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

From: David P. Ferguson (david.ferguson1_at_cox.net)
Date: 11/07/04 

Date: 6 Nov 2004 17:46:23 -0800

D.McAnally@i'm_a_gnu.uq.net.au (David McAnally) wrote in message news:<cl5qvj$9sp$1@bunyip.cc.uq.edu.au>...
> David Ferguson, PAY ATTENTION to the following two paragraphs since you
> have not understood the points that are made in those paragraphs.
>
> As I have commented before, you seem to have the idea that if two sets
> A and B are constructed in different manners, then they are distinct sets,
> EVEN IF THEY HAVE EXACTLY THE SAME ELEMENTS.

First of all E and E* do not have the same elements. The set E* has
twice as msnay elements as does set E

Set E can be associated with the set of points on the number line:
"0, 2, 4, 6 ...

Set E* can be associated with the set of points on the number line:
"0, 1, 2, 3, 4, 5, 6 ...

There is no doubt whatsoever that there are two elements of set E* for
every element of E.

I am impressed by your apparent conviction. You, however, are in the
position one would be in if one were defending the earth centered
solar sytem versus the sun centered solar system. Your crytal spheres
rotating in crystal spheres is outdated and does not describe reality.
The fact is and allways shall be that a set which has two elements for
each element of another set has twice as many elements as the smaller
set.

The best thing I can do here is print out your comments and decide if
they are worth a reply here or if you should read my forthcomming book
"Sets and Numbers"
david.ferguson1@cox.net

> throughout your work, and it is wrong. If A and B have exactly the same
> elements, then they are equal. It does not matter that their
> constructions are different. A set is *uniquely* determined by its
> elements. Other considerations don't enter into it.
>
> Given two sets A and B, you show that A is a subset of B by proving that
> every element of A is also an element of B. You prove that A is equal to
> B by proving that every element of A is also an element of B and that B is
> also an element of A. You prove that A and B are not equal, either by
> naming an element of A which is not an element of B, or by naming an
> element of B which is not an element of A, or by PROVING that there is an
> element of A which is not an element of B, or an element of B which is not
> an element of A (an unsupported claim that there is such an element is NOT
> sufficient).
>
> david.ferguson1@cox.net (David P. Ferguson) writes:
>
> >D.McAnally@i'm_a_gnu.uq.net.au (David McAnally) wrote in message news:<cj1p8j$rg1$1@bunyip.cc.uq.edu.au>...
> >> david.ferguson1@cox.net (David P. Ferguson) writes:
> >>
> >> As I have commented before, you seem to have the idea that if two sets A
> >> and B are constructed in different manners, then they are distinct sets,
> >> EVEN IF THEY HAVE EXACTLY THE SAME ELEMENTS. This is an idea perpetuated
> >> throughout your work, and it is wrong. If A and B have exactly the same
> >> elements, then they are equal. It does not matter that their
> >> constructions are different. A set is *uniquely* determined by its
> >> elements. Other considerations don't enter into it.
> >>
> >> >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<=>2 N
> >> >E*= {0, 2, 4, 6, 8...}: E* 1<=>1 N
> >>
> >> Which is simple enough to do. The elements 0 and 2 of E correspond to
> >> the element 0 of E*. The elements 4 and 6 of E correspond to the element
> >> 2 of E*. The elements 8 and 10 of E correspond to the element 4 of E*.
> >> And so on. For each natural number n, the elements 4n and 4n+2 of E
> >> correspond to the element 2n of E*, and so this is a 2-to-1 correspondence
> >> between E and E*. And if you do not believe that this is a 2-to-1
> >> correspondence, then demonstrate this by giving an answer to one of the
> >> following four questions (since these are the ONLY possible ways in which
> >> what I have written can fail to be a 2-to-1 correspondence):
>
> > You claim that: E_ = 0 2 4 6 8 10 12 14 16 ...
> > E 2<=>1 E* | | | | | | | | |
> > E* = 0 0 2 2 4 4 6 6 8 ...
>
> Yes, I do.
>
> > Thus you claim: E has more elements than E*.
>
> No, I don't.
>
> If A and B are nonempty finite sets, and there is a 2-to-1 correspondence
> between A and B, then A has more elements than B. The proof of this fact
> uses the finiteness of the sets.
