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every vector space is known to have a basis (assuming the axiom of choice). This is attributed to Georg Hamel (http://de.wikipedia.org/wiki/Georg_Hamel). Moreover, any two bases have the same cardinality. In the finite dimensional case this is often proved by the Steinitz exchange lemma. An interesting note on the history of this lemma can be found here. In particular, the author points out that the exchange lemma isn't originally due to Steinitz but was published already by Grassmann in 1862.

However, my question is: Who has at first proved (or published a proof of) the uniqueness of dimension in the general case ?

I checked Hamel's paper (an online link to this paper is included in the wikipedia article), but he only proves the existence of a basis without discussing its cardinality.

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    $\begingroup$ Your question is about "Cardinality of the basis ..." or "Size of the basis ...", $\hspace{2 in}$ although your title would also be interesting. $\:$ $\endgroup$
    – user5810
    Commented Mar 17, 2012 at 22:56
  • $\begingroup$ My question is about the fact that any two bases have the same cardinality. $\endgroup$
    – KBuck
    Commented Mar 17, 2012 at 23:15
  • $\begingroup$ Interesting question. Somehow in the typical linear algebra class the uniqueness of the size of a basis is only proved for finitely generated vector space. This due to the fact that at that level students don't know enough set theory for the general case. And once we teach them the set theory, we usually say "everything goes through just as in the finite dimensional case", which is true, of course. Maybe this is how history went: The infinite dimensional case was always regarded as obvious once you know the finite dimensional case and enough set theory. $\endgroup$
    – Stefan Geschke
    Commented Mar 18, 2012 at 8:23
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    $\begingroup$ Isn't the infinite case much easier than the finite case (and in fact almost trivial)? Given two sets $A,B$ of basis elements just write each element of $A$ in terms of finitely many elements in $B$, and since this spans we must use every element of $B$ somewhere, so $B$ is a union of finite sets indexed by $A$, hence if they're both infinite, $|B| \leq |A|$. Similarly $|A| \leq |B|$. Maybe when set theory wasn't well-developed this proof was less obvious, but to me it seems likely that it's a sufficiently easy result it might not have been formally published but just remarked somewhere. $\endgroup$
    – Tom Lovering
    Commented Mar 18, 2012 at 13:56
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    $\begingroup$ The reference in Howard-Rubin points to L. Löwig who published a proof in this paper: matwbn.icm.edu.pl/ksiazki/sm/sm5/sm513.pdf. Also, note that it already follows from the Boolean primer ideal theorem (or ultrafilter lemma) which is strictly weaker than the full axiom of choice $\endgroup$
    – godelian
    Commented Mar 18, 2012 at 15:08

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Since this has already been bumped, I'll copy godelian's reference into an answer:

Löwig, H. "Über die Dimension linearer Räume." Studia Mathematica, 5(1):18-23, 1934. http://matwbn.icm.edu.pl/ksiazki/sm/sm5/sm513.pdf.

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