H. Cartan - Differential Calculus. Query? In H. Cartan - Differential Calculus (1971) p. 29 he investigates differentiating a bi-linear function $f: E_1 \times E_2 \to F$ where $E_1, E_2, F$ are Banach spaces and $E_1 \times E_2$ the product (presumably Cartesian).
He claims $E_1 \times E_2$  to be a Banach space with the obvious rules of addition and scalar multiplication.
I think this might be OK if $E_1, E_2$ are one-dimensional, but not otherwise. To be algebraically complete mustn't he instead use the tensor product $E_1 \otimes E_2$ ?
Since for $E_1, E_2$ at least two dimensional with bases $\{u_1, u_2\}, \{v_1, v_2\}$ there is a very clear counterexample....
$(u_1, v_1), (u_2, v_1), (u_1, v_2), (u_2, v_2) $ are elements of $E_1 \times E_2$.
But then $(u_2, v_1) + (u_1, v_2) + (u_2, v_2) $ is not of the form $(u, v)$ and so not in $E_1 \times E_2$, i.e. $E_1 \times E_2$ is not algebraically closed under addition.

I may have mixed up some concepts in the above.
It seems that $E_1 \times E_2$ with addition and scalar multiplication as noted by @JohnHughes is the direct sum of $E_1, E_2$ and nothing to do with the tensor product.
And then as noted by @JoonasIlmavirta $(u_2, v_1) + (u_1, v_2) + (u_2, v_2) = (u_1 + 2.u_2, v_1 + 2.v_2)$.
 A: I suspect that the definition of addition is that
$$
(u_1, v_1) + (u_2, v_2) = (u_1 + u_2, v_1 + v_2),
$$
where the first addition (on the right-hand side) is the addition on $E_1$, and the second is addition on $E_2$.
For scalar multiplication, I'd guess the definition is
$$
r(u, v) = (ru, rv)
$$
where the first entry uses scalar multiplication from $E_1$, and the second uses scalar multiply from $E_2$.
Presumably the norm on $E_1 \times E_2$ is just something like the sum of the individual norms. Completeness then follows from something like the triangle inequality: if $\{(u_i, v_i)\}_i$ is a Cauchy sequence, then each of $\{u_i\}_i$ and $\{v_i\}_i$ is too, so both converge, hence the paired-sequence converges to the paired limits.
A: 
Since for $E_1, E_2$ at least two dimensional with bases $\{u_1, u_2\}, \{v_1, v_2\}$ there is a very clear counterexample....
$(u_1, v_1), (u_2, v_1), (u_1, v_2), (u_2, v_2) $ are elements of $E_1 \times E_2$.
But then $(u_2, v_1) + (u_1, v_2) + (u_2, v_2) $ is not of the form $(u, v)$ and so not in $E_1 \times E_2$, i.e. $E_1 \times E_2$ is not algebraically closed under addition.

The sum you gave is
$$
(u_2, v_1) + (u_1, v_2) + (u_2, v_2)
=
(u_2+u_1+u_2,v_1+v_2+v_2)
$$
and this is an element of $E_1\times E_2$ because $u_2+u_1+u_2\in E_1$ and $v_1+v_2+v_2\in E_2$.
The vector is indeed of the form $(u,v)$, where $u=u_2+u_1+u_2$ and $v=v_1+v_2+v_2$.
This has nothing do to with Banach spaces.
This is all about the concept of a product of two vector spaces as another vector space.
A: A Banach space is a normed space which is complete, i.e. every Cauchy sequence w.r.t. the space norm is convergent. The fact that there exist bilinear forms on $E_1 \times E_2$ which are not pure tensors has nothing to do with $E_1 \times E_2$ being a Banach space.