>
> If A and B are infinite sets, then the proof is no longer valid, since it
> uses assumptions which are false (i.e. the proof uses the false assumption
> that A and B are finite). This means that it is not permissible to
> conclude that if A and B are infinite sets, and if there is a 2-to-1
> correspondence between A and B, then A has more elements than B (UNLESS
> YOU FIND ANOTHER PROOF THAT DOES NOT USE FINITENESS OF THE SETS).
>
> The fact of the matter is that it is quite possible for two infinite
> sets A and B to be such that there is a 2-to-1 correspondence between A
> and B, and that A and B have the same number of elements. In fact, under
> the Axiom of Choice, if two infinite sets A and B are such that there is a
> 2-to-1 correspondence between A and B, then A and B must have the same
> number of elements.
>
> > Examining E and E* we can see that either E C E* or E* c E. (or both)
> > If both then E = E*
>
> True. Here, I note that you use 'c' just to denote 'subset', not 'proper
> subset'.
>
> > When the Fact Is: E_ = 0 0 2 2 4 4 6 6 8 ...
> > E 1<=>2 E* | | | | | | | | |
> > E* = 0 2 4 6 8 10 12 14 16 ...
>
>
> This last observation is COMPLETELY IRRELEVANT to whether E is a subset of
> E*, E* is a subset of E, or E = E*. What is the point of introducing this
> irrelevancy into the discussion?
>
> >Rather than repeating an elt we can use "o" as a "null image"
>
> There is no such object as a "null image", and the "null image" is not an
> element of any set.
>
> > So that:In fact: E c E*. => Ex elts of E* which are not elts of E
>
> The first statement does not imply the second, unless you take the 'c' to
> mean 'proper subset'. E is a subset of E*, but not a proper subset.
>
> >If we begin with:
> > : E_ = 0 o 2 o 4 o 6 o 8
> > E* = 0 2 4 6 8 10 12 14 16 ...
>
> > I can obtain: By: E1 = 0 2 o o 4 o 6 o 8 o 10 o 12 ...
> > E2 = 0 2 4 o o o 6 o 8 o 10 o 12 ...
> > E3 = 0 2 4 6 o o o o 8 o 10 o 12 ...
> > E4 = 0 2 4 6 8 o o o o o 10 o 12 ...
> >
>
> What sort of correspondences between the EQUAL sets E1, E2, E3, E4, are
> these supposed to be?
>
> > Efinal = 0 2 4 6 8 ... o o o ...
>
> What sort of limiting procedure are you thinking of, here? It is nothing
> that is recognized in set theory (just as the artificial structures that
> you insist on imposing on sets are not recognized in set theory).
>
> > E* = 0 2 4 6 8 ... Xe Xe+2 Xe+3 ...
>
> What are Xe, etc? They are not defined as yet, so it your responsibility
> to define them. Furthermore, it is evident that Xe is not a natural
> number. Xe is certainly not of the form 2n for any natural number n, and
> therefore it is not an element of E*. As a further note, you have
> artificially created these new elements of E* that don't actually exist,
> just in order that the sets E and E* should conform to some utterly
> unjustified prejudice of yours. In order to make E and E* conform to your
> prejudice (a prejudice which you have not tried to justify), you have had
> to resort to introducing elements into E* which do not exist. Just what
> natural number n is such that Xe = 2n?
>
> >Thus E c E* and Card(E*) = 2 x Card(E)
>
> E is not a proper subset of E*. E and E* are equal. You are correct
> that the cardinality of E* is two times the cardinality of E, but the
> cardinality aleph_0 is such that two times aleph_0 is equal to aleph_0.
>
> > To support this I need only observe that:
> >
> > E* = [E'= {x|x = 2e: e elt E}] + [E" = {x|x = e'+2: e' elt E'}]
>
> > E' = 0 4 8 12 16 ...
> > E" = 2 6 10 14 ...
>
> > It should be clear that E* has twice as many elts as does either E'
> >or E"
>
> True. But this does not mean that E* has more elements than E' or E".
> In fact, E*, E' and E" have exactly the same cardinality.
>
> > And thus E* has twice as many elts as does E !
>
> True. It is also true that the cardinalities are equal.
>
> >You have ignored the fact that E* has twice as many elements as does
> >E.
