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.


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

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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