I About the existence of Hamel basis for vector spaces

cianfa72
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Does an Hamel basis always exist in any specific instance of a vector space?
It is well known that a vector space always admits an algebraic (Hamel) basis. This is a theorem that follows from Zorn's lemma based on the Axiom of Choice (AC).

Now consider any specific instance of vector space. Since the AC axiom may or may not be included in the underlying set theory, might there be examples of vector spaces in which an Hamel basis actually doesn't exist ?
 
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It can be shown that if the Axiom of Choice is false, then there are vector spaces with no Hamel basis. The exclusion of the Axiom of Choice means the exclusion of the proof based on the Axiom of Choice and this does not mean there are vector spaces with no Hamel basis. Every vector space has a Hamel basis.
 
Gavran said:
It can be shown that if the Axiom of Choice is false, then there are vector spaces with no Hamel basis. The exclusion of the Axiom of Choice means the exclusion of the proof based on the Axiom of Choice and this does not mean there are vector spaces with no Hamel basis. Every vector space has a Hamel basis.
Ah ok. So your claim is that every vector has an Hamel basis even though it can't proved without assuming the Axiom of Choice (AC). Basically the exclusion of the AC rules out just the proof based on it but not the conclusion (i.e. every vector space has an Hamel basis).
 
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It is well known that a vector space always admits an algebraic (Hamel) basis. This is a theorem that follows from Zorn's lemma based on the Axiom of Choice (AC). Now consider any specific instance of vector space. Since the AC axiom may or may not be included in the underlying set theory, might there be examples of vector spaces in which an Hamel basis actually doesn't exist ?
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