
Riga, LV-1459
Latvia
P/S. I would like to be in the member list of QED. Please, let
me know, how can I achieve this.