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Suppose I am given the following axioms for a vector space $V$ over the field $F$:

For addition of vectors

  • Closure
  • Associativity
  • Existence of an additive identity
  • Existence of additive inverse
  • Commutativity

For multiplication by scalars

  • Closure
  • Associativity
  • The two distributive laws

From this I would like to prove that the statement $\forall v \: 1v = v$, where $v \in V$ and $1$ is the multiplicative identity of $F$, cannot be deduced from the above axioms.

One possible method would be to exhibit a model, that is, a set of objects, a field, with operations of addition and multiplication by the elements of the field defined, where all the above axioms are satisfied but the statement $\forall v \: 1v = v$ is not. Is there some method by which I could prove the unprovability of the statement without exhibiting a model?

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The only thing along this line I can imagine is an induction on proofs, which would be pretty intractable with so many axioms. There are very easy counterexamples-models-to this. What motivates the desire for a syntactic proof? –  Kevin Carlson Aug 21 '12 at 4:24
    
@Kevin : The desire for a syntactic proof is just based on curiosity; a proof by a counter example model is what immediately came to mind, was just curious as to what other possible ways this could be achieved. –  Monty Gill Aug 21 '12 at 4:47
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If you prove the statement is unprovable, then by the completeness theorem you've already done enough work to exhibit a model. But it is much easier just to exhibit the model. –  Qiaochu Yuan Aug 21 '12 at 5:00

1 Answer 1

up vote 3 down vote accepted

For your particular example for Vector Spaces, the only thing I can think of to prove that an axiom is unproveable from the rest is to exhibit a model for which some collection of axioms hold and some other collection does not.


When your theory is sufficently strong, there exists another method to prove that a statement is unproveable. The Godel Incompleteness statements states that certain type of theories can not prove its own consistency. Peano Arithmetics, Second Order Arithmetics, Various systems of ZFC, etc are example of such theories for the incompleteness theorem holds.

An actual use of this is the proof that ZFC can not prove that there exists (weakly) inaccessible cardinals. (You can look up the definition of inaccessible cardinals and other large cardinal properties.) The idea is that if ZFC proved the existence of in inaccessible cardinal, then since inaccessible cardinal are "very large" in some sence, you can use it to prove the consistency of ZFC by producing a model of $ZFC$. This would contradict the incompleteness theorem; hence, $ZFC$ can not prove the existence of inaccessible cardinals. (Also many set theorist believe that $ZFC$ can not prove that inaccessible cardinals must not exist.)


Something a bit more down to earth: The fact that you can not lose the Hydra Game is unproveable in Peano Arithmetic. Basically, the Hydra Game is that you have a Tree. At each step $n$, you cut some node of the tree. You go down one node, and duplicate what left on that node $n$ times. To win the game you need to cut off all the "heads". See this website for a better description : http://math.andrej.com/2008/02/02/the-hydra-game/

You can try to prove using sufficiency strong mathematics that whatever you do, you will eventually win. However, Peano Arithemtics can not prove this result. The fact that you can not lose the Hydra Game can be used to prove the consistency of Peano Arithemtics. Again by the incompleteness theorem, Peano Arithmetics can not prove that you can never lose the Hydra Game.

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Given me a fair bit to think about, thanks. –  Monty Gill Aug 30 '12 at 2:18

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