Talk:Vector Space has Basis between Linearly Independent Set and Spanning Set

Axiom of Choice template
Why do we need a blurb here about the semi-controversial nature of the axiom of choice? Isn't it enough just to note that the axiom of choice is, in fact, required? (and by the way, we need a proof that the existence of bases for all vector spaces implies AoC)Dfeuer (talk) 03:36, 1 December 2012 (UTC)


 * 1) Because it's cool
 * 2) Well volunteered, but please respect our house style in future. --prime mover (talk) 08:13, 1 December 2012 (UTC)


 * The structure of this proof is somewhat closer to how this site works. Specifics:
 * a) Unfortunately we can't use predefined macros so as to shorten the source code because there are conditions by which MathJax fails to render the interpretations in MediaWiki. So reluctantly we have to disallow this construction. Damn shame, because it would smarten and streamline considerably.
 * b) Every statement has its own line. Compound sentences are to be split into their components. Don't fear the "return" key.
 * c) Every instance of parentheses (except for equation numbering) requires its own \left ... \right bracketing, and within that bracketing, use of { ... } around the thing being bracketing. Rigorous adherence to this ensures that all parentheses are i) guaranteed to match, ii) guaranteed to automatically resize, and iii) having complicated stuff inserte into a cut-and-pasted instance of the bracketed expression, without anything needing a rewrite.
 * d) Because of its ambiguity, we discourage the use of $\subset$. Either $\subseteq$ or $\subsetneq$ are acceptable, despite the widespread convention in certain branches of set theory that traditionally use $\subset$.
 * e) Every concept needs its link.
 * Basically, that's it. --prime mover (talk) 20:40, 1 December 2012 (UTC)

Refactor request
Could whoever wants this thing refactored please propose a name for the fact that if V is a vector space, L is a subset of S, S is a subset of V, L is linearly independent, and S spans V, then there is a basis, B, for V such that L is a subset of B, which is a subset of S? Perhaps "bounded basis theorem?" The "downward basis theorem" (using only S without L) is historically significant because DB implies AoC was proven a couple decades before "every vector space has a basis" implies AoC was shown. Dfeuer (talk) 21:48, 2 December 2012 (UTC)

Stub?
I note your stub suggesting that the article is incomplete. But as far as this site is concerned, such a statement constitutes another theorem and therefore is to be entered as a new page, a new theorem and a proof. This page can then link to that one via an "also see" which is used to reference results and/or definitions which are immediately relevant. --prime mover (talk) 22:34, 2 December 2012 (UTC)
 * I didn't add the stub template. I just modified it once or twice to reflect the improvements that had been made to the article. Dfeuer (talk) 00:32, 3 December 2012 (UTC)

Rename
How about Vector Space has Basis between Linearly Independent Subset and Spanning Subset? Or we might as well do away with the rename request. &mdash; Lord_Farin (talk) 11:30, 27 September 2014 (UTC)


 * I'm unconcerned. TheSpleen (whose rename suggestion it was) hasn't been seen for some time, we may be able to assume he's no longer around. --prime mover (talk) 12:46, 27 September 2014 (UTC)


 * I've opted for the easy way out. &mdash; Lord_Farin (talk) 14:11, 27 September 2014 (UTC)