Are arithmetic operations open maps? Viewing $+,-,\times , \div$ as function $R \times R \rightarrow R$,
 I 'proved' they are open maps by showing that for each function, the images of elements of basis for box topology (of form $(a,b) \times (c,d)$) are open.
I need verification if it is right (and right approach) and whether there is some generalization or theorem concerning this.
 A: Yes, the operations you list are all open maps. Proving that the image of each element of a basis of the domain maps to an open set is sufficient, since every open set is a union of the basis elements, so the image of any open set is a union of the images of basis elements, each of which was open.
I don't think there's any better or more general approach than handling each individually. The only other way I can think of, is to note that maps $$f_1(a,b)=(a+b,b)$$
$$f_2(a,b)=(a-b,b)$$
have that $f_1$ and $f_2$ are continuous functions and inverses of each other. Thus, they are both homeomorphisms are therefore are open maps. However, if you compose either with the projection $(a,b)\mapsto a$ (which is clearly open), you get the addition and subtraction maps respectively, showing they are open.
For multiplication, you could work with
$$g_1(a,b)=(ab,b)$$
$$g_2(a,b)=(a/b,b)$$
which are continuous except at $b=0$. You at least get that division is a homeomorphism on its domain this way, and you can, noting that multiplication is commutative, get that multiplication is an open map on $\mathbb R\times \mathbb R\setminus \{(0,0)\}$. It's not clear to me that there is any way to escape checking manually that the multiplication map takes open neighborhoods of the origin to open sets.
