Re: Godel proved maths inconsistent not incompleteness theorem

On Mar 14, 12:05 pm, Chris Menzel <cmen...@xxxxxxxxxxxxxxxxxxxx>
wrote:
On Fri, 14 Mar 2008 02:37:37 -0700 (PDT), Charlie-Boo
<shymath...@xxxxxxxxx> said:





...
You haven't answered the question either (the source of the problem
with the Russell Paradox.)

The generally acknowledged source is implicit in ZF -- though you would
have to understand elementary ZF to appreciate the point. The source --
in the sense of the principle most central to its derivation -- is the
unrestricted principle of comprehension: That, for any formula "F(x)" in
the language of set theory with "x" free there is a set y such that, for
all x, x in y iff F(x). This principle was replaced by the schema of
separation, which does not permit one to prove the existence of sets ex
nihilo.

A wff can reference relations (sets) e.g. (all x)P(x) refers to
relation P.

What are you talking about? Let, e.g., P be "x=0", i.e., "x is the
empty set". So you are saying that "(all x)x=0" refers to the relation
"x is the empty set", i.e., the set {0}? It doesn't *refer* to anything
at all; it is simply the false statement that everything is identical to
the empty set.

May a wff contain a reference to something instead of the relation and
use a non-relation there? If f(some relation) is a wff then is
f(something not a relation) necessarily a wff as well?

Your question is ill-formed as stated. It is allegedly a question about
set theory. So formulate your question in the language of set theory.
Can you even do that?

(FWIW, on the simplest way of cashing out your question, the answer is
trivially "yes" in ZF. Let s1 be a relation (i.e., a set of n-tuples,
for some n>0). Let s2 be a set that is not a relation. Then, if f is a
wff with a free variable x, then "f(s1)" and "f(s2)" are obviously both
wffs, where they are the result of substituting "s1" and "s2" for "x" in
f, respectively. Somehow I don't think this is what you have in mind.)

This is all pretty simple and if you'd put your weapons down for a
second, you can see the point,

So you answer that a wff can contain a reference to a non-relation?

Once again, my answer is that your question is so ill-formed that one
can only guess at what you mean.  Spelled out the only way that I could
think of, the answer was trivial.

It's a question of definitions, not a problem to claim trivial in your
pointless meager efforts to debase the intellectual capabilities of
those who disagree with you.

Then (all x)P(x) is a wff even if P is not a relation?

Well, there you go again.  You *think* you are asking a question but it
is completely ill-formed.  You are badly confusing syntax and semantics.
What does or does not count as a wff is wholly a matter of the grammar
of the language in question.

But whether it defined a set or not depends on the parts of the wff.

 If you have a question about a specific
expression, that is, a specific string of symbols in the language of set
theory, I can tell you in two seconds whether or not it is a wff.

(x e y) v (y e x)

 In
particular, if P(x) is a wff (containing, presumably, free occurrences
of "x"), then, by the grammar for first order languages, we know
immediately that "(all x)P(x)" is a wff.  If P(x) is not a wff, then we
know immediately that "(all x)P(x)" is also not a wff.  This is just
elementary formal syntax.

So you include non-relations.

Your question also seems to involve the issue of whether or not, for a
given wff P(x) containing free occurrences of "x", there is a set of
things of which P(x) is true.  That, of course, is settled by the axioms
of ZF.  But your question as it stands is just gibberish.  Really, the
best thing you could do for yourself is study basic mathematical logic
and set theory.

Study it? I'm trying to fix it. (Sometimes I think the only solution
is to throw it out and replace it with CBL. But I know that users
don't like to have their systems replaced. First they worry the new
system will be hard to learn and use, and at some point they wonder if
the information in the existing system will be available in the new
one.)

If we assume that, then naturally a wff does not necessarily define a
set since it can refer to non-sets.  If a wff doesn't refer to non-
sets, then the wff defines a set.  There's no paradox or
contradiction.

If you were actually to study set theory, you would see very early on
why your assertion here is just a confused mess.

This isn't Set Theory, it's CBL. Maybe that's your problem.

When we see in a wff (all x) ... must the ... be a reference to a
relation e.g.  P(x) above or can wffs contain a non-relation for the
... ?

More incoherent gibberish. When we see a wff "(all x)...", the "..." is
obviously itself a wff -- this is just a trivial fact about formal
languages. Wffs don't ever "contain" relations or non-relations, they
are syntactic entities that contain other pieces of syntax.  Some of
those pieces of syntax might themselves indicate relations in the
intended semantics of the language. You appear to be badly confusing
syntax generally with semantics; or something.

The fact of the matter is, naive set theory is right,

If by naive set theory you mean the theory consisting of the axiom of
extensionality and every instance of the principle of of comprehension,
that theory is provably inconsistent.  Though it does appear that you
have trouble distinguishing "inconsistent" from "right", so your claim
is understandable.

they were just confused as to how to define wffs after it was
discovered that there were things that are not sets.  You are not
showing any better ability to grapple with the question, even with a
guide.

Mmm hmm. Zermelo, von Neumann, Gödel, et al were confused. And you're
not. You *might* want to rethink that.

I am addressing the quesion of WHICH wffs define sets.  

That is pretty much exactly the question addressed by ZF.  The problem
with naive set theory answered that question too permissively.  ZF found
a reasonable way to answer it more carefully while preserving the
useful, legitimate mathematics that had been done with the naive theory.

Ok, does (x e y) v (ye x) define a set?

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".

"The answer depends on what the meaning of the word "is" is. If it
means is and always has been . . ." - Pres. Clinton to special
prosecutor Ken Star.

"Everthing he said that I said about Rosie O'Donnell is a lie." -
Barbara Walters on Donald Trump on Barbara Walters on Rosie O'Donnell.

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

C-B
.

Relevant Pages

  • Re: Godel proved maths inconsistent not incompleteness theorem
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  • Re: Godel proved maths inconsistent not incompleteness theorem
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  • Re: Godel proved maths inconsistent not incompleteness theorem
    ... they said, ok, a wff is not always a set. ... profound ignorance of the history and content of set theory is more ... in the sense of the principle most central to its derivation -- is the ... are syntactic entities that contain other pieces of syntax. ...
    (sci.logic)
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