Suppose $S\subseteq V$ is a spanning set(possibly infinite) of a non-zero vector space $V$.

Let $s'=\{s\subseteq S|$$s$ also spans V$\}$. Suppose $ l \subseteq s'$ is a totally ordered non-empty subset of $s'$.

How do I show that span$(\cap l)$ =V.

It looks true to me. If we take a $c\in l$. It would contain all the vectors of those sets(whose spans are also V) that are subsets of it.

And all supersets would also contain vectors of $c$. But I just can't seem to show vigorously that the intersection spans V. (i.e. can't find least element.)

Any ideas on how to prove it or is it false?

P.S. I'm trying to prove that every spanning set has a basis. i.e. Zorn's Lemma, Dual order etc

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    $\begingroup$ As an aside, the usual way of proving that every spanning set has a basis is to consider the partially ordered set of all linearly independent subsets of the spanning set, and showing that a maximal element of this set will necessarily include the spanning set in its span, and hence will itself span. $\endgroup$ – Arturo Magidin Jan 19 '19 at 3:32
  • $\begingroup$ @ArturoMagidin I see. So the proof is similar to showing that any linearly independent set can be extended to a basis except now we are just working within a spanning set. Thanks. PS. just out of curiosity is the above statement still true? $\endgroup$ – Jhon Doe Jan 19 '19 at 3:38
  • $\begingroup$ I’m fairly sure that it is. BTW, you can actually prove both statements (extension and paring down) at the same time as the existence of a basis, by sbowing that if $L\subseteq S$, $L$ is linearly independent, and $S$ spans, then there is a basis $B$ with $L\subseteq B\subseteq S$. You get the statement for linearly independent subsets by taking $S=V$, and the statement for spanning sets by taking $L=\varnothing$, and the existence of bases by taking $L=\varnothing$ and $S=V$. $\endgroup$ – Arturo Magidin Jan 19 '19 at 4:13
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    $\begingroup$ Actually, I think the statement you are trying to prove is false. $\endgroup$ – Arturo Magidin Jan 19 '19 at 4:36

I believe the statement you are trying to prove is false.

Consider $V=\mathbb{R}$ as a vector space over $\mathbb{R}$. Let $S$ be the interval $(-1,1)$ (which is certainly a spanning set; note that a subset of $V$ is a spanning set if and only if it contains at least one nonzero number). Now let $l=\{(-\frac{1}{n},\frac{1}{n})\mid n\in\mathbb{N}_{\gt 0}\}$. This is a totally ordered subset of the collection of all subsets of $S$ that span $V$. However, $\cap l = \{0\}$, which is not a spanning set.

  • $\begingroup$ Hence the obvious proof that every spanning set contains a basis, by using Zorn's lemma to show it contains a minimal spanning set, doesn't work. Which explains why "maximal independent set" seems to be so much more popular than ""minimal spanning set"... $\endgroup$ – David C. Ullrich Jan 19 '19 at 14:17

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