Open set in the product of two vector spaces? Suppose we have an $R$ vector space $E$, such that $E$ is a topological vector space. Then by definition, we know that the addition $E \times E \to E$ is continuous.
What I'm wondering is how to visualize this? In my book, a function is continuous iff the preimage of open sets are open. 
I can't seem to wrap my head around what an open set in $E \times E$ is supposed to look like. 
So my question is an example of an open set in $E \times E$? And I guess as a follow up: if I were to graph it, what does this mapping look like?
 A: Example: $\Bbb R \times \Bbb R$ as an $\Bbb R$ vector space. 
If $U,V$ are open intervals in $\Bbb R$, then their product $U \times V$ is open in $\Bbb R \times \Bbb R$. 
A more intuitive example (maybe): The open balls of radius $r \in \Bbb R$ around $x \in \Bbb R \times \Bbb R$:
$$
B(x,r) = \{ y \in \Bbb R \times \Bbb R : |x - y| < r \}
$$
are all open. These are easy to visualize. Literally picture a disk around a point $x$ in the plane with radius $r$ that has no boundary. That is, for any point $y$ such that $|x-y| = r$ exactly we can approach $y$ arbitrarily closely using points in $B(x,r)$, but we can never quite hit $y$.
A: Welcome to MSE. You might use pencil and paper, reading this answer, to be able to draw graphs. 
As in the other answer, I will take $E=\Bbb R$ to make visualization easier. 
Yes, if $U,V$ are open in $\Bbb R$ then $U\times V$ is open in $\Bbb R\times\Bbb R$. But you might want to visualize some other open sets, in particular the preimage of an open set under addition. Note that the line $x+y=1$ is the preimage of the singleton $\{1\}$ under addition. (It is a line that could be written as 
$y=-x+1$, slope $-1$ and $y$-intercept $1$.) Now take a neigborhood of $1$, say the open interval $(\frac12,\frac32)$. The preimage is the strip strictly between the lines $y=-x+\frac12$ and $y=-x+\frac32$ (slopes $-1$ and $y$-intercepts $\frac12$ and $\frac32$). Notice that this open strip is not of the form $U\times V$ for any $U$ and $V$, but it is the union of sets of this form, and it illustrates what the preimage under addition of an open interval looks like. 
Re the follow up, if you were to graph addition 
$+:\Bbb R\times\Bbb R\to\Bbb R$ just think of the function $z=f(x,y)=x+y$. Its graph is the plane $z=x+y$ 
or equivalently $z-x-y=0$. Three points that belong to this plane (to make it easier to visualize) are: (i) the origin $(0,0,0)$, (ii) the point $(x,y,z)=(1,0,1)$, 
and (iii) the point $(0,1,1)$ (in the usual $xyz$-coordinate system). 
Think again of the open interval $(\frac12,\frac32)$, think now this interval is on the $z$-axis, and you are trying to find its preimage. Take horizontal planes 
$z=\frac12$ and $z=\frac32$, and intersect the region between these two planes with the graph of $+$, i.e. with the plane $z-x-y=0$, and call the resulting strip on the graph $S$. Then project $S$ onto the $xy$-plane, then you get exactly the strip in the $xy$-plane that, as discussed earlier, is the preimage of $(\frac12,\frac32)$ under addition (strictly between lines $y=-x+\frac12$ and $y=-x+\frac32$ in the $xy$-plane, i.e. in $\Bbb R\times\Bbb R$.)
