It is known that the statement "every set can be totally ordered" is strictly weaker than Axiom of choice. How does one go about proving without using AC?
Proving "every set can be totally ordered" without using Axiom of Choice
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set-theory
axiom-of-choice
forcing
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0You mean "without" in the title. – 2011-08-16
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4One uses the technique of *forcing* to produce a model where every set is linearly orderable and yet some sets are not well-orderable. Are you familiar with forcing in general? Otherwise it would not be possible to explain the argument in any meaningful way. However, it is *not* a theorem of ZF that every set is linearly orderable (this can also be shown by forcing). – 2011-08-16
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0@Stefan: Thanks, edited it. – 2011-08-16
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0To extend Andres' comment, this is not something that can be explained sufficiently without some background in set theory, much like "algebraic groups" cannot be explained without some group theory background. – 2011-08-16
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0@Andres: Thanks for the prompt response. I only have a cursory familiarity with forcing, but if you can direct me to a reference, I would really appreciate it. – 2011-08-16
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7There's a difference between "is strictly weaker than AC" and "is provable in ZF". – 2011-08-16
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1@user3714: You can read T. Jech *The Axiom of Choice* which covers both forcing and AC related forcings, or his newer book *Set Theory, Third Millennium Ed.*. In the meantime I am trying to write an explanation. – 2011-08-16
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0@Arturo: Thanks, I will modify the question. – 2011-08-16
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2@user3714: And, there is also a difference between "is strictly weaker than AC" and "one can prove it without using AC". The Axiom of Countable Choice is strictly weaker than AC, but one cannot 'prove it' in ZF or "without using choice". What one can do is show that there are models of ZF in which the Axiom of Countable Choice holds, but the Axiom of Choice does not, but that if AC holds then so does countable choice. Similar with your proposition: showing it is "weaker than" AC just means that there are models in which "every set can be linearly ordered" holds but AC is not, but AC implies it – 2011-08-16
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0@user3714: To phrase it correctly one can say "X is independent of Y", where in this case X is the ordering principle and Y is the axiom of choice. Of course, one would have to say which theory you are using. Clearly in ZFC they are equivalent. – 2011-08-16
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1Also, you might want to read [this thread on MO](http://mathoverflow.net/questions/37272/are-all-sets-totally-ordered) – 2011-08-16