Is $A=\{ (x,y)|x>0, 0I know, if let say we omit the term "$x>0$" in set A , called it A', we can apply the theorem "a criterion for a continuity mappings in terms of Open sets" we can define $f:R^2 \to R$
$f(x,y)= x^2 + y^2$
then since $f$ is Polynomial then it's continous and we know that $(0,\infty)$ is open set on R so we can conclude that $f^{-1}(0,\infty)$ which is the set A' is open .Firstly, what do they mean by taking inverses $f^{-1}$ , it does not even one to one function for example the image 1 in R can be belongs to $(1,0)$ and $(-1,0)$ in $R^2$.
And also i am doubt wheter we can do the same thing for the set A ? because we have some restriction in there (i.e $x>0$). Can someone explain ? my guess is cannot, since it does not even one to one function for example the image 1 in R can be belongs to $(1,0)$ and $(-1,0)$ in $R^2$ . And $(-1,0)$ is the outside of set $A$.

The function $f(x,y)=x^2+y^2$ is continuous from $\Bbb R^2$ to $\Bbb R.$ And $(0,5)$ is open in $\Bbb R.$ So $B=f^{-1}(0,5)$ is open in $\Bbb R^2.$
The function $g(x,y)=x$ is continuous from $\Bbb R^2$ to $\Bbb R.$ And $\Bbb R^+$ is open in $\Bbb R.$ So $C=g^{-1}(\Bbb R^+)$ is open in $\Bbb R^2.$
So $A=B\cap C$ is the intersection of two open sets, so $A$ is open.
