Re: Request for Peer Review - Refutation of Cantor Theorem Conclusion


Arturo Magidin wrote:
In article <1147141298.077804.85150@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Scott <ToaTerra@xxxxxxxxx> wrote:
I critiqued your work. In response, you ignored everything I wrote and
simply proceeded to put forth yet another argument.

I had thought we had already discussed this. You make counter-claims, I
agree with them, and proceed to improve my argument with your valuable
feedback.


Why are you asking for "review", if all you do is ignore it?

I don't ignore it. Personally, I don't like just arguing for the sake
of arguing. You prove a point. I accept it and move on.


This is no longer bordering on the insulting: you are, in fact,
insulting me by continuing to waste my time.

I find your feedback very useful.


What you perhaps attempted to say is that since Cantor's argument
is valid whether or not f is surjective, then from the mere fact that
Cantor's argument is valid one cannot deduce that f is surjective or
that f is not surjective. In other words, you are trying to say that
if A->P and not(A)->P, then we cannot, in general, deduce either A or
not(A) from P. That much is correct.

This is an interesting concept and I will have to think about it.
Perhaps there is an easy way to say if A->P and ~A->P then you cannot
deduce either A or ~A.

No. You again misread, misinterpret, and misstate what is quite
simple.

I was just attempting to copy your statement. I assumed you would
recognize it and know it had the qualifier "in general" attached to it.
My mistake.

However, my original goal was that Cantor's
Theorem proves there is a mapping function

Once again, you fail to define your terms.

What is "a mapping function"?

A mapping function is a function that maps one set to another; ie, f:A
-> B.


What does it mean to say that "Cantor's Theorem proves there is a
mapping function"?

That (exists x)(f(x)=D_f)


As far as I can tell, that sentence is, yet again, complete and
unadulterated nonsense. Any meaning I can come up with that makes "a
mapping function" sensible would make the statement false: Cantor's
Theorem does not prove that there ->is<- a function with some
properties; quite the opposite! It proves that there is NO function
with certain properties. So what could you possibly mean?

Perhaps an analogy would be useful. Suppose I asked "Is there a quarter
in the box on the table?" If you look at the table and there is no box,
then the answer is always no. This is the case of ~(exists
x)(f(x)=D_f). If no function exists, then Cantor's proof does not
generate a contradiction and never will.

Now suppose you see the box on the table and you open it to see if
there is a quarter inside. In order to do so, you must accept that fact
that the box exists. What Cantor is saying is that if there is no
quarter inside, then it can be concluded that the box does not exist.
The box, just like f(x), is an indirect means to a contradiction.

Suppose (exists x)(f(x)=D_f). What should the expected behavior of
Cantor's proof be given f(x)=D_f? If (exists x)(f(x)=D_f) then the
expected behavior is a contradiction. Two possible contradictions are
thus possible:

1. f(x)=D_f exists and we get a contradiction from D_f; ie, D_f = { x
in S : x notin D_f }; or,

2. f(x)=D_f does not exist and we get a contradiction from the
non-existence.

Since we get a contradiction both ways, we cannot determine from
Cantor's proof which one is causing the contradiction.



Proposition #1: (forall y)(forall x)(y = { x in S : y subset S and x
notin y } ) -> ( x in y <-> x notin y).

This is nosense yet again. What is S? S is not quantified. This is not
a sentence. This cannot be a proposition.

S is a set. We already agreed on that in a previous post. Yes, S should
have been quantifier. My mistake. Adjusted (with Patricia's feedback):

Proposition #1: (forall S)(forall y)(exists x)(y = { x in S : S <>
emptyset and y subset S and x notin y } ) -> ( x in y <-> x notin y).


Second, the sentence you write to describe "y" is not a valid
one-free-variable formula. There are THREE variables in "y subset S
and x notin y", of which only x should be free. y cannot appear in
that description. What you have is nosense (which is, quite clearly,
your specialty).

I do not believe a set must only have one variable. D_f = { x in N : x
notin f(x) } has two: x and N.


Proof: Let S be any set. Let x by any element of S. The set y = { x in
S : y subset S and x notin y } exist by the Axiom Schema of Separation


(Note: The "let x.." is incorrect and should be stricken.)

There is only one free variable: x. S is defined as a set, just as N is
in D_f. y is fixed as the given set.

No. You cannot invoke Separation here because separation requires a
formula which has only x free. Here y is playing two roles, both as
the NAME of a set to be constructed, and as part of the
construction. YOU CANNOT DO THAT and claim this is a set.

