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Every vector space admits a Hamel basis by AC.
That is there are maximally linear independent sets.
But how to prove their cardinalities necessarily agree?

..I couldn't really find any reference.

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marked as duplicate by Najib Idrissi, Asaf Karagila set-theory Jun 15 '16 at 13:56

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  • $\begingroup$ The easy answer is: this is just like the finite dimensional case, only with transfinite recursion to construct the bijection between the two bases, rather than just an induction up to $n$ or whatever. $\endgroup$ – Asaf Karagila Jun 15 '16 at 13:59
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If there is a finite Hamel base, the vector space is finite dimensional and we can assume to be known that any basis has the same number of elements. Suppose that $\mathcal B$ is an infinite Hamel basis for the $F$-vector space $V$. Here $F$ is the field of the scalars and I assume in this answer that $F$ has the cardinality of $\mathbb R$. Any vector is a finite linear combination of element of $\mathcal B$ in a unique way. Note that this imply that $V$ has the cardinality of $F$, because there is an injection. $$ V\to \bigcup_{n=1}^\infty\left(F\cup F^2\ldots\cup F^n\right) $$ So, there are two possibilities: $\mathcal B$ is infinite but countable or $\mathcal B$ has the continuum cardinality of $F$. For the case of topological VS The first possibility is ruled out here; No infinite-dimensional $F$-space has a countable Hamel basis.

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    $\begingroup$ The result you cite is for topological vector spaces. The space of sequences of reals (or whatever field) with finite support is decidedly a space with a countable basis. $\endgroup$ – Milo Brandt Jun 15 '16 at 13:43
  • $\begingroup$ How do you construct your injection? I think you miss substantially lot of expressions.. $\endgroup$ – C-Star-W-Star Jun 15 '16 at 13:46
  • $\begingroup$ Many thanks to Milo Brandt for having highlighted my oversight! The remark is absolutely correct. I will have to re-edit or erase the "answer". $\endgroup$ – guestDiego Jun 15 '16 at 13:52
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    $\begingroup$ I posted some results along these lines in this 23 July 2000 sci.math post (see also my follow-up here), which you're welcome to make use of to rewrite your answer, if you wish. $\endgroup$ – Dave L. Renfro Jun 15 '16 at 14:07

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