Re: Cantor's proof that #(Evens) = #(Naturals) is inconsistent
From: David McAnally (D.McAnally_at_i'm_a_gnu.uq.net.au)
Date: 10/23/04
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Date: 23 Oct 2004 09:54:34 GMT
Throughout, I will use "TE" to denote "there exists" and "FA" to denote
"for all".
stush@rocketmail.com (Stush) writes:
>I think what you are missing is that the symbol "4" in N is not the
>same as the symbol "4" in E.
False. 4 is exactly the same as an element of N, and as an element of E.
The statements are that 4 is an element of N, and that the exact same
object 4 is also an element of E.
>For example:
>A = { 1,2,3,4 }
>B = { 2,4,6,8 }
>Clearly A and B have the same card even though 6 is not A.
What does, whether or not 6 is an element of A, have to do with questions
of cardinality? It seems strange that you think that there is any
connection between
(A) the cardinality of a set,
and
(B) whether or not a specified object is an element of the set.
The ONLY connection that exists between (A) and (B) is that when the
cardinality of a set is 0, then it is guaranteed that no object is an
element of the set. Specifically, for any object x, and any nonzero
cardinal m, there exists a set A of cardinality m such that x is an
element of A. For any object x, and any nonempty set A, there exists a
set B bijective with A such that x is an element of B.
This makes the sentence of your above, "Clearly A and B have the same card
even though 6 is not A." bizarre at best, and a clear indication that you
have missed the important facts about the cardinality of a set.
>In the case
>of A and B I can define one of several bijections e.g.
>f: A->B
>f(1) = 4
>f(2) = 2
>f(3) = 8
>f(4) = 6
>Here 3 in A is the same as 8 in B.
Garbage. The element 3 of A is a completely different object to the
element 8 of B. The element 3 of A CORRESPONDS to the element 8 of B by
the bijection which you have defined.
You should learn to be more precise with the language that you use.
>If we write i in A as x_i and f(i)
>in B as x_i,
You should not use the same symbol x_i to denote two different things at
the same time. Because you have written the element i of A as x_i, then
you have defined x_1 = 1, x_2 = 2, x_3 = 3, x_4 = 4. You cannot maintain
this assignment of values and simultaneously define x_i = f(i) for all
elements i of A. That would make x_1 = 4, x_2 = 2, x_3 = 8, x_4 = 6,
which contradicts the already established assignments x_1 = 1, x_3 = 3,
x_4 = 4. The fact that you want to simultaneously use BOTH assignments
is demonstrated by your next statement (immediately below). Instead, you
should have written the element f(i) of B as y_i.
>we have
>A = { x_1,x_2,x_3,x_4 }
>B = { x_1,x_2,x_3,x_4 }.
Since you derived this using two contradictory assignments of x_1, x_3
and x_4, then you have not demonstrated that A = B. You have merely
demonstrated that either you do not know the rules for assignment of
values to variables, or you do not acknowledge the rules.
>This can be easily extended to infinite sets to dervive E=N, N=NxN,
>N=Q, etc.
Garbage. Your so-called "method of proof" does nothing more than
demonstrate bijections between E and N, between N and NxN, between
N and Q, etc. E does not equal N. N does not equal NxN. N does not
equal Q.
>HOWEVER the point of this post is not to debunk your work.
Which might be a good thing, considering the number of false statements
that you have made in your posting.
>I often
>wondered if you removed "renaming" morphisms from the study of
>infintely sets,
They are not "renaming" morphisms. They are FUNCTIONS mapping from one
set to another. Where did this idea of "renaming" morphism come from?
You seem to have missed the entire point of what is happening. A function
f from A to B is a subset of the Cartesian product AxB (so the elements of
f are ordered pairs) such that
(1) FA x in A TE y in B ((x,y) in f),
(2) FA x in A FA y in B FA z in B ([(x,y) in f and (x,z) in f] => y = z).
These two statements can be summarised as the single statement that
FA x in A TE y in B FA z in B ((x,z) in f <=> z = y).
A bijection f between A and B is a subset of AxB such that
(1) FA x in A TE y in B ((x,y) in f),
(2) FA x in A FA y in B FA z in B ([(x,y) in f and (x,z) in f] => y = z),
(3) FA y in B TE x in A ((x,y) in f),
(4) FA x in A FA y in A FA z in B ([(x,z) in f and (y,z) in f] => x = y).
These four statements can be summarised as the two statements that
FA x in A TE y in B FA z in B ((x,z) in f <=> z = y),
FA z in B TE x in A FA y in A ((y,z) in f <=> y = x).
In other words, a bijection is not a "renaming" of the elements of A to
become elements of B, but a subset of AxB satisfying the above conditions,
and an element x of A corresponds to an element y of B under the bijection
f iff (x,y) is an element of f iff
the unique element z of B such that (x,z) is an element of f
is z = y, and
iff
the unique element z of A such that (z,y) is an element of f
is z = x.
>could you derive any useful results.
Considering that the lead-up to this question is completely unsupported in
fact, the question is meaningless, and can be safely ignored.
>Then you could
>say something like "3" is not in E and "3" is in N,
You can say anyway that 3 is an element of N and 3 is not an element of E.
The existence of a bijection between N and E does not force 3 to an
element of E. In fact, for any two nonempty sets A and B, the existence
of a bijection between A and B has NO BEARING ON THE ELEMENTS OF EITHER.
>so |E-{3}| = E
No. E-{3} = E, and |E-{3}| = |E|, but |E-{3}| does NOT equal E.
>but
>|N-{3}| <> N.
This is not true (where '<>' denotes "not equal to" - recall that the
meaning of "not equal to" for '<>' was inspired by an ordered set in which
the order was linear - the ordering of cardinals need not be linear,
except if you invoke the Axiom of Choice). The reason why it is not true
is that the cardinality of N-{3} (i.e. aleph_0) is typically represented
by the set N. It is true that N-{3} is not equal to N. It is also true
that |N-{3}| = |N|.
>In this context |N| = 2 * |E| is true.
|N| = 2 * |E| is ALWAYS true, as I have stated SEVERAL times. It is also
true that |N| = |E|, which I have also stated several times.
>My questions are if you expand on this new definition of card
What new definition of cardinality? I haven't seen a new definition of
cardinality.
Rest snipped until you can express yourself clearly enough to make
yourself understood.
David
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