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Let $V$ be a finite-dimensional vector space. Is it possible that a set $S$ spans $V$, but no subset of $S$ is a basis of $V$?

I think that the answer is no, at least in the case that $S$ is finite. For suppose that $S=\{v_1,\dots,v_n\}$ spans $V$. If $S$ is linearly independent, then we are done. If not, then there is an element $v_i$ such that $v_i\in\DeclareMathOperator{\span}{span}\span(\{v_1,\dots,v_{i-1},v_{i+1},\dots,v_n\})$. If we remove $v_i$ from the set $S$, then we obtain a new set which still spans $V$. By continuing this process of removal, we will eventually end up with a linearly independent set that still spans $V$. (The process must terminate, as the empty set is a linearly independent subset of any vector space.)

My questions are:

  1. Is this line of reasoning correct?
  2. What about in the case that $S$ is infinite?
Joe
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  • I think the reasoning is there, but that your argument could be a bit more rigorous. You have the idea down, though. – Chaotic Good Feb 21 '22 at 22:42
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    Yes that's fine. For the infinite case, you use Zorn's lemma to extract a maximal linearly independent subset; see If S spans a vector space V, does S contain a basis?. – peek-a-boo Feb 21 '22 at 22:44
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    @peek-a-boo: Thanks for the feedback. With regard to the link, I think that covers the case that $V$ is infinite-dimensional. However, I'm curious about the case that $V$ is finite-dimensional, but $S$ is infinite. – Joe Feb 21 '22 at 22:57
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    @Joe The link covers all cases: finite or infinite sets spanning finite or infinite-dimensional vector spaces (though, of course, one of these four cases is not possible!). – Theo Bendit Feb 21 '22 at 23:03
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    As Theo mentions, the above link already addresses all the cases in one swoop. Anyway, for the specific case of $V$ being finite dimensional (by which I mean "has a finite spanning set"), one of the first few theorems regarding dimension is that "the dimension is well-defined" and that any set with fewer vectors than the dimension cannot be spanning, and any set with more vectors than the dimension cannot be independent. For the proofs of these (somewhat tedious induction-based results) see any good linear algebra text. – peek-a-boo Feb 21 '22 at 23:44
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    With this in mind, suppose $S$ is an infinite spanning set and $V$ is finite-dimensional. Then, it must contain some finite spanning set $S'$, so now you can use your argument to show $S'$ (and hence $S$) contains a basis. Unfortunately, I can't think of a "cleaner" argument right now. – peek-a-boo Feb 21 '22 at 23:45
  • Assume the dimension is equal $k.$ Then $S$ contains at most $k$ linearly independent set of vectors. As $S$ spans $V,$ there are exactly $k$ such vectors. Is it clean enough ? – Ryszard Szwarc Feb 22 '22 at 09:21

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