Basis for Tensor Product of Infinite Dimensional Vector Spaces If V and W are vector spaces over a common field with bases $V_B = ${$v_i : i \in I$} and $W_B = ${$w_j : j \in J$}, then is {$v_i \otimes w_j: i \in I, j \in J$} a basis for $V \otimes W$ ?
I have several references for the finite dimensional case but they seem to involve either a dimensional argument or reference to dual spaces, neither of which would seem to be appropriate in the infinite dimensional case.
The references that I can find for infinite dimensional spaces focus on "orthonormal bases" (usually in the context of Hilbert Spaces). 
So, I'd appreciate any help answering 


*

*is this actually true if {$v_i : i \in I$} and {$w_j : j \in J$} are the Hamel (algebraic) bases of infinite dimensional spaces and what is the proof ? 

*Does it make any difference if only one space is infinite dimensional ? 

*Is there any difference between countable and larger infinities ?


Addendum 15 March 2015:
Why not simply define the tensor product as $F(V_B \times W_B)$, i.e. the free vector space generated by the set of ordered pairs of basis vectors ? There seems to be a bilinear mapping from $V \times W$ and it appears to satisfy the universal property. 
Have I missed something here ?
 A: I give a "conceptual" answer.
Let $k$ be a field, and $V, W$ vector spaces over $k$. Recall that $k$-vector space $V\otimes_k W$ has the following universal property : if we note $\pi : V\times W\to V\otimes_k W$ the canonical $k$-bilinear map $(v,w)\mapsto v\otimes w$ and $\textrm{Bil}_k (V,W;P)$ the set of $k$-bilinear applications $V\times W\to P$ for a $k$-vector space $P$, then we have : for each $k$-vector space $P$ the map $\textrm{Hom}_k (V\otimes_k W,P)\to \textrm{Bil}_k (V,W;P)$ definied by $f\mapsto f\circ\pi$ is a bijection.
As I think that any answer to your question should emphasize what is true in general, what is proper to the situation (vector spaces = modules over a field) and what things are independent. I will prove the following three general lemmas, with the following hypothesis : from now on, $k$ is not a field but a commutative ring with unit, and $V,W$ are modules over $k$.
Lemma 0. The elements of the form $v\otimes w$ for $v\in V$ and $w\in W$, called elementary tensors (or also simple tensors), generate $V\otimes_k W$ over $k$.
Proof. I give a proof of this lemma using only the universal property of $V\otimes_k W$ and not any of its constructions (as quotient of $A^{(W\times V)}$ for instance). Note $Q$ the sub-$k$-module of $V\otimes_k W$ generated by elementary tensors. Note $\pi' : V\times W\to Q$ the map defined by $(v,w)\mapsto v\otimes w \in Q$. I pretend that $Q$ and $\pi'$ satisfies the same universal property that $V\otimes_k W$ and $\pi$ do. Let $P$ be a $k$-module. The map $\textrm{Hom}_k (Q,P)\to \textrm{Bil}_k (V,W;P)$ defined by $f\mapsto f\circ\pi'$ is a injective. Indeed : if $f\circ \pi' = g\circ \pi'$ for $f,g\in \textrm{Hom}_k (Q,P)$ then it means that $f$ and $g$ are equal on elementary tensors, and as elementary tensor generate $Q$ by definition of $Q$, we have $f=g$. Now, let's show that this map is surjective, so take $b\in \textrm{Bil}_k (V,W;P)$. By the universal property of $V\otimes_k W$ we can find $f\in \textrm{Hom}_k (V\otimes_k W,P)$ such that $b = f\circ\pi$, which obviously imply that $b = f' \circ \pi'$, where the $k$-linear map $f' : Q\to P$ is the restriction of $f$ to the submodule $Q$ of $V\otimes_k W$, that is, the $k$-linear map $f' : Q\to P$ defined by $q\mapsto f(q)$ for $q\in Q$. This shows the surjectivity of the map $\textrm{Hom}_k (Q,P)\to \textrm{Bil}_k (V,W;P)$, which is finally injective and surjective, that is, bijective. This shows that $Q$ and $\pi'$ satisfies the same universal property that $V\otimes_k W$ and $\pi$ do. Thanks to this, we will now show that $Q = V\otimes_k W$. Note $i : Q \to V\otimes_k W$ the