Re: Godel proved maths inconsistent not incompleteness theorem

On Fri, 14 Mar 2008 23:34:22 -0700 (PDT), Charlie-Boo
<shymathguy@xxxxxxxxx> wrote:

On Mar 14, 10:28 pm, Chris Menzel <cmen...@xxxxxxxxxxxxxxxxxxxx>
wrote:
On Fri, 14 Mar 2008 12:26:58 -0700 (PDT), Charlie-Boo
<shymath...@xxxxxxxxx> said:
[....]

To say that a wff is not a set is throwing the baby out with the
bathwater.

Well, no wff *is* a set (at least, not in the way you mean it).

You have now joined the ranks of Bill Clinton by adding to the list
[fix, avoid, refer] the word "is".

Well, I feared I'd confuse you if I added that qualification.  What
*you* mean when you say "a wff is a set" is that the wff *defines* a set
in the sense indicated above.  Your way of putting it again confuses
syntax and semantics.

It's just a choice of words - defines or is.

"Just a choice of words". Right. People are saying they
don't understand this and don't understand what you mean
by that. You insist on using words to mean something other
than what they mean when other people use them - how
can you expect people to understand what you're saying?
Complaining about people saying they don't understand
your meaning when you're speaking your own private
language that just happens to look a lot like English is
a little, let's say silly.

Mostly I say "defines"
and perhaps occasionally I say "is", so you try to say there's a
problem with "confusing syntax with semantics"?

If you're so picky, how can you say both "No wff *is* a set." and "In
that case wffs are sets."?

He explained that. Quite clearly, I thought.

In fact, if we're speaking strictly and using the standard
definitions, every wff _is_ a set. But the fact that every wff
_is_ in fact a set has nothing to do with what _you_ mean
when you talk about whether or not a wff is a set. The
reason being that when you say a wff is a set that's not literally
what you mean - you mean that the wff _defines_ a set.

If you really want to know the sense that he's talking about,
in which a wff _is_ a set: By definition, a wff is a finite sequence
of symbols which satisfies certain properties. Hence a wff is
a set, because by definition every finite sequence is a set.

Every finite sequence is a set because by definition a finite
sequence is a function with domain {1,2,...n} for some n,
and every function is a set.

Every function is a set because by definition a function is
a set of ordered pairs which satisfies a certain property.

So every wff _is_ a set. And this has nothing to do with
what _you_ mean when you talk about wffs "being" sets,
because for example the wff "x=x" _is_ a set, although
it certainly does not _define_ a set, hence in your private
language it is not a set. The string of symbols ")fx="
is also a set, although it is not a wff.

What *I* meant by adding the qualification above
is that, if one does the metatheory of a language in pure ZF, then the
lexical elements of the language can themselves be identified with
arbitrary pure sets.  So, in that special case, wffs are literally sets.
But that's an irrelevant theoretical nicety that is only a red herring
here.

As for your "list", I'm afraid it simply reflects your inability to get
the point.





Do you mean: "To deny that every wff defines a set is to throw the
baby out with the bathwater"?

No, the problem itself is that not every wff defines a set.

That's a clear assertion. It is
also wrong. To affirm that every wff (with a free variable) defines
a set leads to inconsistency. Inconsistency is the bathwater. The
baby is the useful mathematics that can be done in, or with the aid
of, set theory. ZF does a remarkably good job of tossing out the
bathwater and hanging on to the baby. Again, if you'd commit six
months to actually learning the basics, you could discover this for
yourself.

We can do better by telling which wffs are sets and which are not.

Again, if you mean "by telling which wffs define sets and which do not",
that (among other things) is pretty much exactly what axiomatic set
theory does for you.

Where did Zermelo, von Neumann, Gödel, et al address that?

In ZF set theory and VonNeumann/Gödel/Bernays set theory, of course.

What is the procedure to determine if a given wff defines a set?

Well, I doubt there is a general procedure.

Finally!!!!!!!!!!!!!!!!!!!

I'm not enough of a set
theorist to be able to come up with an example off the top of my head,
but it strikes me there must be, say, definable properties P of ordinals
such that it is not known, and perhaps provably unknowable in ZF,
whether there is an upper bound on the ordinals that have P.  (Anybody?)
If a wff A(x) defines such a P, then there is no procedure for
determining whether A(x) defines a set, just as there is no general
procedure for determining whether a given wff of first-order logic is a
logical truth.  Thanks to Gödel we know there are limitations of this
sort that we just have to live with.

I'm relative certain you'll claim CBL has such a procedure.  No
surprise, though, CBL can do anything if we wish hard enough, right?

Yuh, right.

C-B

David C. Ullrich
.

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