[FOM] paper announcements

Timothy Y. Chow tchow at alum.mit.edu
Wed May 20 17:02:19 EDT 2009

Nik Weaver <nweaver at math.wustl.edu> wrote:
> E.g., if you start with Peano arithmetic and randomly
> add ten new formulas as axioms, the chance that what
> you get is sound is only 2^{-10} but the chance that
> what you get is consistent is close to 1.  Or maybe it's
> exactly 1, depending on how we define "random".  There
> should be a way of making this assertion rigorous.

That would be interesting.  Rephrasing what I said before, I would say 
that you are claiming that

1. a random consistent extension of PA is not sound; and
2. we expect (the arithmetical part of) ZFC to behave like a random 
   consistent extension of PA.

Part 1 seems plausible, though I would like to see a rigorous version.

Part 2, however, requires more argument.  ZFC is highly structured, and so 
it is not clear that we have any reason to believe that it behaves like a 
random consistent extension of PA.

We know from many different areas of mathematics that random objects can 
often easily be proved to have certain properties (with high probability), 
but this fact does not allow us to conclude that a specific, structured 
object is *likely to have that property*.  A random countable graph 
contains every finite graph as a subgraph; does that mean that if I hand 
you a specific countable graph G about which I know very little, then it 
is highly likely that G contains every finite graph as a subgraph?  Not 
necessarily.  If G was not generated by a random process but was highly 
structured, then in the absence of some specific argument that G looks 
like a random graph, we must simply say that we *don't know*.  Absence of 
evidence that a specific, structured G is not random does not constitute 
positive evidence that G is random, let alone that G is highly likely to 
have properties that random graphs have.

Similarly, ZFC was not generated by anything like a random process, so in 
the absence of specific evidence that ZFC ought to behave like a random 
consistent extension of PA, we can't say that it is highly likely.  All we 
can say is that we don't know.

Similar comments apply to your argument from proof theory.


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