>
> Because the fact was irrelevant to the point that I was making (which was
> that E has twice as many elements as E*). In fact, if you assume the
> Axiom of Choice, then for infinite sets A and B, the following two
> statements are EQUIVALENT:
>
> (1) A has twice as many elements as B;
>
> (2) B has twice as many elements as A.
>
> Furthermore, both statements are equivalent to the statement that A and B
> HAVE EXACTLY THE SAME NUMBER OF ELEMENTS.
>
> >The set you label "E" is actually a proper subset of an ordinal
> >derivative of the set: "H4"
>
> Define your term "ordinal derivative".
>
> > If N = {0, 1, 2 ...|AE}
>
> What is this "AE" after the vertical line supposed to mean? Are you
> completely sure that what you are doing is set theory?????
>
> > H1 = {0/1, 1/1, 2/1 ...|AE}
> > H2 = {0/2, 1/2 2/2, 3/2, 4/2 ...|2AE}
>
> What does "2AE" mean?
>
> > H2 = H (H was referenced in previous posts)
> > (H1, H2, H3, etc represent a refinement of previous notation:
> > "H = {0/2, 1/2, 2/2, 3/2, 4/2, 5/2, 6/2 ...}
>
> >IT IS CRITICAL THAT YOU RECOGNIZE THE UNEXPECTED FACT THAT "H2" (or
> >"H") HAS TWICE AS MANY ELEMENTS AS DOES THE SET N.
>
> THIS IS TRUE. IT IS ALSO TRUE THAT THE CARDINALITY OF H2 AND THE
> CARDINALITY OF N ARE *EQUAL*.
>
> > HE = {0/2, 2/2, 4/2, 6/2, 8/2 ...} = N
> > and since:
> > HO = {1/2, 3/2, 5/2, 7/2, 9/2 ...} and HO 1<=>1 HE
> > Card(H) = 2Card(N)
>
> This is true. It is also true that Card(H) = Card(N).
>
> >As points on the NUMBER LINE and as appropriate rational numbers:
>
> This statement is IRRELEVANT to questions of cardinality of a set.
>
> >There exists!
>
> > H4 = {0/4 1/4 2/4 3/4 4/4 5/4 6/4 7/4 8/4 9/4 10/4 11/4 ...|H4 for n
> >elt N}
>
> What is the phrase "H4 for n elt N" after the vertical line supposed to
> mean, and what is its relevance to the definition of the SET H4?
>
> > oH4 = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 ...}
>
> Define oH4. Name explicitly an example of an element of N which is not an
> element of oH4. Name explicitly an example of an element of oH4 which is
> not an element of N. If you can't name either, then you have no basis for
> claiming that oH4 and N are not equal.
>
> > N = {0, 1, 2, 3 ...}
> > H2 = {0/2, 2/4, 4/4, 6/4, 8/4, 10/4, 12/4 14/4,
> >16/4 ...}
> > H2 = {0/2, 1/2, 2/2, 3/2, 4/2, 5/2, 6/2 ...}
> > E* = {0, 2, 4, 6,
> >8 ...}
> > E = {0, 2,
> >4 ...}
> >Your E:{0, 2, 4, 6, 8, 10, 12 14,
> >16 ...}
> >
> > Your E 2<=>1 E* Your E 4<=>1 E
> > and STILL: E* 2<=>1 E.
>
> ALL of these statements are TRUE. There IS a 2-to-1 correspondence
> between E and E*. There IS a 4-to-1 correspondence between E and E.
> There IS a 2-to-1 correspondence between E* and E. I can give you
> examples of such correspondences. This means that E has four times as
> many elements as itself. This fact about E is WELL-KNOWN to everybody but
> you.
>
> >> (1) Name an element of E that I have missed.
>
> > You have included in E elements not in E.
>
> No, I haven't. EVERY single object that I listed as an element of E was
> an even natural number. Since E is the set of even natural numbers, then
> every single object that I listed as an element of E IS an element of E.
> I note that you were too cowardly to actually NAME an object which is not
> an element of E, but which I claimed was an element of E. I asked you
> to NAME such an object. Instead, you make the claim that I have
> included in E objects which are not elements of E, and for some
> unfathomable reason, you expect us to just accept your word for it. If I
> have claimed that an object is an element of E when it is not an element
> of E, then NAME it, and justify BOTH that it is not an element of E and
> that I claimed that it is an element of E. If you cannot do this, then
> withdraw your false allegation against me.
