Is the product of two vector spaces of sequences a vector space? Let $K$ be a field and let $K^\mathbb{N}$ be the $K$-vector space of sequences of elements of $K$. Also, let $A, B$ be two $K$-linear subspaces of $K^\mathbb{N}$.
Is it true that $AB := \{a b : a \in A, b \in B\}$ is a $K$-linear subspace of $K^\mathbb{N}$ ? Here $ab$ denotes the term-wise product of sequences, that is, if $a = (a_1,a_2,\dots)$ and $b=(b_1, b_2, \dots)$ then $ab = (a_1b_1, a_2b_2,\dots)$.
A guess the answer is: "No, it's not" but I haven't find any counterexample. Thanks for any help.
 A: Yeah, this one is tricky.
It is not hard to see that the product contains the zero sequence and that it is closed under scalar multiplication. The tricky part is whether it contains the sum of two such products. And your intuition is correct: it need not. 
To construct an example, at least along the lines I did, we need to find four sequences, $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$, and $\mathbf{d}$ such that:


*

*$\mathbf{a}$ and $\mathbf{b}$ are linearly independent.

*$\mathbf{c}$ and $\mathbf{d}$ are linearly independent.

*$\mathbf{ac}$, $\mathbf{ad}$, $\mathbf{bc}$, and $\mathbf{bd}$ are linearly independent. 


Then we let $A=\mathrm{span}(\mathbf{a},\mathbf{b})$, and $B=\mathrm{span}(\mathbf{c},\mathbf{d})$. 
Every element of $A$ is uniquely expressible in the form $\alpha\mathbf{a}+\beta\mathbf{b}$; every element of $B$ can be uniquely written in the form $\gamma\mathbf{c}+\delta\mathbf{d}$. And so the elements of $AB$ are uniquely expressible in the form
$$(\alpha\gamma)\mathbf{ac} + (\alpha\delta)\mathbf{ad} + (\beta\gamma)\mathbf{bc} + (\beta\delta)\mathbf{bd}.\tag{1}$$
for some real numbers $\alpha,\beta,\gamma\delta$. 
Now, let 
$$\begin{align*}
\mathbf{x} &= \mathbf{bc} \in AB\\
\mathbf{y} &= \mathbf{ad} \in AB
\end{align*}$$
The sum of these two is $$\mathbf{x}+\mathbf{y} = \mathbf{ad}+\mathbf{bc}.$$
If this element were in $AB$, then we could express it as in equation (1). But this would require $\alpha\gamma=0$, $\alpha\delta=1$, and $\beta\gamma=1$. The first equation requires $\alpha=0$ or $\gamma=0$, but the other two equations exclude either possibility. Thus, $\mathbf{x}+\mathbf{y}$ cannot lie in $AB$.
So... can we find such sequences $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$, and $\mathbf{d}$? Yes: take $\mathbf{a}$ the indicator function of the multiples of $2$, $\mathbf{b}$ the indicator function of the multiples of $3$, $\mathbf{c}$ the indicator function of the multiples of $5$, and $\mathbf{d}$ the indicator function of the multiples of $7$. Then $\mathbf{a}$ and $\mathbf{b}$ are linearly independent; $\mathbf{c}$ and $\mathbf{d}$ are linearly independent; and the four products correspond to the indicator functions of the multiples of $10$, $14$, $15$, and $21$, which are linearly independent.
This gives an example of two subspaces $A$ and $B$ of the vector spaces of real sequences with the property that $AB$ is not a subspace of the space of real sequences.
