Axiom of choice is debatable. it leads to pardoxes like well ordering theorem which is intuitively false.
Inceasingly people working in fields like tructive analysis or compter science tend to believe it is false
It is mostly used for results for a class of objects. If a particular instance is given the result can be proved to be true without thjs axiom
some results like every field has an algebraic cloosure strictly not necessary.
one can take a field and a specific polynomial and construct its splitting field. So Galois theory can be done.
yes tychonoff theorem will be false and we better live with this fact.
existence ofa complete orthonormal set will not be true. But when one computes fourier coefficients all but countable many are zero.
HAhan banch theorem for separable spaces will it hold?
anyway we can do Mathematics mostly under separability assumtion
why carry aan axiom whose one consequence well ordsring theorem which has to be false and also a consequnce leads to banach-tarski paradox..
The axiom only simplifies reasoning in that we can assume maximal ideals exist or dual space of Banach space is nonempty etc.
with the exception of Tychonoff theorem we really use the axiom as a convenient blanket for a class of objects.
may be we need to add extra assumptions to theorems but better than carry a wrong axiom
i used to believe the axiom in the sense that when product of three sets has more elements than the product of two sets the arbitrary cartesian product of nonempty sets must be nonempty.
But while product of two sets is defined using notion of ordered pair product of more sets is defined using the notion ofa mapping which depends on the notion of product of two sets. herein lies the point which changed my inutiion.
Dear Peter,
Do you think that the axiom of choice and the axiom of countable choice
are the same?
Dear Anil,
< Which theorems do not follow from countable axiom of choice? >
I think it is better to reverse the question:
What theorems that are follow from the axiom of countable choice?
Agree?
Best regards
No a vector space has a Hemel basis is uselesstheorem there is no use of Hemel.basis of an infinite dim space and it has to be uncountable . Equally useless is rhevgheorem that evetyve field has algebraicclosure to so famous theory given a polynomial one needs to construct its splitting field in specific instance C is slgebraically closed field containing re. Complete orthogonal set exists not much needed as all but countable many Fourier covers must vanish. Maximal.ideals exist is a theorem one can not resist and one will be sad tychnoff theorem may be discarded with all its consequence for functional analysis one has dual of C(a.b) and la is dual of l.p. . So is that a.c. is really needed a good use of countable axiom of choice is for a cluster of there exists a sequence converging to it . Why carry an axiom whose consequence well ordering theorem is counterintuitive non constructive and so should be treated as false . YES I WAS AN INTUITIVE LELEIVER OF AC AT SOME TIME BUT AS I have explained in question arbitrary Cartesian product is not defined using tuples but mappings . So my intuition got corrected we do not want beach tarski paradox either and even if axiom of choice is true in the firm stated in my other question arbitrary product of indexed family of no memory sets is no memory one can not form a non measurable set unless we have explicit indexing ofvequivalence classes under partition
I type fast on mobile and the text is distorted by mobile editor . It is neither a problem with English not advance thing but mobile editorvwhich replaces words on its own . I think My knowledgeable person can make out what is meant. I don't care much sboutvtypos and English usage
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Asked 4 days ago
Dr. Anil Anant Pedgaonkar
2.67The Institute of Science, Mumbai
Which theorems do not follow from countable axiom of choice?
Axiom of choice is debatable. it leads to pardoxes like well ordering theorem which is intuitively false.
Inceasingly people working in fields like tructive analysis or compter science tend to believ
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Galois Theory
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4 days ago
Peter Breuer
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AC is true in the constructivist universe L, so one may as well use it. If you are sufficiently suspicious of weird stuff that all you are going to believe in is stuff you can construct with your own hands, then you get AC for free, so well done and go with the
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3 days ago
Igael Azoulay
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Perhaps the invalidity conclusion is just hardy, particularly when dealing with infinities. I lean for reviewing the way the axiom is called in the proofs leading to unconstistencies.
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2 days ago
Issam Kaddoura
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Dear Peter,
Do you think that the axiom of choice and the axiom of countable choice
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2 days ago
Wulf Rehder
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No a vector space has a Hamel basis is useless theorem. There is no use of Hamel.basis of an infinite dim space and it has to be uncountable . Equally useless is the theorem that every field has algebraic closure in Galous theory. Given a polynomial one can t construct its splitting field .In specific instance C is slgebraically closed field containing real field . Complete orthogonal set exists is not much needed as all but countable many Fourier coefficients must vanish. Maximal.ideals exist is a theorem one can not resist and one will be sad . Tychnoff theorem may be discarded with all its consequence. For functional analysis one has dual of C(a.b) and lq is dual of l.p. . A SEOARABKE banach space has nibemoty dual and one dies nogneed Ac for this a.c. iwhat is needed really needed a good use of countable axiom of choice . Why carry an axiom whose consequence well ordering theorem is counterintuitive non constructive and so should be treated as false . YES I WAS AN INTUITIVE beELEIVER OF AC AT SOME TIME BUT AS I have explained in question arbitrary Cartesian product is not defined using tuples but mappings . So my intuition got corrected We do not want banach tarski paradox either and even if axiom of choice is true in the form stated in my other question arbitrary product of indexed family of non empty sets is non emoty. One can not form a non measurable set AS NO EXPLICIT INDEXING OF EQUIVALENCE CLASSES UNDER PARTITION

A vectorcsoace has Hamel.vasis is useless theorem. Anyway it has to be uncountable for finite dim spaces simple proofs without a.c.. One does not needbthr fact that algebraic closure exists. Given A polynomial splitting field exists and only this is needed in famous theory. RACH SEOARABLE Banach space has no memory dual space and explicit dials formed classically for many Banach spaces . SO WHY XARRY AN AXIOM WHICH HAS COUNTERINTUITIVE well ordering theorem and Banach Tarski paradox as consequences . Yes tychnoff them and existencemaximal ideals may cease to be true but we rather cope with these facts than accepting unrealistic a.c. as axiom
. I was once the believer but it seems it is really unnecessary for important maths throw it . In applications all spaces are separable as world is finite . PEOPLE WILL.THROW THIS AND MANY EXVESSES OF 20TH CENTURY INNTRASH LIKE THEY THROWN computational excesses of 29th CENTURY maths
It on my means Hemel.vasis for infinite time exists is not much important and yes it is not I work in analysis a lot we do not need this result if at all Hemel basis exists for an infinite some Banach space over R it is uncountable only this much I said
I recommend the Peter Breuer account.
A concern for me, not related to the question, is the observation about Computer Science. I can't imagine CS dealing with a non-Choice universe at all, so I don't understand that particular claim. Fundamentally, CS deals with denumerable ground sets. Is there some way of thinking that ties unsolvability to Choice? I don't see the necessity of that.
Peter Breuer, insofar as your long remark was addressed to me, I am not a (strict) constructivist. However, with respect to models of computation that have operational interpretations, I think it is sufficient to stick to (well-founded?) denumerable domains (in the FOL= sense of domain of discourse). I'm not making a claim about mathematical subject matter, only about suitable computation models that illustrate general-purpose computation and its limitations, insofar as Church-Turing completeness is reached. That's behind my employment of the structure ‹ob› introduced at https://github.com/orcmid/miser/blob/master/oMiser/obtheory.txt
Tapas now people have understood thdcquestion. It is straightforward for a person who is knowing countable axiom of choice. And very significant one. MANY PEOPLE HAVE ZBSWERED IT
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