>
> >Note the difference above
> >between "your E" and "E".
>
> There is no difference. If you claim that there is a difference, then
> PROVE that there is a difference by naming explicitly an element of one
> which is not an element of the other.
>
> >And between "your E" and E*.
>
> There is no difference. If you claim that there is a difference, then
> PROVE that there is a difference by naming explicitly an element of one
> which is not an element of the other.
>
> >"Your E" is a subset of oH4 NOT of "oN" = N.
>
> PROVE it by naming explicitly an element of "My E" which is not an element
> of N. If you cannot name such an element, then withdraw your claim.
>
> >As you say! Two sets are
> >the same set if they have the same elements. oH4 is a superset of oN
> >But we know that: "If set A is a superset of set B then set A and set
> >B cannot be the same set.
>
> Only if A is a PROPER superset of B.
>
> >Your questions (1) -> (4), seem to have lost all relevance!
>
> What garbage. You have not answered any of the questions (1) to (4).
> You have just thrown around a whole lot of false statements that you have
> not even bothered to attempt to justify, then you expect us to accept your
> false statements for some reason which is completely unfathomable. When
> challenged to explicitly name objects with certain properties, you do not
> bother to do so. Instead, you go back to the fantasy world of 'set
> theory' which you have created for yourself, and which bears absolutely no
> resemblance to genuine set theory, and you use this fantasy world of 'set
> theory' to construct fantasy objections which have no validity in the
> genuine world of mathematics.
>
> Questions (1) to (4) were important, because they were the ONLY means that
> you had at your disposal to prove that I did not give a 2-to-1
> correspondence between E and E*. You DID NOT ANSWER any of the questions.
> Instead, you made completely unjustified claims, and we were
> presumably supposed to accept your patently false assertions without
> question.
>
> >> (2) Name an element of E that I have counted more than once.
>
> >YOUR "E" is not E = {e|e even elt of N}
> >YOUR "E" is E** = {e|e even elt of oH4}
>
> So what? Explicitly NAME one element of "My E" which is not an element of
> E, or withdraw any suggestion that there are elements of "My E" which are
> not elements of E.
>
> >> (3) Name an element of E* that I have missed.
> >>
> >> (4) Name an element of E* that I have counted more than once.
> >>
> >> If you can't answer any of these questions, then you can't claim that I
> >> don't have a 2-to-1 correspondence between E and E*.
>
> You have NOT answered ANY of these questions. Therefore you have not
> given any LEGITIMATE objection to my claim that I gave a 2-to-1
> correspondence between E and E*.
>
> >> You have, more than once, made the claim that E and E* are not equal.
> >> This can only happen if one of them has an element which the other does
> >> not. Since you are so insistent that E* has more elements than E, then
> >> name explicitly an element of E* which is not also an element of E. And
> >> be prepared to defend your claim that what you name is an element of E*,
> >> and be prepared to defend your claim that it is not an element of E.
> >>
> >> >---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"
> >>
> >> It seems, David P. Ferguson, that YOUR little knowledge is a disadvantage
> >> to you, and to those who try to discuss with you.
> >>
> >> >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 -...
> >>
> >> Why have you left in all the dashes? Exactly what are they supposed to
> >> represent? They are certainly not element of the sets. So why include
> >> them? After all, a set is COMPLETELY and UNIQUELY determined by its
> >> elements.
> >>
> >> >Neither C1, nor C2, nor C3 nor C4 are order ISOMORPHIC to N.
> >> It seems, David P. Ferguson, that YOUR little knowledge is a disadvantage
> >> to you, and to those who try to discuss with you.
> >>
> >> >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 -...
> >>
> >> Why have you left in all the dashes? Exactly what are they supposed to >represent? They are certainly not elements of the sets. So why include
> >> them?
>
> >Isn't it painfully obvious?!
>
> Only that you are trying to impose a special structure onto sets. Sets do
> not have ANY structure. Sets ONLY HAVE ELEMENTS.
>
> >I am defining a PARTITION!
>
> So what. Defining a partition is irrelevant to questions of cardinality,
> except for the obvious facts that the sum of the cardinalities of the
> equivalence classes is equal to the cardinality of the whole set, and that
> the cardinality of any of the equivalence classes is less than or EQUAL TO
> the cardinality of the whole set. Specifically, C1, C2, C3 and C4 do NOT
> come complete with the spaces that you included. If you think that they
> do come with the spaces, then you do not understand sets.
