How to show that some set is open? I know my question will sound stupid, that it should be simple, and I know there are already a lot of questions related to this topic, but I've spent hours on it and I still don't get how to show that a given set is open or not. I'm completely stuck and I don't exactly know how I'm supposed to start and proceed. I already looked for other posts about the topic (like this one, for example, for which I did not manage to understand the answers that were given) but it didn't help.
Here is a set for which I'm supposed to determine if it is open or not:
$$U := \{(x_1 , x_2) \in \mathbb{R}^2 : x_2 > \sqrt{|x_1|} \}$$
What I know (and what I would like to use):


*

*Let $(X,d)$ be a metric space. A set $U \subset X$ is open if, for all $x \in U$, there exists $r>0$ such that $B(x,r) \subset U$.

*For a metric space $(X,d)$ with $x \in X$, an open ball $B(x,r)$ is defined as $\{y \in X : d(x,y) < r \}$

*The metric $d$ is not specified, so I guess it's the Euclidean metric: for $x = (x_1, x_2)$ and $y=(y_1, y_2)$, we have $d(x,y) = \sqrt{(x_1 - y_1)^2+(x_2 - y_2)^2}$.


So, if I understand all this properly, what I'm supposed to show is that, for all $x \in U$, there exists (or not) $r$ such that $B(x,r) \subset U$, i.e. such that the elements $(y_1, y_2)$ of $B(x,r)$ are such that $y_2 > \sqrt{|y_1|}$. And, since these elements are in $B(x,r)$, they are such that $\sqrt{(x_1 - y_1)^2+(x_2 - y_2)^2} < r$.
So, if I get this right, the elements $y$ of $B(x,r)$ should be such that $y_2 > \sqrt{|y_1|}$ and $\sqrt{(x_1 - y_1)^2+(x_2 - y_2)^2} < r$. But then what? I don't understand what I'm supposed to do with all this. How can I "mix" those things together so that I have, at the same time, a convenient $r$  and $y_2 > \sqrt{|y_1|}$?
I'm missing something, but what? Could anyone give me some hints or indications that would help me to get started? I hate to ask for help in such situations, because it looks like I did not even try, but the thing is that I really don't know where to start. Any help would be, therefore, greatly appreciated.
 A: Your region is above the line in the graph below.  Intuitively, any point above the line is some distance away from it.  You can draw a circle of half that radius around the point and stay within your region, so the region is open.  It is a little work to find the shortest distance from a given point to the line.  If your given point is $(x_3,x_4)$, let us assume $x_3 \gt 0$.  The distance to a general point $(x_1,\sqrt{x_1})$on that side of the curve will be $\sqrt{(x_3-x_1)^2+(x_4-\sqrt {x_1})^2}$.  You can minimize this over $x_1 \gt 0$ to find the distance.

A: I've been playing with this and it seems like the shortest distance between a point in $U$ and the curve $y = \sqrt{|x|}$ is a root of a cubic polynomial. If you would like to avoid solving a cubic polynomial, what you can do instead is try to show that the complement of $U$ is a closed set.
That is, if you have a sequence $(x_n,y_n)$ with $y_n \le \sqrt{|x_n|}$ such that $x_n \to x_0, y_n \to y_0$ then show that $y_0 \le \sqrt{|x_0|}$. This works because the square root function is continuous.
A: There are various tools available for showing that  a set is open. 
If  for every $p\in U$ there exists an open set $N(p)$ such that $p\in N(p)\subset U,$ then $$U=\cup \{ \{p\}: p\in U\}\subset \cup \{N(p):p\in U\}\subset U.$$ So $U=\cup \{N(p):p\in U\}$ is a union of a family of open sets, so $U$ is open.
We may also show that $U$ is the intersection of a finite family of  sets, and show that each member of the family is open. This is often useful when $U$ is "complicated".
In your Q  we have $U=A\cap B$ where $A=\{(x,y): y^4>x^2\}$ and $B=\{(x,y): y>0\}.$ So if $A$ and $B$ are open then $U$ is open.
We will show that if $p\in A$ then there exists  $r>0$ such that the open ball $N(p)=B(p,r)\subset A$, and we will do this similarly for the set $B.$
(I)....  For $(x,y)\in A$ let $y^4-x^2=r.$ We have $ r>0.$ There exists $s>0$ such that $y'^4>y^4-r/2$ whenever $|y'-y|<s.$ There exists $t>0$ such that $x'^2<x^2+r/2$ whenever $|x'-x|<t.$ So let $r=\min (s,t).$ 
We have $r>0$. Now whenever $\sqrt {(x'-x)^2+(y'-y)^2}\;<r$, we have $|x'-x|<r\leq s$ and $|y'-y|<r\leq t$. So $y'^4-x'^2>(y^4-r/2)-(x^2+r/2)=0.$ So $A$ is open.
(II).... For $(x,y)\in B$ let $r=y/2.$ We have $r>0.$ If $\sqrt {(x'-x)^2+(y'-y^2)}\;<r$ then $|y'-y|<r$, and we have $(|y'-y|<r=x/2 \land y>0)\implies  y'\geq y-r=y/2>0 $. So $(x',y')\in B.$ So $B$ is open. 
