The span of $n$ vectors $v_1, \ldots, v_n$ is the smallest vector space containing them I'm trying to understand a proof that given a vector space $V$ over the field $F$ and $n$ vectors $v_1, \ldots, v_n$, $\mathrm{span}(v_1, \ldots, v_n)$ is the smallest subspace containing them.
I'm fine with most of the proof, including how it is proved in Axler, but the lecture notes I'm working through include a comment I don't fully understand.
I'm ok with showing that $\mathrm{span}(v_1, \ldots, v_n)$ is a subspace, that it contains $v_1, \ldots, v_n$, and that if $S$ is some other subspace of $V$ containing $v_1, \ldots, v_n$, then $\mathrm{span}(v_1, \ldots, v_n) \subset S$.
The comment made in the lecture notes is: how do we know that this smallest subspace containing $v_1, \ldots, v_n$ actually exists? Axler does not provide a proof of this fact, suggesting that it isn't necessary because by definition, "smallest" means that any other subspace containing those vectors contains their span as a subset.
Is this something I need to show for a complete proof? If so, I cannot figure out how exactly I should go about proving this. The idea, I think, is taking the intersection of all subspaces containing $v_1, \ldots, v_n$, and I know that the arbitrary intersection of subspaces is a subspace, but is it necessary to show this?
 A: Here, $v_1, v_2, \ldots, v_n$ are $n$ vectors.
Let us fix some notation first.
We define $T = \mbox{span}\left\{ v_1, v_2, \ldots, v_n \right\}$.
It is easy to verify that $T$ is a subspace of $V$.
The question is this: How to show that $T$ is the smallest subspace of $V$ containing the $n$ vectors $v_1, v_2, \ldots, v_n$ ?
Indeed, let us take any subspace $S$ of $V$ that contains the $n$ vectors $v_1, v_2, \ldots, v_n$.
Since $S$ is a subspace of $V$ and $v_1, v_2, \ldots, v_n \in S$, $S$ must contain all the linear combinations of the $n$ vectors  $v_1, v_2, \ldots, v_n$.
In other words, for any set of $n$ scalars $\alpha_1, \alpha_2, \ldots,
\alpha_n$,
$$
v = \alpha_1 v_1 + \alpha_2 v_2 + \cdots + \alpha_n v_n \in S.
$$
This essentially means that $T \subset S$, since $v$ is a general element in the subspace $T$.
Since we have shown that $T$ is a subset of any subspace $S$ of $V$
containing the vectors $v_1, v_2, \ldots, v_n$, we have established that $T = \mbox{span}\{ v_1, v_2, \ldots, v_n \}$ is the smallest subspace of $V$ containing the vectors $v_1, v_2, \ldots, v_n$. $ \ \ \ \ \ $$\blacksquare$
A: The confusion you are having is that you are checking the equivalence of one definition with itself.
If you define $span\{v_{1},v_{2},...,v_{n}\}$ as the smallest subspace containing $v_{1},...v_{n}$ then by definition it is the intersection of all subspaces containing $v_{1},...v_{n}$. You already say that you can prove that arbitrary intersection of subspaces is a subspace and hence it follows by definition that any subspace containing $v_{i}$ is the intersection. Hence $\displaystyle\text{span}\{v_{1},v_{2},...,v_{n}\}=\bigcap\{\text{S is a subsapce of V containing}\,v_{1},...v_{n}\}$.
Now if you define the span as the set of all linear combination of $\{v_{i}\}$ then you can show that this set is actually the smallest subspace containing $\{v_{1},v_{2},...,v_{n}\}$ and hence equals the intersection.
That is you need to check the equivalence of two definitions :-

*

*$\displaystyle\text{span}\{v_{1},..,v_{n}\}=\{\sum_{i=1}^{n}c_{i}v_{i}\,\,,c_{i}\in\mathbb{F}\}$
2.$span\{v_{1},...v_{n}\}$ is the is the smallest subspace containing $v_{1},...v_{n}$.
Infact you can define this for arbitrary collection of vectors $\{v_{i}\}_{i\in I}$ where $I$ is some arbirary indexing set.
Then you can again check the equivalence of two definitions:-

*

*$\displaystyle\text{span}\{v_{i}\}_{i\in I}=\{\sum_{i\in I}c_{i}v_{i}\,\,,c_{i}\in\mathbb{F},c_{i}=0\,\text{for all but finitely many}\,i\in I\}=\{\sum_{\text{finite}}c_{i}v_{i}\,\,,c_{i}\in\mathbb{F}\}$.


*$span\{v_{i}\}_{i\in I}$ is the smallest subspace of containing $\{v_{i}\}_{i\in I}$ which is equal to $\displaystyle\bigcap\{\text{S is a subsapce of V containing}\,\{v_{i}\}_{i\in I}\}$
So you see that if you prove $1\implies 2$ then it solves your "existence" crisis. And if you prove $2\implies 1$ then it means that if such a space exists then it is unique and is given by $1$.
