Infinite products of integers and the axiom of choice
A restricted axiom of choice
Brief history of the negation of the axiom of choice
The story begins with Cantor who was working on the functions theory and noticing there could not exist any bijection between N and R.
A tremendous work made him able to give shape to the set theory, being encouraged by Dedekind and criticised by Kronecker . On about 1900, it happened to him to hit on two
problems.
The first is a paradox which make it impossible for the set of all sets to exist.
The second , called continuum problem, is about the proof
that there does not exist a cardinal number between that of N and of R. Indeed, Cantor was reasoning by taking for obvious what will later be called the Axiom of choice as he was taking for granted a total order relation between cardinal numbers.
This last issue made him exhausted.
Zermelo was trying to make the theory of Cantor formalized...
So on 1904, he defines the Axiom of choice as being the assumption that the Cartesian product of an infinite family of (non empty) sets is always different from the empty set. Zermelo and Russell gave at about the same time another statement for the Axiom which is that there exists a function which associates to each set of the infinite family one element of the set.
The Axiom was called at the beginning the multiplicative Axiom.
Controversies started between mathematicians, particularly, the French mathematicians Hadamard, Borel and Lebesque, while Zermelo was going on with his formalization work.
The polish mathematician Sierpinski undertook the identification of the theorems which need the axiom.
The German mathematician Fraenkel used the axioms of Zermelo to define as early as 1922 a model where the negation of the axiom of choice is an axiom.
Polish mathematicians like Tarski, Mostowski, Lindenbaum studied around the thirties the negation of the axiom of choice. However, tragic deaths (of young set theorists) happened after Banach pointed out unrealistic consequences of the axiom.
The French mathematician and logician Herbrand died; Ramsay, a disciple of Russell, also.
Lindenbaum died during the second world war and some papers of Mostowski were lost at that time.
Meanwhile , the reference to Zorn's lemma was taking the place of the reference to the axiom.
Mostowski was the one who studied most the particular cases of the axiom :
denumerable or non denumerable family and above all, the case where all the sets have the same number of elements n, whose notation is Cn.
He made about it a work remarkably difficult to access.
In the fifties, the Swiss Specker and the French Fraiss?studied also the negation of the axiom of choice. But the coming of forcing on 1963 produced a fashion phenomena
( with some lobbies )which relegated Fraenkel Mostowski,
as they were called henceforth, as of secondary mportance.
However, the use of the axiom for the definition of infinite sums and products was not drawing attention, although it was mentioned in the ?ncipia Mathematica?? Russell and Whitehead before the first world war.
However,this use of the axiom was evoked in a remarkable paper, from Sierpinski, on 1918, which did not get the echo it deserved.
Gauntt, in the seventies, was the only one trying to complete the work of Mostowski on the Cn. May be a contribution from Truss should be also mentioned.
The continuum problem is not an interesting one, because, with the axiom of choice not being true, there is no total ordering between the cardinal numbers.
Mr Andreas Blass proved on 1999 that we can assume only
CC(n) , n from 2 to m, m included, with CC(n) the countable axiom of choice for families of sets of n elements.
Notation might be CC(2 through m).
Adib Ben Jebara adib.jebara@planet.tn
Ben Jebara evoke the possibility that Fermat was a harbinger (forerunner) for the theories using the negation of the axiom of choice through the existence of infinite products of integers.
Fermat was reading Aristotle who wrote about the existence of the actual infinite and Fermat's intuition could have enabled him to distinguish between the infinite products of integers existing and the others.
Indeed, let us assume that only some of the infinite products of integers, made with an integer repeated infinitely, exist and others not and we can
give a proof,but a problematic one, of Fermat's last theorem.
It is problematic because of an extrapolation principle which enable to go from the equation
zn=xn+yn to the equation z.z....z...=x.x....x...+y.y...y....
and that problem was underlined by MR ANDREAS
BLASS, Professor at the University of Michigan (Ann Arbor).
PLEASE READ THE FOLLOWING PROOF WHICH IS VERY SHORT.
THE PROOF IS PROVIDING A CLUE FOR
-THE AXIOM OF CHOICE NOT TRUE
-A SHORT PROOF FOR FERMAT'S THEOREM
-A POSSIBILITY FOR THE PROOF MENTIONED BY FERMAT.
