As per this question, it is possible to extend any l.i. subset of vectors $A \subset V$ of a vector space to a (Hamel) basis, $B \subset V$. Is it likewise possible to delete vectors from a spanning subset $S\subset V$ ($span(S) = V$)? The argument is trivial in finite-dimensions, but it's not clear to me that this is possible in infinite dimensions. Even a sketch of the proof would be greatly appreciated.
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Let $B$ be a maximal independent subset of $S$, which exists by Zorn’s Lemma. Then we can show that $span(B) = span(S) = V$. For given $s \in S$, suppose $s \notin span(B)$. Then $B \cup \{s\}$ is a larger independent subset of $S$ than $B$ is; contradiction.
Also, note that we cannot prove this theorem without Zorn’s Lemma (or some statement of the equivalent axiom of choice). In fact, over ZF, the claim that every vector space has a basis is equivalent to Zorn’s Lemma. Clearly, your claim implies that all vector spaces have bases, as we can take $S = V$.
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$\begingroup$ Ah, so to be clear you're saying that Zorn's lemma is not just equivalent to saying that $span(S) = V$ has a basis, but it's also equivalent to saying that I can choose that basis as coming from elements of $S$? $\endgroup$– EE18Apr 4 at 19:38
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$\begingroup$ Yes. We just proved that Zorn’s Lemma implies we can pick a basis from $S$. Trivially, we can thus find some basis for $V$. Far less trivially, we can then use this principle to prove Zorn’s Lemma. $\endgroup$ Apr 5 at 22:02