Re: Cantor's proof that #(Evens) = #(Naturals) is inconsistent

From: David P. Ferguson (david.ferguson1_at_cox.net)
Date: 09/23/04 

Date: 23 Sep 2004 11:44:59 -0700

Virgil <ITSnetNOTcom#virgil@COMCAST.com> wrote in message news:<ITSnetNOTcom#virgil-D1E26E.20395406092004@comcast.dca.giganews.com>...
> In article <2cf2d21.0409060304.6f73ba4@posting.google.com>,
> david.ferguson1@cox.net (David P. Ferguson) wrote:
>
> > Virgil <ITSnetNOTcom#virgil@COMCAST.com> wrote in message
> > news:<ITSnetNOTcom#virgil-272CE0.23002205092004@comcast.dca.giganews.com>...
> > > In article <2cf2d21.0409052023.5b1949c1@posting.google.com>,
> > > david.ferguson1@cox.net (David P. Ferguson) wrote:
> > >
> > >
> > > >
> > > > In my initial post and in several posts thereafter I demonstrated
> > > > conclusively that the set E* has two elements for every element of set
> > > > E. this cannot be disputed.
> > >
> > > The difficulty is that it is equally true that set E has two elements
> > > for every element of E*, and this cannot be disputed either.
> >
> > Can you support your claim the there are two elements of set E for
> > every element of E*? I dispute it.
>
> If both sets are countably infinite and n is any natural number other
> than zero, there are trivial proofs that there are n elements of either
> set for each element of the other.
>
> In fact one can prove that there are n members of N corresponding to
> every member of N itself.
>
> Any natural n greater that 1 partitions N into n congruence
> classes,modulo n, proper subsets of N consisting of all those members
> of N whose differencec from some fixed member of N is a multiple of N,
> each of which is order isomorphic to N itself, so there are n naturals,
> the k'th member of each equivalence class corresponding to the k'th
> natural number. Q.E.D.
>

The task was to show that there are two elements of E for every
element of E*
E = {0, 2, 4, 6, 8...}: E 1 &#61683;2 N
E*= {0, 2, 4, 6, 8...}: E* 1&#61683;1 N
---you respond---
Case I
1 2 3 4 5...
| | | | |
3 7 11 15 19...
2 6 10 14 18...
1 5 9 13 17...
0 4 8 12 16...

Your comment is an excellent example of the adage
"Knowledge is an advantage but a little knowledge can be a
disadvantage"

The partition of N into congruence classes is given:
Table I
N_ = 0 1 2 3 4 5 6 7 8...
C1 = 0 - - - 4 - - - 8...
C2 = - 1 - - - 5 - - -...
C3 = - - 2 - - - 6 - -...
C4 = - - - 3 - - - 7 -...

Neither C1, nor C2, nor C3 nor C4 are order ISOMORPHIC to N.
For two sets to be truly ISOMORPHIC they must have the same
cardinality.
The cardinality of E is Xo/2 [C (E) = Xo/2] and
The cardinality of E* is Xo [C (E*) = Xo]

The information in table I can be condensed into:
Where "sX" means "successor of X"
N = {0, 4, 8 ... 1, 5, 2, 6 ... 3 7 ... }
     | | | | | | | | |
N = {0, 1, 2, Xo/4, sXo/4 ... Xo/2, sXo/2 ... 3Xo/4, s3Xo/4 ...}

Again: What does this have to do with your claim that E 2to1 E* rather
than E 1to2 E*?

Since E is a proper subset of N and E* = {x|x = 2n: n elt N}
E = {0 2 4 6 8 10 12 16 ...}
     | | | | | | | |
E*= {0 4 8 12 16 20 24 28 ...}

so that:

E = {0 2 4 6 8 10 ...}
     | | | | | |
E*= {0 2 4 6 8 10 12 14 16 18 20 ...}
     | | | | | | | | | | | ...}
N = {0 1 2 3 4 5 6 7 8 9 10 ...}

Clearly; E 1to2 E*

Expressed in terms of (en, e*n) we have:
E = {0 2 4 6 8 10 ... x - - - - ...}
     | | | | | | ... |
E*= {0 2 4 6 8 10 ... x x+1 x+2 x+3 x+4 ...}

The most accurate label for x is Xo/2. This is because there is the
sequence of relations:] (here read "<" as subset symbol or as "less
than" by context) )

Seq = [{0} < {0, 2) < (0, 2, 4) < (0, 2, 4, 6) ...< (0, 2, 4, 6, 8, 10
...} = E
< {0, 1, 2, 4, 6 ...} < (0, 1, 2, 3, 4, 6 ...} < {0, 1, 2, 3, 4, 5
...} = N]

Note that E = (0, 2, 4, 6 ...} has as many predecessors in the
sequence as it has successors in the sequence.

Now we note the correspondence:
[(0) (0, 2) (0, 2, 4) ... E (0, 1, 2, 4, 6 ...) (0, 1, 2, 3, 4, 6
...} ... N]
  | | | | | |
     |
  1 < 2 < 3 < x < x+1 < x+2 <
 ... 2x}

We know that C (N) = Xo so C (E) = Xo/2
wE ALSO KNOW THAT c (E*) = Xo

so E 1<=>2 E* Q.E.D.

I think my Q.E.D. is better than yours. but that is just my opinion.

There is much more but this will have to do for now>
Also:
I believe that your PROOF that E 2to1 E* should at least relate to E
and E*.

david.ferguson1@cox.net 09 23 04

> >
> > If the correspondence f(e) = e in f:E -> E* insures that every element
> > of E is an element of E* that there are elements of E* which have no
> > correspondent in E guarantees that the subset E of E* is a proper
> > subset of E* since f(e) = e
> > >
> > > To show that E is a proper subset of E* requires proof that (1) every
> > > member of E is also a member of E* and proof that (2) there is at least
> > > one member of E* that is not a member of E.
>
> I see no evidence that David P. Ferguson has provided the necessities of
> such a proof. Ergo, he has not established his claim.

Relevant Pages

  • Re: abundance of irrationals!)
    ... >> less than sqrthas no largest member. ... > The set of all finite naturals is not infinite, ... >> I WILL claim that your incomplete definition of cardinality ... > Sets defined by mapping functions from the naturals to the reals ...
    (sci.math)
  • Re: Orlow cardinality question
    ... > to be finite if the set of naturals up to and including it is a finite ... > set by the Cantor definition, is, by the Cantor definition, an infinite ... > member of the set. ... Even worse, I posit that the first may come after the last, and the last before ...
    (sci.math)
  • Re: Orlow cardinality question
    ... >> can't have a finite set without a largest member. ... I have proven thrice that the infinite set must ... Einstein said the moon is made of cheese. ... >> Is N+1 a member of the set of naturals? ...
    (sci.math)
  • Re: Orlow cardinality question
    ... > member of the set. ... In the set of NATURALS, ... > Too bad they have to ignore the basic axioms of the naturals. ... >> broad statements about my proofs with them omitted get pretty ...
    (sci.math)
  • Re: An infinite debate
    ... Tony Orlow wrote: ... the sequence of naturals, it cannot become infinite at any point in its ... largest member is k and which contains k members. ...
    (sci.math)