Linear Algebra Basics I'm having my first linear algebra classes in college right now, and a few difficulties with the symbolism used. Missing some basics so to say.
So I have a few small questions I will just ask here:


*

*what does the notation $C^\infty(a,b)$ as in $W = \{f \in C^\infty(a,b) \;/ \;\dfrac{d^2}{dx^2}f=0\}$ mean?

*what does the notation $(u,v)\longmapsto u+v$ in the context of $V \times V \longrightarrow V$ mean?

*what exactly is an 'additive unit'?

*in the proof: 
Prop.: If $W_1, W_2$ are two subspaces of V, then $W_1 \cap W_2$ is a subspace
(1) Since $0\in W_1$, $0\in W_2$, we have $0\in W_1 \cap W_2$.
(2) Want: If $u, v \in W_1 \cap W_2$ then $u+v \in W_1\cap W_2$.
If $u,v \in W_1 \cap W_2$ then $u,v \in W_i,\; i = 1,2$. 
Since $W_i$ is a subspace we have $u,v \in W_i$.
Thus $u+v \in W_1 \cap W_2$.
(3) ...
How can he say that $W_i$ is a subspace, when that is what he is trying to prove? (or which part did i get wrong?)
Also: how does the third part of the proof (if $u\in W,\alpha \in \mathbb{R}$ then $\alpha u \in W$) look, or why did my prof only write the first two?
Any help is very appreciated !
Thank you :)
 A: *

*$C^\infty(a,b)$ is usually the functions on the open intervall (a,b) which are infinitely many times differentiable.

*$V\times V\to V$ is a map, let's call it $f$. For a map you need to know where an element is mapped to. The notation $(u,v)\mapsto u+v$ means that $f((u,v))=u+v$. Here $(u,v)$ is an element of $V\times V$ whereas $u+v$ is an element of $V$.

*An additive unit (of $V$) is an element $o\in V$ such that $v+o=v$ for all $v\in V$.

*For the proof: You have to distinguish between the assumptions and the claim. The assumptions are that $W_1$ and $W_2$ are already vector spaces. The claim is that the intersection $W_1\cap W_2$ is a vector space. So in the proof of (2) as you have that $W_1$ is already a subspace (by assumption) and $u,v\in W_1$ you also have $u+v\in W_1$. Similarly $u+v\in W_2$ since $W_2$ is a subspace. Hence $u+v\in W_1\cap W_2$. 

*(3) then follows similarly: As above $W_1$ and $W_2$ are already subspaces, hence $\alpha u\in W_1$ and $\alpha u\in W_2$. Hence $\alpha u\in W_1\cap W_2$.