>
> >IN: Table I*
> > N = 0 1 2 3 4 5 6 7 ...
> > C1* = 0 4 8 12 16 20 24 28 ...
>
> C1* is equal to C1. If you don't believe this, then EXPLICITLY name one
> element of C1* which is not an element of C1.
>
> > C2* = 1 5 9 13 17 21 25 29 ...
>
> C2* is equal to C2. If you don't believe this, then EXPLICITLY name one
> element of C2* which is not an element of C2.
>
> > C3* = 2 6 10 14 18 22 26 30 ...
>
> C3* is equal to C3. If you don't believe this, then EXPLICITLY name one
> element of C3* which is not an element of C3.
>
> > C4* = 3 7 11 15 19 23 27 31 ...
>
> C4* is equal to C4. If you don't believe this, then EXPLICITLY name one
> element of C4* which is not an element of C4.
>
> > C1*, C2*, C3*, C4*, together, do NOT comprise the set N
>
> Yes, they do. Name explicitly one element of N which is not an element of
> C1, C2, C3 or C4. Name explicitly one element of C1, C2, C3 or C4 which
> is not an element of N.
>
> > Card(C1* U C2* U C3* U C4*) = 4 x Card(N)
>
> This is true. But 4 x Card(N) = Card(N), so Card(C1* U C2* U C3* U C4*)
> = Card(N).
>
> > C1* U C2* U C3* U C4* has FOUR times as many elements as does set N.
>
> N has FOUR times as many elements as N. C1* U C2* U C3* U C4* has exactly
> the same number of elements as N.
>
> >Even though there DO exists sets C1* thru C4*; they are all subsets,
> >not of N, but of a superset of N.
>
> No. They are subsets of N. PROVE your claim by EXPLICITLY naming one
> element of C1*, C2*, C3* or C4* which is not an element of N.
>
> >Just as E* is a subset not of N but
> >of a superset of N.
>
> PROVE your claim by EXPLICITLY naming one element of E* which is not an
> element of N.
>
> >IN:
> >> >Table I
> >> >N_ = 0 1 2 3 4 5 6 7 8...
> 54>> >C1 = 0 - - - 4 - - - 8...
> >> >C2 = - 1 - - - 5 - - -...
> >> >C3 = - - 2 - - - 6 - -...
> >> >C4 = - - - 3 - - - 7 -...
>
> > Card(C1 U C2 U C3 U C4) = Card(N)
>
> You just proved that immediately above, when you proved that
> Card(C1* U C2* U C3* U C4*) = Card(N).
>
> > C1, C2, C3, C4, together, DO comprise the set N
>
> And yet, you claimed the negation of this statement when you wrote that
> "C1*, C2*, C3*, C4*, together, do NOT comprise the set N". I can prove
> that you claimed the negation because C1* is equal to C1, C2* is equal to
> C2, C3* is equal to C3, and C4* is equal to C4.
>
> > Thus: Table I represents a VALID PARTITION of N
> > Whereas your Table I* does NOT!
>
> But there is NO DIFFERENCE between corresponding sets. C1* IS equal to
> C1. C2* IS equal to C2. C3* IS equal to C3. C4* IS equal to C4.
>
> >THAT IS WHY ALL OF THE DOTS AND DASHES ARE INCLUDED. They are
> >necessary requirements for a consistent representation of a valid
> >partition of a set.
>
> But the dots and dashes are NOT parts of the sets. A set does not come
> endowed with an order, or with spaces where elements have been taken out,
> or anything like that. That is a fantasy which you have constructed for
> yourself, and which would be better for you to drop, if you don't want to
> keep making silly false claims.
>
> >This is just as in the case where:
> > The sum 3.45 + 23.6 is CORRECTLY represented by
> > _3.45
> > +23.6_
> > = 27.05
> > And the sum 3.45 + 23.6 and is INCORRECTLY represented by:
> > 3.45
> > +23.6
> > = ?????
