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## Main Question or Discussion Point

Is the existence of basis in all vector space equivalent to the axiom of choice?

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Is the existence of basis in all vector space equivalent to the axiom of choice?

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Hurkyl

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mathwonk

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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...

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matt grime

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HallsofIvy

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matt grime

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How is it a refinement of the relation between existence of basis and axiom of choice?

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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.

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

Because B is a basis for V, then so is D. But, because |D| = |B| < |C|, C cannot be a basis.

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

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Hurkyl

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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.

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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.

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matt grime

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http://planetmath.org/encyclopedia/EveryVectorSpaceHasABasis.html [Broken]

answers the original question

answers the original question

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Hurkyl

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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?)

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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 HAFHurkyl said:I'm pretty sure you don't need the axiom of choice to prove that if you have two bases, their cardinalities are equal.

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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.Hurkyl said: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?)

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