Inner product is a function from…to…? Example, if $v,w \in \mathbb{R}^2$, then the inner (dot) product defined by 
$$f(v,w) = \left< v,w \right>$$
is bilinear, so is $f$ a function from $\mathbb{R}^2 \times \mathbb{R}^2 \to \mathbb{R}$ or just $\mathbb{R}^2$? I thought that $\mathbb{R}^2 \times \mathbb{R}^2 = \mathbb{R}^4$? The reason I conjectured the cartesian product is because it takes in two vectors.
My brain is farting tonight, that's why I am asking such a basic question
 A: You have every right to "brain-fart" here; this stuff is confusing.
Yes, you are correct in writing that $f$ is a function from $\Bbb R^2 \times \Bbb R^2 \to \Bbb R$, since we are taking two parameters, each of which is (a vector) in $\Bbb R^2$.
Now, if we were to break this down further, we would note that $\Bbb R^2 = \Bbb R \times \Bbb R$, so that we have
$$
\Bbb R^2 \times \Bbb R^2 = (\Bbb R \times \Bbb R) \times (\Bbb R \times \Bbb R)
$$
and it is natural to identify this with 
$$
\Bbb R^4 = \Bbb R \times \Bbb R \times \Bbb R \times \Bbb R
$$
However, the Cartesian product is not "associative" in this way, and so these to sets are considered distinct.  One consists of pairs of the form $((x_1,x_2),(x_3,x_4))$, and the other consists of quadruples of the form $(x_1,x_2,x_3,x_4)$. Yes, there is a natural connection between them, but they are not quite "identical".
A: An inner product by definition is a function from $ V \times V \to \Bbb F $ where $F$ is the field over which $V$ is defined. 
Hence the function $f$ is a function from $ \Bbb R^2 \times \Bbb R^2 \to \Bbb R $. 
The equation $ \Bbb R^2 \times \Bbb R^2 = \Bbb R^4 $ does not make any sense. The notation $ \Bbb R^2 \times \Bbb R^2 $ denotes the set of all duples $ [(x, y), (x', y')] $ where $ (x, y), (x', y') \in \Bbb R^2 $ that is, $x, y, x', y' \in \Bbb R$
A: $f$ is a function from $\mathbb R^2 \times \mathbb R^2$ to $\mathbb R$.
