How to really understand the tensor algebra? If $V$ is a vector space over $F$, then we define $T^r_0(V)=V^{\otimes r}$, then we define the algebra of contravariant tensors to be
$$T(V)=\bigoplus_{r=0}^\infty T^r_0(V)$$
together with the tensor product. But I'm really confused with that. The set $T(V)$ by construction, is the set of all sequences $T=(T_i)$ with $T_i=0$ for all but finitely many indexes $i\in \Bbb N$ with addition defined by $(T+S)_i = T_i+S_i$ and $(kT)_i=kT_i$.
The book I'm working with doesn't tell how the tensor product is defined in $T(V)$, but I think it is defined by $(T\otimes S)_i = T_i\otimes S_i$ which is the most natural thing to do. But I can't really understand how we should think about this space, it seems like to be putting together all kinds of contravariant tensors, but I can't really see why do we need this space, and how should we work with it.
I tried showing that this is really an algebra, but I'm confused if I did it right. I did the following: let $T,S\in T(V)$ and $a,b\in F$, then $(T\otimes (S+R))_i=T_i\otimes (S+R)_i,$ then from multilinearity of the tensor roduct on each factor $(T\otimes(S+R))_i=T_i\otimes S_i+T_i\otimes R_i$ wich means that $(T\otimes (S+R))_i = (T\otimes S)_i + (T\otimes R)_i$ and finally
$$(T\otimes (S+R))_i = ((T\otimes S)+(T\otimes R))_i\Longrightarrow T\otimes(S+R)=T\otimes S + T\otimes R,$$
the same for distribuitivity on the other side and for multiplication by scalars.
So, is this proof right? And in general, how should we really understand and think about the tensor algebra of a vector space?
Thanks very much in advance!
 A: Your proof looks correct to me, and I don't think you said anything else wrong either.
I prefer not to think of $T(V)$ in terms of sequences like this. For me, an element of $T(V)$ is a linear combination of tensors of possibly mixed degree; to convert a sequence into this, just add up the terms, which you can do because all but finitely many are zero.
Then the tensor product can essentially be defined to be distributive. You already know how to take the tensor product of two pure tensors, so to take the product of two combinations, just expand by the distributive law and then use the definition for pure tensors.
As an analogy, this is kind of like thinking of the polynomial ring $\mathbb{C}[x]$ as:
$$\mathbb{C}[x]=\bigoplus_{i=0}^\infty\mathbb{C}\langle x^i\rangle$$
where $\mathbb{C}\langle x^i\rangle$ is the one-dimensional vector space spanned by $x^i$. Then we can define $x^i\cdot x^j=x^{i+j}$ and extend this product to $\mathbb{C}[x]$ uniquely by insisting it is multilinear.
A: Let $V$ be a vector space over a field $F$. The structure
$$ T(V)=\bigoplus_{i\geq 0} V^{\otimes i}$$
is just a set that can be endowed with an algebra, or coalgebra structure. For these reasons, $T(V)$ is known also as the "tensor module".
 The algebra structure is defined by an associative product $\bigotimes$ with unit $1\in F$, where
$$(v_1\otimes \dots \otimes v_q)\bigotimes (w_1\otimes\dots\otimes  w_r):=
v_1\otimes \dots \otimes v_q \otimes w_1\otimes\dots\otimes  w_r\in V^{\otimes q+r}, $$
for all $(v_1\otimes \dots \otimes v_q)\in V^{\otimes q}$, $(w_1\otimes\dots\otimes  w_r) \in V^{\otimes r}$ and $q,r\geq 0$. By definition, the product $\bigotimes$ is called concatenation. Associativity is easily proven.
EDIT. This is a very good reference http://math.unice.fr/~brunov/Operads.pdf
In chapter 1 you can find the definitions of the tensor module / algebra and coalgebra.
A: There is a very simple underlying idea: we want to multiply vectors.
The algebra $T(V)$ gives us the most general way that we can multiply vectors while still retaining the rules of arithmetic, and contains nothing beyond what is needed to do so.
It's not hard to check that every element of $T(V)$ is, in fact, obtained by using scalars, vectors, addition, and multiplication. A little harder is to prove -- but very important to understand -- is the universal property that if $R$ is any ring over $F$, and $f : V \to R$ is an $F$-linear map, then there is a unique $F$-ring homomorphism $g : T(V) \to R$ that satisfies the property $g(v) = f(v)$ for all vectors $v$. (uniqueness is easy: it's the fact that it exists that is nontrivial)
