Prove the set $S=\{(x,y)\subset \mathbb R^2:x>0,y\geq 0\} $ is Neither Open nor Closed I want to prove the set $S=\{(x_1,x_2)\subset \mathbb R^2:x_1>0,x_2\geq 0\} $  is 
neither open nor closed.
I know that if the compliment $S^c$ is closed then $S$ is open. I also know that the compliment of $S$ is $S=\{(x_1,x_2)\subset \mathbb R^2:x_1\leq 0,x_2< 0\} $ and we want to show that $S \subset B(x_1,x_2)$ but I am so confused as how to do this.
I am actually confused if the compliment of S is open/closed too.. Otherwise I would know how to write my proof..
I am confused by the fact that $:x_1>0$ and $x_2\geq 0$ and perhaps slight bad understanding. I would be extremely grateful for any help. 
 A: Consider point $(1,0) \in S$. Note that if we take the $\varepsilon$-ball around this point, it will always include the point $(1,-\frac{\varepsilon}{2}) \in S^c$. This shows that $S$ is not open.
Now take point $(0,1) \in S^c$. Note that if we take the $\varepsilon$-ball around this point, it will always include the point $(\frac{\varepsilon}{2},1) \in S$. This shows that $S^c$ is not open, so $S$ is not closed.
A: The set is not open because it contains boundary points; those points in the set that lie along the $x_2$ axis cannot be surrounded by an open ball that lies entirely within the set. 
The set is also not closed because it's complement is likewise not open.
A: $S$ cannot be closed because the point $(x,y) = (0,1)$ is a limit point of $S$ not contained in $S$, as every open ball of radius $\epsilon > 0$ around this point necessarily contains a point of $S$ (which is not hard to prove).  Yet, $S$ cannot be open because its complement has a limit point $(x,y) = (1,0)$ which again by the same reasoning as above, is not contained in $S^c$.
