From: [email protected] (Cattabriga)
Subject: Re: R: R: R: Why Cantor Was Wrong
Date: 09 Aug 1999 00:00:00 GMT
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On 06 Aug 1999 19:00:33 GMT
Dennis Paul Himes  wrote:

>"Cattabriga"  wrote:
>> (Notation)
>> x neq y denotes y is not equal to y.
>> x eq y denotes y is equal to y.
>> x subseteq y denotes x is a subset of y.
>> (x subset y denotes x is a proper subset of y.)
>> ran g denotes the range of g
>> 
>> The so-called Cantor's 'proof' (Elements of Set Theory, H. B.
>> Enderton, Academic Press, Inc., pag. 132) 
>
>  There are a number of things wrong with the following, 
>  mostly due to inconsistent notation.


We could discuss about notation till the end of time, but let us
make everything simple.

Cantor's 'theorem' constructs a subset b of a that is not in ran g.
But every time that b is constructed a subset c of a that
is in ran g is constructed too.

 VERSION II -------------------------------------------------------
 
 
 Let g:a -> P(a). We will construct a subset b of a
 that is not in ran g. Specifically let
 
 i)              b = { x in a | x notin g(x)}.
 
 Then b subseteq a, but for each x in a
 
 ii)               x in b <-> x notin g(x).
 
 Hence b neq g(x), but
 
 iii) c = a - b,
 
 i.e.         
 
 iv)            c = { x in a | x in g(x)}.
 
 Then c subseteq a, and for each x in a
 
 v)                 x in c <-> x in g(x).
 
 Hence c eq g(x).
 
 Whenever b is constructed, c is constructed too.
 ------------------------------------------------------------------ 


iii) is the relative complement of b in a. Its existence is ensured
by the axiom of subsets. In fact we are always allowed to assert
AxAwEyAz(x in y <-> x in w & x notin z) whenever the set w is given,
and z is a subset of w. This set y is the relative complement of z in
w.


-



> To put it another way "c eq g(x)" doesn't make sense, 
> since "x" is not defined in that statement.

Following your reasoning "b neq g(x)"  doesn't make sense too,
since "x" is not defined in that statement.  







                
Paola Cattabriga
_________________________________________________________________

http://www.serdata.it/cattabriga/
_________________________________________________________________ 
"The process of negation is a necessary instrument of theoretical 
research; only its unconditional application allows the
completeness and the realization of logic." 
                                              David Hilbert 1931 
_________________________________________________________________
                
  


From: "Cattabriga" 
Subject: R: R: R: R: Why Cantor Was Wrong
Date: 13 Aug 1999 00:00:00 GMT
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In Message-ID: <[email protected]>
[email protected] (theodore hwa) wrote

: I mean, from
:              Ax(x in b <-> x in n & )    (*)
: we draw that
:               b = n   ?

:Yes.  Suppose b were non empty.  Let m be an element of b.  Then by (*)
:m is in n, a contradiction.  (Since the left side of the biconditional
:(*) holds, so does the right side.)  Therefore, b is empty.

This is puzzling.

              left side      right side    
              
           Ax(x in b <-> x in n & )    (*)


Since ~Ex(x in n),  Ax(x in n) is always false, then by
the true table for & the right side of the biconditional
is always false. Hence the left side is always false.


In (Foundation of Set Theory,  Fraenkel, Bar-Hillel, Levy,
North-Holland)

+)  EyAx(x in y <-> x in a & ~(x = x)) 

 then for any a we obtain a subset of a which contains no member.

Any empty set is always defined by a contradiction.



:Secondly, your claim then a - b is in ran g is false.  Let a be the
empty
:set,

then a has been defined by a contradiction, and following classical
rules for ZF  a is a subset of something already defined.

for example 

+)  (x in a <-> x in l & ~(x = x)) 


 so that when we get  (x in c <-> ~(x in a & x notin g(x)))
 
 (for example from a first order theorem like
  Ax[(x in b <-> x in a & x notin g(x)) <-> 
                     (x notin b <-> ~(x in a & x notin g(x)))] 
 and the definition of b as an instance of the axiom of subsets)
                     
 and then  Ax(x in c <-> x notin a or x in g(x))),                 
 
 we could conclude that there can be cases such that both 
 
 (x notin a) and (x in g(x)) hold, and being   a   a subset of l
 
 the elements that are not in a (when a is the empty set) are in l,
 
 so that we obtain for c = a - b that the ran g is in l.
  
 
 
As regard to proof for contradiction of the so-called
Cantor's theorem please consider my VERSION I in: 
 
 
Subject:      Why Cantor Was Wrong
Author:       Cattabriga 
Date:         27 Jul 99 08:35:02 -0400 (EDT)

 
 
 
 
 
 
 Paola Cattabriga



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http://www.serdata.it/cattabriga/
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