How to prove that the set is closed. I was calculating the closure of the set $$A = \{(x,y) \mid 0<x^2 -y^2 \leq 1\}$$
My claim is to prove $\overline{A} = A \cup \{(x,y) \mid |x| = |y|\}$.
I have received a answer here. Maybe I did not understand the answer well.
But, I want to prove first that $B = A \cup \{(x,y) \mid |x| = |y|\}$ is closed. To prove $B$ is closed, I want show $B^c$ is open.
Let $(x,y) \in B^c$, that is $x^2 -y^2 < 0$ or $x^2 -y^2>1$. Then how to prove that there always exists a $r>0$  such that $B_d((x,y), r) \subset B^c$.  If I can prove this, I can conclude that $B^c$ is open.
Please help me.
 A: Let $f(x,y)=x^2-y^2$ and note that $f$ is continuous. If $f(x,y)=\alpha>1$, take $r>0$ such that\begin{align}(x',y')\in B_r\bigl((x,y)\bigr)&\implies|f(x',y')-f(x,y)|<\alpha-1\\&\implies f(x',y')>1\end{align}(since $f(x,y)=\alpha)$. So,$$B_r\bigl((x,y)\bigr)\subset\{(x,y)\in\Bbb R^2|f(x,y)>1\}\subset B^\complement.$$A similar argument shows that if $f(x,y)<0$, then, for some $r>0$, $B_r\bigl((x,y)\bigr)\subset B^\complement$.
A: let us write $\{(x,y)\in\mathbb{R}^2\mid x^2-y^2=0\}$ instead of $\{(x,y)\in\mathbb{R}^2 \mid |x| = |y|\}$, because then we get
$$ B=A\cup \{(x,y)\in\mathbb{R}^2 \mid |x| = |y|\}= A\cup \{(x,y)\in\mathbb{R}^2\mid x^2-y^2=0\} = \{(x,y)\in\mathbb{R}^2 \mid 0\leq x^2-y^2\leq 0\}.$$
Now, if you like to show, that $B$ is closed, observe that $$f\colon\mathbb{R}^2\to\mathbb{R},\ (x,y)\mapsto x^2-y^2$$ is continuous and $[0,1]$ is closed in $\mathbb{R}$. Hence $B=f^{-1}([0,1])$ is closed as the preimage of a closed set under a continuous mapping. Further $B^c=f^{-1}(\mathbb{R}\setminus [0,1])=f^{-1}(]-\infty,0[\ \cup\ ]1,\infty[)$ is open as the preimage of an open set under a continuous mapping.
