Re: Bijections (was OM = SIGMA1)
Le 14-nov.-07, à 17:23, Torgny Tholerus a écrit :
Bruno Marchal skrev:
0) Bijections
Definition: A and B have same cardinality (size, number of elements)
when there is a bijection from A to B.
Now, at first sight, we could think that all *infinite* sets have the
same cardinality, indeed the cardinality of the infinite set N. By
N,
I mean of course the set {0, 1, 2, 3, 4, ...}
What do you mean by ...?
Are you asking this as a student who does not understand the math, or
as a philospher who, like an ultrafinist, does not believe in the
potential infinite (accepted by mechanist, finistist, intuitionist,
etc.).
I have already explained that the meaning of ...' in {I, II, III,
, I, II, III, , I, ...} is *the*
mystery.
A beautiful thing, which is premature at this stage of the thread, is
that accepting the usual meaning of ... , then we can mathematically
explained why the meaning of ... has to be a mystery.
By E, I mean the set of even number {0, 2, 4, 6, 8, ...}
Galileo is the first, to my knowledge to realize that N and E have the
same number of elements, in Cantor's sense. By this I mean that
Galileo realized that there is a bijection between N and E. For
example, the function which sends x on 2*x, for each x in N is such a
bijection.
What do you mean by each x here?
I mean for each natural number.
How do you prove that each x in N has a corresponding number 2*x in E?
If m is the biggest number in N,
There is no biggest number in N. By definition of N we accept that if x
is in N, then x+1 is also in N, and is different from x.
then there will be no corresponding
number 2*m in E, because 2*m is not a number.
Of course, but you are not using the usual notion of numbers. If you
believe that the usual notion of numbers is wrong, I am sorry I cannot
help you.
Bruno
Now, instead of taking this at face value like Cantor, Galileo will
instead take this as a warning against the use of the infinite in math
or calculus.
--
Torgny Tholerus
http://iridia.ulb.ac.be/~marchal/
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Re: Bijections (was OM = SIGMA1)
Bruno Marchal skrev:
Le 14-nov.-07, 17:23, Torgny Tholerus a crit :
What do you mean by "..."?
Are you asking this as a student who does not understand the math, or
as a philospher who, like an ultrafinist, does not believe in the
potential infinite (accepted by mechanist, finistist, intuitionist,
etc.).
I am asking as an ultrafinitist.
I have already explained that the meaning of "...'" in {I, II, III,
, I, II, III, , I, ...} is *the*
mystery.
Do you have the big-black-cloud interpretation of "..."? By that I
mean that there is a big black cloud at the end of the visible part of
universe, and the sequence of numbers is disappearing into the cloud,
so that you can only see the numbers before the cloud, but you can not
see what happens at the end of the sequence, because it is hidden by
the cloud.
For
example, the function which sends x on 2*x, for each x in N is such a
bijection.
What do you mean by "each x" here?
I mean "for each natural number".
What do you mean by "each" in the sentence "for each natural number"?
How do you define ALL natural numbers?
How do you prove that each x in N has a corresponding number 2*x in E?
If m is the biggest number in N,
There is no biggest number in N. By definition of N we accept that if x
is in N, then x+1 is also in N, and is different from x.
How do you know that m+1 is also in N? You say that for ALL x then x+1
is included in N, but how do you prove that m is included in "ALL x"?
If you say that m is included in "ALL x", then you are doing an illegal
deduction, and when you do an illegal deduction, then you can prove
anything. (This is the same illegal deduction that is made in the
Russell paradox.)
then there will be no corresponding
number 2*m in E, because 2*m is not a number.
Of course, but you are not using the usual notion of numbers. If you
believe that the usual notion of numbers is wrong, I am sorry I cannot
help you.
I am using the usual notion of numbers. But m+1 is not a number. But
you can define a new concept: "number-2", such that m+1 is included in
that new concept. And you can define a new set N2, that contains all
natural numbers-2. This new set N2 is bigger than the old set N, that
only contains all natural numbers.
--
Torgny Tholerus
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Re: Bijections (was OM = SIGMA1)
Hi,
Le Thursday 15 November 2007 14:45:24 Torgny Tholerus, vous avez écrit :
Bruno Marchal skrev:
Le 14-nov.-07, à 17:23, Torgny Tholerus a écrit :
What do you mean by each x here?
I mean for each natural number.
What do you mean by each in the sentence for each natural number? How
do you define ALL natural numbers?
There is a natural number 0.
Every natural number a has a natural number successor, denoted by S(a).
There is no natural number whose successor is 0.
Distinct natural numbers have distinct successors: if a ≠ b, then S(a) ≠ S(b).
You need at least the successor axiom. N = {0 ,1 ,2 ,3 ,... ,N ,N+1, ..}
All natural numbers are defined by the above.
How do you prove that each x in N has a corresponding number 2*x in E?
If m is the biggest number in N,
By definition there exists no biggest number unless you add an axiom saying
there is one but the newly defined set is not N.
Quentin Anciaux
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