canonical injection : it is the map sending an element of $Q$ to the same element but viewed as element of $V\times_k W$. Rephrased, what we want to show is that $i$ is surjective. Now $\pi' : V\times W \to Q$ is a $k$-bilinear map, so that by the universal property of $V\otimes_k W$ and $\pi$, there exists a unique $k$-linear map $\varphi : V\otimes_k W \to $ such that $\varphi\circ \pi = \pi'$. But then $(i\circ\varphi)\circ\pi = i\circ(\varphi\circ\pi) = i\circ\pi' = \pi = \textrm{Id}_{V\otimes_k W}\circ\pi$, and by the universal property $i\circ\varphi = \textrm{Id}_{V\otimes_k W}$, showing that $i$ has a right inverse, and is therefore surjective. $\square$
Lemma 1. If $(v_i)_{i\in I}$ (resp. $(w_j)_{j\in J}$) is a familly of generating elements of $V$ (resp. of $W$) over $k$, then the familly $(v_i\otimes w_j)_{(i,j)\in I\times J}$ is a familly of of generating elements of $V\otimes_k W$ over $k$.
Proof. As $\pi$ is $k$-bilinear, it is clear that any elementary tensor will be linear combination of $v_i\otimes w_j$'s over $k$, and by the previous lemma, the elementary tensors generate $V\otimes_k W$, showing thereby that the elements $v_i\otimes w_j$'s generate V\otimes_k W$
From now on we suppose that $k$ is a field.
Lemma 2. If $(v_i)_{i\in I}$ (resp. $(w_j)_{j\in J}$) is a familly of independant elements of $V$ (resp. of $W$) over $k$, then the familly $(v_i\otimes w_j)_{(i,j)\in I\times J}$ is a familly of independant vector of $V\otimes_k W$ over $k$. $\square$
Proof. Suppose we have a dependance relation $$\sum_{(i,j)\in I\times J} c_{i,j} v_i\otimes w_j = 0$$ where $(c_{i,j})_{(i,j)\in I\times J}$ is a familly a elements in $k$ such that $c_{i,j}\not=0$ only for finitely many indexes $(i,j)$'s. Complete $(v_i)_{i\in I}$ (resp. $(w_j)_{j\in J}$) in a basis $(v_i)_{i\in \overline{I}}$ (resp. $(w_j)_{j\in \overline{J}}$) of $V$ resp. (of $W$) over $k$. Take $(i_0,j_0)\in I\times J$ and let's show that $c_{i_0,j_0} = 0$. For this, consider the bilinear function $V\times W\to k$ sending $(v,w)=\left( \sum_{i\in \overline{I}} \lambda_i v_i, \sum_{j\in \overline{J}} \mu_j w_j \right)$ to $\lambda_{i_0} \mu_{j_0}$. Take by the universal property $f : V\otimes_k W \to k$ to be the linear map associated to this bilinear form. Applying the linear function to the above dependance equation gives $c_{i_0,j_0} = 0$. $\square$
Lemma 2 and 3 applied to the case of a field do answer your question.
Remark. No cardinality hypothesis were made.
A: Yes it is true independent of the cardinality of the bases of the vector spaces. Use the universal property: Let $T$ be any vector space over the given field.
A bilinear map $V \times W \to T$ is uniquely determined by the images of the pairs $(v_i,w_j)_{i \in I,j \in J}$, so we get
$$Bil(V \times W, T) = Abb(\{(v_i,w_j), i \in I, j \in J\},T)= Abb(\{v_i \otimes w_j, i \in I, j \in J\},T)$$
By the universal property $V \otimes W$ is the vector-space with the property
$$Hom(V \otimes W,T) = Bil(V \times W, T)$$
, so we obtain
$$Hom(V \otimes W,T) = Abb(\{v_i \otimes w_j, i \in I, j \in J\},T)$$
, which precisely states that $\{v_i \otimes w_j, i \in I, j \in J\}$ is a basis of $V \otimes W$.
Let us precise this:
Given a map in $f \in Abb(\{v_i \otimes w_j, i \in I, j \in J\},T)$, which means that we are given $f(v_i \otimes w_j)$ for all $i \in I, j \in J$, we have to show that this extends uniquely to a map $F:V \otimes W \to T$.
We define the bilinear map $\hat f: V \times W \to T$ by $\hat f(v_i,w_j) := f(v_i \otimes w_j)$. By the universal property, this gives us a unique map $F:V \otimes W \to T, F(v \otimes w)=\hat f(v_i,w_j)=f(v_i \otimes w_j)$. Hence this is the desired unique map.
The uniqueness of $F$ is trivial anyway, because $\{v_i \otimes w_j, i \in I, j \in J\}$ is clearly a set of generators. We only need to show the existence of $F$ here (And this corresponds to the linear independence).
