The role of linear combination in definition of a subspace. I am an undergraduate who just finished taking my first course in Linear Algebra. I am wanting to get a deeper understanding of the subject out of interest and for its wide reaching applications (primarily to physics).
To do so, I have been working through Hoffman and Kunze's 2nd Edition of Linear Algebra. I am on the second chapter, which concerns vector spaces.
Kunze defines a vector space in the way I was taught this past semester: axiomatically. He then defines subspaces as such:
"Let V be a vector space and F be a field. A subspace of V is a subset W of V which is itself a vector space over F with the operations of vector addition and scalar multiplication on V."
The first theorem presented after this definition states:
"A non-empty subset of W of V is a subspace of V iff for each pair of vectors a, b in W and each scalar c in F the vector ca + b is again in W."
Now, the definition of a subspace that I am familiar with is something like "a subspace of a vector space V is a subset of the vectors in V closed under vector addition and scalar multiplication. This translates into closure under linear combination."
I am quite confused as to why Kunze uses "...ca + b is again in W" instead of closure under linear combination which would look like ""...ca + db is again in W."
My thought is that "...ca + b is again in W" actually implies that "...ca + db is again in W". Thus, it is unnecessary to say closure under linear combination. I have thought through a sketch of this proof which goes something like:
Let W be a non-empty subset of vectors in vector space V on field F such that for vectors a, b and for an arbitrary scalar c in the field F ca + b is contained in W. I.e. let W be a subspace as defined by Kunze. Then, any scalar multiple of a is itself a vector in W because we can set b to be the 0 vector which necessarily must be in a subspace. If we then now set b to be the said scalar multiples of a that we just found out are elements of W, we can now input our original arbitrary vectors b in ca and produce linear combinations of vectors a, b as we set out to do.
However, one of the chapter's emphasis is that it's useful to consider a set of objects and linear combinations of said objects. So, why not emphasize it here?
Further adding to my confusion is Kunze's theorem 3:
"The subspace spanned by a non-empty subset S of a vector space V is the set of all linear combinations of vectors in S."
Here, Kunze actually refers to linear combinations of vectors in S. Now, if my hypothesis about "...ca + b is again in W" implying that "...ca + db is again in W" is correct, why wouldn't Kunze just say "...is the set of all combinations of vectors in S of the form ca + b"?
This leads me to believe I am misunderstanding something. Any help would be much appreciated.
Explicit edit: I thought it might be helpful to provide Kunze's proof for the above stated theorem 1:

Below is Kunze's definition of a subspace spanned by a subset, theorem 3, and theorem 3's proof.

 A: You also need to show that $\textbf a, b \in W \Rightarrow c\textbf a + \textbf b \in W$ implies $\textbf 0 \in W$, but it's simple (and essentially proved in the proof of theorem 1).
For non-empty $W \subseteq V$, the following are equivalent:

*

*$W$ is vector space itself (with the same operations)


*$W$ is closed under vector addition and scalar multiplication


*$W$ is closed under taking linear combinations


*if $\textbf a, \textbf b \in W$ and $c \in F$ then $c \textbf a + \textbf b \in W$


*if $\textbf a, \textbf b \in W$ and $c,d \in F$ then $c \textbf a + d \textbf b \in W$
The 4th variant is usually shortest to write, while the 2nd is simplest to check.
For the theorem 3, it's important to consider all linear combinations, not just of form $c\textbf a + \textbf b$, because when you talk about building a closure (not of checking if something is closed) you may need to use this operations more than once. And set $\{c\textbf a + \textbf b | \textbf a, b \in S, c \in F\}$ in general isn't a subspace spanned by $S$.
For example, if $V = \mathbb R^2$ and $S = \{(0, 1), (1, 0)\}$, then subspace spanned by $S$ is entire $V$, but vector $(2, 2)$ isn't of form $c\textbf a + \textbf b$.
