Clarifying proof of Kunze’s theorem 3 in chapter 2 I am an undergraduate (took one course in Linear Algebra) working through the 2nd edition of Hoffman and Kunze's Linear Algebra.
Theorem 3 in chapter 2 states: "The subspace W spanned by a non-empty subset S of a vector space V is the set of all linear combinations of vectors in S."
I am having trouble understanding the proof of this theorem. I have written up what I think is a sufficient proof (it is basically Kunze's with the "left out" steps written fully out).
To my understanding, this is approached as a set equality proof. The fact that $$\textbf{L} \subset \textbf{W}$$ is clear to me.
What's less clear is how $$\textbf{W} \subset \textbf{L}$$
Any help would be much appreciated. As previously mentioned, I have written up what I think is a sufficient proof and will provide it as an answer.
 A: Note: I defer to Kunze for the proof that $\textbf{L}$ is a subspace because it was not the point of confusion and thus importance to me.
My proof: Suppose $\textbf{W}$ is a subspace spanned by the non-empty set $\textbf{S}$. Let $\textbf{v}, \textbf{u} \in \textbf{W}$ be vectors.
$\textbf{W}$ is a subspace containing $\textbf{S}$
​$\implies \textbf{W}$ is closed under scalar multiplication
$\implies$ if $\textbf{v} \in \textbf{W}$, then $c\textbf{v} \in \textbf{W}$ where $c \in \textbf{F}$ and if $ \textbf{u} \in \textbf{W}$, then $d\textbf{u} \in \textbf{W}$ where $d \in \textbf{F}$
Being a subspace also $\implies \textbf{W}$ is closed under vector addition
$\implies c\textbf{v} + d\textbf{u} \in \textbf{W}$
$\implies \textbf{W}$ contains all linear combinations of vectors in $\textbf{S}$.
Thus, the set of all linear combinations of vectors in $\textbf{S}$ denoted $\textbf{L}$ is contained in $\textbf{W}$. More concisely, $\textbf{L} \subset \textbf{W}$.
Now, $\textbf{L}$ obviously contains $\textbf{S}$. It turns out that $\textbf{L}$ is also a subspace of $\textbf{V}$. This is proven nicely in (Kunze, 37). Thus, $\textbf{L}$ is a subspace of $\textbf{V}$ containing $\textbf{S}$. In other words, it is one of the intersected sets which "contributes" to the subspace $\textbf{W}$ spanned by $\textbf{S}$.
Now, to reiterate, $\textbf{W}$ and $\textbf{L}$ are subspaces of $\textbf{V}$ containing $\textbf{S}$. And, $\textbf{W}$ is by definition the intersection of all subspaces containing $\textbf{S}$. Thus, $\textbf{W} \subset (\textbf{W} \cap \textbf{L})$. $\textbf{W} \cap \textbf{L} = \textbf{L}$ because $\textbf{L} \subset \textbf{W}$. Thus, $\textbf{W} \subset \textbf{L}$.
We have proven that $\textbf{L} \subset \textbf{W}$ and $\textbf{W} \subset \textbf{L}$​. Thus, $\textbf{W} = \textbf{L}$, i.e. the statement of the theorem.