>
> So what? That is because of the MEANINGS of the symbols. In order to add
> the numbers represented by the decimals above, you have to add them in the
> manner that you described above BECAUSE OF THE DISTRIBUTIVE LAW OF
> MULTIPLICATION OVER ADDITION. The meaning of the 2 in 23.6 is 2 times 10,
> the meaning of the 3 is just 3, and the meaning of the 6 is 6 divided by
> 10. The positions of the digits are important for their meaning in the
> number.
>
> In the case of sets, all that is important about a set is its elements,
> AND IT DOES NOT MATTER HOW YOU ARRANGE THE ELEMENTS. C1* and C1 have
> exactly the same elements, so they are the same set. It is COMPLETELY
> IRRELEVANT that you have no gaps in how you represent C1* and you have
> included gaps in how you represent C1. The gaps are irrelevant, and have
> ABSOLUTELY NO BEARING on the equality or otherwise of C1 and C1*. Any
> "positioning" of the elements of a set is irrelevant to the meaning of the
> element. An element of a set has exactly the same meaning in the set, no
> matter where you put it, or what you do with it. Unlike the digits in a
> decimal expansion of a real number, all the elements of a set are on an
> equal footing, no matter what you do with them.
>
> Similarly, the gaps that you included in how you represent E have
> absolutely no bearing on the equality or otherwise of E and E*.
>
> >> PROVE IT.
> >>
> >> >For two sets to be truly ISOMORPHIC they must have the same
> >> >cardinality.
> >>
> >> There exists a bijection between C1 and N. Therefore they have the same
> >> cardinality. The bijection is given by 0 <-> 0, 4 <-> 1, 8 <-> 2, and so
> >> on, so that for all n in N, 4n <-> n. If you do not accept that this is
> >> a bijection between C1 and N, then answer one of the following questions:
> >>
> >> (1) Name an element of C1 which I have missed.
> >>
> >> (2) Name an element of C1 which I have counted more than once.
> >>
> >> (3) Name an element of N which I have missed.
> >>
> >> (4) Name an element of N which I have counted more than once.
> >>
> >> And if you can't answer any of these questions, then you have no excuse to
> >> say that this is not a bijection.
>
> No answer here, I see.
>
> >> There exists a bijection between C2 and N. Therefore they have the same
> >> cardinality. The bijection is given by 1 <-> 0, 5 <-> 1, 9 <-> 2, and so
> >> on, so that for all n in N, 4n+1 <-> n. If you do not accept that this is
> >> a bijection between C2 and N, then answer one of the following questions:
> >>
> >> (1) Name an element of C2 which I have missed.
> >>
> >> (2) Name an element of C2 which I have counted more than once.
> >>
> >> (3) Name an element of N which I have missed.
> >>
> >> (4) Name an element of N which I have counted more than once.
> >>
> >> And if you can't answer any of these questions, then you have no excuse to
> >> say that this is not a bijection.
>
> No answer here, I see.
>
> >> There exists a bijection between C3 and N. Therefore they have the same
> >> cardinality. The bijection is given by 2 <-> 0, 6 <-> 1, 10 <-> 2, and so
> >> on, so that for all n in N, 4n+2 <-> n. If you do not accept that this is
> >> a bijection between C3 and N, then answer one of the following questions:
> >>
> >> (1) Name an element of C3 which I have missed.
> >>
> >> (2) Name an element of C3 which I have counted more than once.
> >>
> >> (3) Name an element of N which I have missed.
> >>
> >> (4) Name an element of N which I have counted more than once.
> >>
> >> And if you can't answer any of these questions, then you have no excuse to
> >> say that this is not a bijection.
>
> No answer here, I see.
>
> >> There exists a bijection between C4 and N. Therefore they have the same
> >> cardinality. The bijection is given by 3 <-> 0, 7 <-> 1, 11 <-> 2, and so
> >> on, so that for all n in N, 4n <-> n. If you do not accept that this is
> >> a bijection between C4 and N, then answer one of the following questions:
> >>
> >> (1) Name an element of C4 which I have missed.
> >>
> >> (2) Name an element of C4 which I have counted more than once.
> >>
> >> (3) Name an element of N which I have missed.
> >>
> >> (4) Name an element of N which I have counted more than once.
> >>
> >> And if you can't answer any of these questions, then you have no excuse to
> >> say that this is not a bijection.
>
> No answer here, I see.
>
> >> These 4 bijections are also order isomorphisms.
