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 $A$.
 A: 
Firstly, what do they mean by taking inverses $f^{-1}$

$f^{-1}$ does not always refer to the inverse of a function. For any function $f:A\to B$, and any set $C\subseteq B$, the set $f^{-1}(C)$ (called the "preimage" of $C$" is defined as $$\{a\in A| f(a)\in C\}$$
this definition is valid even if $f$ is not invertible.

And also i am doubt wheter we can do the same thing for the set A

Sort of. You can define $A_1=\{(x,y)| x>0\}$ and $A_2=\{(x,y)|0<x^2+y^2<5\}$.
Then, if you define $f_1$ as $f_1(x,y)=x$, and $f_2$ as $f_2(x,y)=x^2+y^2$, then it is relatively easy to show:

*

*$A_1=f^{-1}((0,\infty))$, therefore, $A_1$ is open.

*$A_2=f^{-1}((0,5))$, therefore, $A_2$ is open.

*$A=A_1\cap A_2$, therefore, $A$ is open.

A: As comments have suggested, the method used to demonstrate that the set is open depends on what methods that either:

*

*you have been taught are valid

*you are comfortable using.

Let $S$ denote the set of all $(x,y)$ where:

*

*$x > 0$

*$x^2 + y^2 < 5$.

Then, the most basic method is to take any element $s$ in $S$ and explicitly construct the neighborhood $N$ around this element $s$ such that $N \subseteq S.$
Here, I am assuming that you are accustomed to regarding an open circle of a fixed positive radius $R$ around a point $(x_1,y_1)$ as a neighborhood $N$ around the point $(x_1,y_1)$.
So: for any $(x_1,y_1)$ in $S$:
Let $a = (x_1/2)$. 
Choose $b$ to be any (small) positive real number such that 
$\displaystyle \frac{5 - x_1^2 - y_1^2}{2} + x_1^2 > (x_1 + b)^2$ 
and 
$\displaystyle \frac{5 - x_1^2 - y_1^2}{2} + y_1^2 > (y_1 + b)^2$.
This will imply that $(x_1 + b)^2 + (y_1 + b)^2 < 5.$
Let $R = \min(a,b) \implies R > 0$.
Now, consider the subset of $S$ given by 
the set $T$ equal to the set
$x_1 - R < x < x_1 + R$ and $y_1 - R < y < y_1 + R.$
Then, the set $T$ is clearly a rectangular shape that is a subset of $S$.
So, consider the open circle, of radius $R$ around the point $(x_1,y_1)$.  This open circle is clearly a subset of $T$, which is a subset of $S$.
So, given any point $s = (x_1,y_1) \in S,$ a neighborhood $N$ centered around $(x_1,y_1)$, with a positive radius $R$ has been constructed, such that $N \subseteq S$.
Therefore, any element $s \in S$ is an interior point in $S$.
Therefore, the set $S$ is an open set, because all of its elements are interior points.
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.
