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My course notes for an abstract algebra course include this theorem: Let $V$ be a vector space. If $L \subset V$ is a linearly independent subset, and $E$ is minimal amongst all generating sets of $V$ with the property that $L \subseteq E$, then $E$ is a basis.

By "minimal generator of $V$" we mean that for all $\vec{v}\in E$, the set $E\setminus \{\vec{v}\}$ does not generate $V$.

By question is simply why we need $L$ in this theorem? For example, if we assume the above theorem is true, we might as well take $L=\{\}$ to be the empty set. This 'special case' then reads "if $E$ is minimal amongst all generating sets of $V$ then $E$ is a basis" which is much simpler? and more general?

This question here is highly related: Why is this a useful way to prove the characterisation of bases? and indeed, somehow it is useful to keep the $L$ in the theorem, but it really seems pretty useless to me. For example, here is another way I could chuck $L$ into the statement "if $E$ is minimal amongst all generating sets of $V$ then $E$ is a basis" in a way that doesn't seem useful:

"if $L\subset E$ is a linearly independent subset, and $E$ is minimal amongst all generating sets of $V$ then $E \cup L$ is a basis". This is trivially true, and is not useful, I see no reason why the earlier statement doesn't fall into this category.

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  • $\begingroup$ Your second paragraph does not quite capture the correct meaning of the statement "$E$ is minimal amongst all generating sets of $V$ with the property that $L \subseteq E$". Instead, the correct meaning is that for all $\vec v \in E - L$, the set $E \setminus \{\vec v\}$ does not generate $V$. $\endgroup$ Commented Jan 4 at 17:54

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The way this theorem is phrased may be a bit confusing, so I'll put it a bit more explicitly:

  1. Start with any linearly independent set $L$.
  2. Consider the collection of all generating sets which contain $L$.
  3. Then take a minimal element of this collection. It will be a basis which contains $L$.

The key of this theorem is that given a linearly independent set, we can find a basis which specifically includes it. This can be quite useful!

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    $\begingroup$ It would have made sense for the theorem to state this (there is a minimal element ... and it is a basis), but it actually doesn't (assuming the OP stated it faithfully). It says: if E is a minimal element, then it is a basis. In this statement there is indeed nothing gained from mentioning L. $\endgroup$ Commented Jan 5 at 3:57
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    $\begingroup$ @Pilcrow, it is true that the theorem leaves existence aside, but that does not make it (or the inclusion of L) useless. It just means that an additional argument is needed. Maybe that comes afterwards for the case of a finitely generated vector space. $\endgroup$ Commented Jan 5 at 14:19
  • $\begingroup$ @CarstenS On reflection, the inclusion of L in the theorem as stated does add a little bit of logical strength to it (if removing any vector results in a non-generating set, we have a basis vs. if removing any vector outside of the set L results in a non-generating set, we have a basis), but I don't think I have ever used the latter kind of property. $\endgroup$ Commented Jan 6 at 16:03
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Bringing in the set $L$ makes the condition of the theorem weaker, not stronger. The reason is that $E$ only has to be minimal amongst the supersets of $L$, not amongst all sets. A theorem with a weaker precondition is apriori more useful.

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    $\begingroup$ This doesn't seem right. The original statement "for all L ⊂ V [...]" is indeed stronger, not weaker, than the special case of it "let L = {}; then [...]". $\endgroup$ Commented Jan 6 at 6:54
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    $\begingroup$ @DonHatch The entire statement is stronger than the special case, because the condition is weaker. $\endgroup$ Commented Jan 6 at 10:30
  • $\begingroup$ Oh! I misread your first sentence and thought you were saying the whole theorem statement is weaker. Yes, that makes sense now. $\endgroup$ Commented Jan 6 at 11:01

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