> >>
> >> >The cardinality of E is Xo/2 [C (E) = Xo/2] and
> >> >The cardinality of E* is Xo [C (E*) = Xo]
> >>
> >> Whatever Xo/2 is. You have never defined it.
>
> And you STILL haven't defined it.
>
> >> >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 ...}
> >>
> >> DEFINE Xo/4, Xo/2, 3Xo/4. You have not defined any of these objects, and
> >> yet you use them as if they have been defined. DEFINE FIRST, and then use
> >> them ONLY AFTER YOU HAVE DEFINED THEM.
>
> And you have not bothered defining your terminology here.
>
> >> >Again: What does this have to do with your claim that E 2to1 E* rather
> >> >than E 1to2 E*?
> >>
> >> I have demonstrated a 2-to-1 correspondence between E and E*. There is
> >> also a 1-to-2 correspondence between E and E*. The two statements are not
> >> mutually exclusive. Your wording above suggests that YOU think that the
> >> possibilities are mutually exclusive, and if that is the case, then all it
> >> shows is that you make unfounded assumptions and assertions without
> >> bothering to prove them.
> >>
> >> >Since E is a proper subset of N and E* = {x|x = 2n: n elt N}
> >>
> >> so that E* is a proper subset of N (if n is an element of N, then 2n is an
> >> element of of N, and 1 is an element of N, but 1 is not equal to 2n for
> >> any element n of N).
> >>
> >> >E = {0 2 4 6 8 10 12 16 ...}
> >> > | | | | | | | |
> >> >E*= {0 4 8 12 16 20 24 28 ...}
> >>
> >> That second set is not E*. You have a false statement here, and anything
> >> that follows from it is unreliable. Also, 14 is missing from your
> >> enumeration of E.
> >>
> >> >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*
> >>
> >> Yes, there is a 1-to-2 correspondence between E and E*. But any claim
> >> that you make that the existence of such a correspondence makes it
> >> impossible for there to be a 2-to-1 correspondence between E and E* is
> >> just an extra assertion that you are making without proof. Furthermore,
> >> it is an assertion which directly contradicts fact.
> >>
> >> >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 ...}
> >>
> >> What is x? And why are both x and x+1 elements of E*?
> >>
> >> >The most accurate label for x is Xo/2.
> >>
> >> But what 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]
> >>
> >> So? This is a function mapping from 2 omega + 1 to P(N). So what?
> >>
> >> >Note that E = (0, 2, 4, 6 ...} has as many predecessors in the
> >> >sequence as it has successors in the sequence.
> >>
> >> What is the relevance of this observation?
> >>
> >> >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}
> >>
> >> So? You have simply renamed omega as x, and determined an increasing
> >> function from 2 omega + 1 to P(N). But since it was already known that
> >> 2 omega + 1 is countable, you have not given us any surprises.
> >>
> >> >We know that C (N) = Xo so C (E) = Xo/2
> >>
> >> The cardinality of E is aleph_0. The cardinalities of E and N are equal.
> >> There is a bijection between N and E given by n <-> 2n for all natural
> >> numbers n. If you do not believe that it is a bijection, the answer one
> >> of the following questions.
> >>
> >> (1) Name an element of N which I have missed.
> >>
> >> (2) Name an element of N which I have counted more than once.
> >>
> >> (3) Name an element of E which I have missed.
> >>
> >> (4) Name an element of E which I have counted more than once.
> >>
> >> If you cannot answer any of these questions, then you have no right to
> >> claim that there is no bijection, and you have no right to claim that
> >> the cardinalities of E and N are different to each other.
> >>
> >> >wE ALSO KNOW THAT c (E*) = Xo
>
> >> >so E 1<=>2 E* Q.E.D.
> >>
> >> So what? Everybody already knew that there is a 1-to-2 correspondence
> >> between E and E*. That was not where you were stuffing up. Where you
> >> stuffed up was when you used the completely false assumption (which you
> >> asserted without bothering to prove it) that the correspondence forbids
> >> the existence of a 2-to-1 correspondence, or a 1-to-1 correspondence,
> >> between E and E*.
> >>
> >> >I think my Q.E.D. is better than yours. but that is just my opinion.
> >>
> >> But you have only proven what everybody already knew. There is nothing
> >> spectacular in proving what is already well-known.
> >>
> >> >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*.
> >>
> >> Well, *I* have at least provided such a proof.
>
> David
>
> -----

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