I already posted this in Patricia's post, but let me put it here for
completeness (from the textbook):

"Proposition: There is no set b such that (forall a)(a subset b -> a in
b).
Proof: Suppose that b is a set. So there is a level V such that b
subset V. Let a = { x in b : x notin x }. Then (1) a = { x in V : x in
a and x notin x }, which exists by the axiom schema of separation. If a

in b then a in a <-> a notin a, which is absurd. So a is a
subcollection of be but not a member of it."

The place marked (1) defines a set in terms of itself. Here is a set
with three variables: x, V, and a.(Even the first case is defined in
two variables.) I do the same thing: x is free, S is defined by
"let...", and y is a name/variable just as a. So it looks to me as if
this is allowed. Is there some subtly I am missing?



We've been over this, Scott. You already tried this formulation
before, and I already pointed out to you that you cannot use the same
variable for both the name of the set you are constructing AND as part
of the condition; that this is an invalid attempt to use
Separation. So what makes you think that saying it ->now<- will make
it any less nonsensical?

From a previous post:

III. Axiom Schema of Separation. If f is a property with parameter p,
then for any X and p there exists a set


Y = { u in X : f(u,p) }


that contains all those u in X that have the property f.

I am defining a property which uses parameter p. Parameter p is a
variable for a set. No where do I see anything that says p cannot be Y.
Thus, translating symbols I have:

y = { x in S : f(x,y) }

What is wrong with this?


No, the Axiom of Power Sets does not give you that y, should it be
defined, would be a subset of x. What the Axiom of Power Sets MIGHT
give you would be that y is an element of P(S).

Upon reviewing the Axiom, you are correct. I guess I am stating it as a
requirement.



Proposition #2: For every set S, for every function f:S->P(S), there
exists D_f, a subset of S, such that there exists a x in S, f(x)=D_f.

This proposition is false. Let S be the empty set, let f be the empty
function. The only possibility for D_f is empty, and I challenge to
find an x in the empty set such that f(x)=D_f.

Agreed. Therefore "For every non-empty set S..."


Proof: Let S and f be given as in the proposition. The set D_f = { x in
S : x notin f(x) } exists by the Axiom Schema of Separation and is a
subset of P(S) by the Axiom of Power Sets.

No. D_f is NOT a subset of P(S), it is a subset of S. No, the fact
that it is a subset of S does NOT depend on the Axiom of Power sets,
though the fact that it is an ELEMENT of P(S) ->does<- depend on that.

Agreed. It should have read "...is a subset of S".

Suppose (exists x)(f(x)=D_f). By deduction, if (exists x)(f(x)=D_f)
then (exists x)(D_f = { x in S : x notin D_f }.

No; this does NOT follow from Deduction. And your description of D_f
in the final clause is also incorrect and also does NOT follow from
separation.

It does not follow from separation because it follows from deduction.
For example:

( ( A -> B ) -> C ) => ( B -> C ) => C

is legal. That is all I am doing.

Perhaps this is better notation? (exists y)(f(y)=D_f) and D_f = { x in
S : x notin f(x) } implies D_f = { y in S : y notin D_f }.

(exists x)(D_f = { x in S : x notin D_f } -> ( x in D_f <-> x notin D_f
).

No. Rather,

(exists x)(f(x)=D_f) --> (exists x)(x in D_f <-> x notin D_f).

Restated: D_f = { y in S : y notin D_f } -> ( y in D_f <-> y notin D_f
).


Very different from the continued nonsense you have.

THUS, we deduce that "(exists x)(f(x)=D_f) is false. You know why? BY
CONTRAPOSITIVE. Because a->b is logically equivalent to
not(b)->not(a); and so

Yes, that is one contradiction that can occur.

The other comes from: suppose a, a -> b, where be is ( y in D_f <-> y
notin D_f ), and we got b. Therefore, a.

[...deleted text...]

I think the rest is just frustration. I am already clear on the first
contradiction - you've proved it before. I am trying to point out a
potentially second contradiction.


(Actually, Proposition #2 is more accurate

Actually, Proposition #2 is false.

and therefore we can deduce
that f(x) is surjective for D_f.

No. Even if Proposition #2 were fixed to make it true (a trivial fix I
will for you to try to figure out), it would not imply surjectivity.
Surjectivity requires that FOR ALL y in P(S) there exists x in S such
that f(x)=y. If Proposition #2 were fixed, it would merely allow you
to deduce that THERE EXISTS y in P(S) such that there exists x in S
such that f(x)=y. You ->do<- know the difference between a for all and
an exists, right?

Yes, thats what the next sentence was trying to say. I realized my
wording was not precise after I posted and after you responded.


However, the proof is not sufficient
to show that f(x) is surjective for all sets but only sets Y such that
D_f subset Y.

.

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