About what Fermat would have written in the margin :
TEXT OF 26 AUGUST 1999 slightly revised later on
ASSUMING zn=xn+yn HAS A SOLUTION FOR n>2AND IS WRITTEN SUCH AS x>y ,
ASSUMING zn=xn+yn for n>2 IMPLIES THE EQUALITY
z.z...z..=x.x...x...+y.y....y...
IN THE MODEL WHERE m.m...m...exists, UP TO x INCLUDED,THE EQUALITY
z.z...z..=x.x...x...+y.y....y...
DOES NOT HOLD FROM ONE SIDE: SOMETHING WHICH DOESN'T
EXIST (the first member of the equation) EQUAL TO SOMETHING WHICH DOES (the second member of the equation).
SO THERE IS A CONTRADICTION WITH WHAT WAS ASSUMED AT FIRST.
AN ANALOGY IS POSSIBLE BETWEEN THE EXTRAPOLATION
USED AT THE BEGINNING OF THE HINT AND THE THEORY OF THE POINT AT THE INFINITE IN DESCRIPTIVE GEOMETRY DEVELOPED
BY DESARGUES, A CONTEMPORARY OF FERMAT.
All this might appear as nonsence for some one
who never heard or noticed that the existence
(well definedness ) of infinite products of integers
depends on the axiom of choice.
About extrapolating finite equations :
The equation z.z...z....=x.x...x...+y.y...y...
is defined by the extrapoling principle as having
as solutions those of the finite
equations starting from power 3.
The axiom which is used, we call the restricted
axiom of choice.
The axiom is countable choice for families of sets
of number of elements n and it is stated :
CC(n) true from 2 up to m, m included.
The question is: why do we start from power 3 (instead of power 3) ?
I REPEAT THE OUTLINE OF THE PROOF FOR THOSE WHO DO NOT UNDERSTAND IT :
ASSUME FERMAT EQUATION HAS A SOLUTION IN ORDER
TO GET A CONTRADICTION.
THEN THE EQUATION WITH INFINITE PRODUCTS HAS
ALSO A SOLUTION.
BUT, CONTRADICTION, IT CANNOT BE BECAUSE OF THE
CASE WHERE AN INFINITE PRODUCT WHICH DOES NOT
EXIST IS EQUAL TO SOMETHING WHICH EXISTS.
No previous knowledge of Fermat theorem is needed.
That a solution of the the equation with
finite products is also a solution of
the equation with infinite products is a
conjecture.
That only CC(2 through m) is really true
is a thesis.
Adib BEN JEBARA adib.jebara@planet.tn
Clarification (lettre a un ami)
letter to a friend (in french)
Une des versions de l axiome du choix est qu un produit cartesien
A1xA2xA3x........xAnx.......
est different de l ensemble vide si les Ai sont non vides.
Prenons le cas particulier ou la famille est denombrable et
ou les Ai ont tous le meme nombre d elements : n.
Il s agit alors d un cas particulier de l axiome qu on peut
noter CC(n) choice for countable family of sets of
n elements.
Supposons alors que la negation de l axiome CC(n) soit vraie.
Le produit cartesien peut etre vide et donc en passant aux
cardinaux de ces ensembles, le cardinal du produit cartesien
n est pas bien defini bien qu il doit etre egal a :
n.n.n.......n.........
Donc n.n...n.....n est pas bien defini non plus.
On a vu que negation de CC(n) implique negation de EP(n)
avec EP pour existence du produit
Ensuite, on introduit CC(2 through m) qui est :
CC(2),CC(3),..............jusqu a CC(m)
Andreas Blass (universite du Michigan), suite a mes questions,
a demontre que on peut supposer que seulement
CC(2 through m) est vrai dans un certain univers.
Enfin, tes questions m interessent car j aimerais pouvoir
ecrire ou te faire ecrire a Blass qu il y a des indices
que des gens peuvent cmprendre cette recherche
sans avoir beaucoup de connaissances specialisees.
negation de CC(n) au dela de CC(m) implique
negation de EP(n) au dela de EP(m)
Maintenant
l equation avec des produits infinis
z.z.......z........=x.x......x....+y.y......y......
Peux tu voir qu elle n a pas de solution ?
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