Re: Conservativity and CH

On Dec 9, 12:50 am, "Robert E. Beaudoin" <rbeaud...@xxxxxxxxxxx>
wrote:
kleptomaniac6...@xxxxxxxxxxx wrote:
On Dec 8, 5:52 am, "Robert E. Beaudoin" <rbeaud...@xxxxxxxxxxx> wrote:
kleptomaniac6...@xxxxxxxxxxx wrote:
On Dec 6, 6:30 am, "Robert E. Beaudoin" <rbeaud...@xxxxxxxxxxx> wrote:
kleptomaniac6...@xxxxxxxxxxx wrote:
I actually have one more question I was wondering about: Is it the
case that if for some axiom T, ZF+T is conservative over ZF for
arithmetical statements, then ZF+notT is also conservative over ZF for
arithmetical statements?
Let's take it for granted that ZF is consistent. (Else ZF entails every
arithmetical statement, so any extension of ZF would be conservative for
arithmetical statements.) Let A be the axiom "there is no measurable
cardinal". Then ZF + A is satisfied by the constructible universe, and
so is conservative over ZF for arithmetical statements. But ZF + ~A
entails Con(ZF), which is arithmetical and not derivable from ZF alone.
Robert E. Beaudoin
Yeah I forgot about the relationship between metamathematical
statements and arithmetical statements!
But how about the following wild conjecture: For any non-arithmetical
statement S which is independent of some system R where the
arithmetical part of R extends minimal arithmetic or Q, for any system
T which extends R and proves S, there is a system T' which extends R
and disproves S and has the same arithmetical part as T.
All the systems at hand are consistent.
If you are willing to accept consistency of ZF plus existence of a
measurable cardinal then the example I already gave refutes your
conjecture: Again let A be the axiom "there is no measurable cardinal"
(a non-arithmetical statement). Assuming consistency of measurable
cardinals this axiom is independent of ZF. Let T be the theory with
axioms ZF + A; since the arithmetical consequences of T are the same as
those of ZF they do not include Con(ZF). But as ZF |- ~A --> Con(ZF)
any theory extending ZF and having ~A as a consequence also has Con(ZF)
as a consequence.

Robert E. Beaudoin

The conjecture was, there would be a theory T' extending ZF that
disproves A and has the same arithmetical consequences as T.

Well yes, and I was saying that any T' extending ZF and disproving A
(i.e. proving existence of measurable cardinals) has Con(ZF) (which is
arithmetical) as a consequence, which T does not.

Robert E. Beaudoin

Ah, of course. I came up with an altered form of the false conjecture,
but I couldn't express it without appealing to the notion of "true
arithmetic".

For any non-arithmetical statement S which is independent of some
axiomatization of set theory R, the arithmetical part of which is a
fragment of true arithmetic which extends minimal arithmetic, there
is, for any theory T which extends R and proves S and has arithmetical
part a fragment of true arithmetic, a consistent theory T' which
extends R and disproves S and has arithmetical part a fragment of true
arithmetic which extends the arithmetical part of T.
.

Relevant Pages

  • Re: Conservativity and CH
    ... Let A be the axiom "there is no measurable ... T which extends R and proves S, there is a system T' which extends R ... (i.e. proving existence of measurable cardinals) ...
    (sci.logic)
  • Re: Conservativity and CH
    ... Let A be the axiom "there is no measurable ... T which extends R and proves S, there is a system T' which extends R ... (i.e. proving existence of measurable cardinals) ...
    (sci.logic)
  • Re: Conservativity and CH
    ... Let's take it for granted that ZF is consistent. ... Let A be the axiom "there is no measurable ... T which extends R and proves S, there is a system T' which extends R ...
    (sci.logic)
  • Re: Conservativity and CH
    ... Let A be the axiom "there is no measurable ... T which extends R and proves S, there is a system T' which extends R ... disproves A and has the same arithmetical consequences as T. ...
    (sci.logic)