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Basis and Axiom of Choice

  1. May 11, 2005 #1
    Is the existence of basis in all vector space equivalent to the axiom of choice?
     
  2. jcsd
  3. May 11, 2005 #2

    Hurkyl

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    It's certainly a consequence of the axiom of choice. I doubt that they're equivalent, but I don't have a proof, or even a concrete idea for answering the question. If you get an answer, let me know. :smile:
     
  4. May 11, 2005 #3

    mathwonk

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    well it requires the ability to choose a maximal chain from any set of independent sets ordered by inclusion, and any set can be the basis of a vector space, so it might be pretty general, i.e. a pretty general form of maximal principel which is equivalent.
     
  5. May 17, 2005 #4
    What do you mean by this Mathwonk?
    I recall my linear algebra textbook saying that if the axiom of choice was used it can be shown that any vector space has a basis but it didn't say how...
     
  6. May 18, 2005 #5

    matt grime

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    How do you picka basis: take one vector, then look at the complementary space to its span. Pick a vector in there, look in the complement to the span of these two and so on. if it's finite dimensional all well and good, but infinite dim means we ave to make an infinite number of compatible choices and it isn't clear that this is possible.
     
  7. May 18, 2005 #6

    HallsofIvy

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    However, we don't seem to be able to answer whether or not the existence of a basis in every vector space implies the axiom of choice.
     
  8. May 18, 2005 #7

    matt grime

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    I would guess that to be false, but it is a guess. How about the refinement "every vector space has a basis, and the cardinaltity of any choice of basis is the same".
     
  9. May 18, 2005 #8
    How is it a refinement of the relation between existence of basis and axiom of choice?
     
  10. May 18, 2005 #9

    Hurkyl

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    Any two bases must have the same cardinality.

    Suppose B and C are both bases for V, and that |B| < |C|.

    Every element of B is a linear combination of finitely many elements of C, and this defines a function f:B --> P(C)


    Now, let D = U f(B), the union of all the sets in the image of f. Because f(b) is finite for each b in B, we have that |D| = |B|.

    Because B is a basis for V, then so is D. But, because |D| = |B| < |C|, C cannot be a basis.
     
    Last edited: May 24, 2005
  11. May 23, 2005 #10
    What is P(C)? The power set of C? I don't understand. And isn't this theorem supposed to rely on Zorn's lemma?
     
  12. May 23, 2005 #11

    Hurkyl

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    Yes, P(C) is the power set.

    The existance of a basis requires the axiom of choice. I'm pretty sure you don't need the axiom of choice to prove that if you have two bases, their cardinalities are equal.
     
  13. May 24, 2005 #12
    how does an element of B being a linear combination of elements of C make a function from B to P(C)? Is it just that if b= a1*c1 + .... +an*cn, then f maps b to {c1,....,cn}, disregarding the coefficients?

    also, why must f(B) be finite? For example, if the two bases are the same, the image of each b in B is a singleton set, but f(B) is not finite.
     
  14. May 24, 2005 #13

    matt grime

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  15. May 24, 2005 #14

    Hurkyl

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    That was a typo! It was supposed to say that f(b) is finite for each b in B. You have the definition of f correct, except for one technical detail -- f(b) only includes those ci for which ai is not zero.

    Is "support" the word I'm looking for? I.E. "f(b) is the support of b in the basis C." (or something like that?)
     
  16. May 27, 2005 #15
    Apparently, you need the ultrafilter principle, which follows from the AC, but is not equivalent to it. So I was right that the uniqueness of cardinality of basis doesn't follow from ZF, but wrong in assuming that it requires choice. See Schechter HAF
     
  17. May 27, 2005 #16
    the support of a function is the closure of the set where the function is nonzero. unless we're talking about topological vector spaces, I don't think we can apply that term.
     
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