Show that $A \subset \mathbb{R}^2$ is open 
Show that  $$A=\{ (x,y) \in \mathbb{R}^2: -1< \max\{x,y\} <2 \}$$ is open

First I know that given $(x,y) \in A $ i try to found $r >0$ such that $B((x,y),r) \subset A$ usually these $r$ is give by the conditions of the set in this case $-1< \max\{x,y\} <2 $ and these implies that $1<||(x,y)||_{\infty}<2$ and therefore $ 0<||(x,y)||_{\infty}-1=r_1$ and $0<2- ||(x,y)||_{\infty}=r_2$ then i try to take $\min \{r_1, r_2\}$ but I not sure to these is correct
Any hint or help I will be very grateful
 A: Define $f(x,y) = \max(x,y)$. Hence $f\in C^0$ and $f^{-1}((-1,2))$ is open.
A: $A=\{ (x,y) \in \mathbb{R}^2: -1< \max\{x,y\} <2 \}=\{ (x,y) \in \mathbb{R}^2: \color{grey}(-1< \max\{x,y\}\color{grey}) \text{ and } \color{grey}(\max\{x,y\}<2\color{grey}) \}=\{ (x,y) \in \mathbb{R}^2: -1< \max\{x,y \}\}\cap \{ (x,y) \in \mathbb{R}^2:  \max\{x,y\} <2 \}$

*

*$\forall (x,y)\in \Bbb R ^2, \max\{x,y\}<2 \iff \color{red}{x<2 \text{ and }y<2}$

*$\forall (x,y)\in \Bbb R ^2, -1<\max\{x,y\} \iff x>-1 \color{green}{\text{ or }}y>-1$
Let $\color{red}{S_0=\{(x,y)\in \Bbb R^2:x<2\text{ and }y<2\}}$
Let $S_1=\{(x,y)\in\Bbb R ^2:x>-1\}$ and $S_2=\{(x,y)\in\Bbb R ^2:y>-1\}$
$ A=\color{red}{S_0}\cap(S_1\color{green}{\cup}S_2)=\color{grey}(\color{red}{S_0}\cap S_1\color{grey})\color{green}{\cup}\color{grey}(\color{red}{S_0}\cap S_2\color{grey})$
Proving that $A$ is open then falls under the usual arguments for the topology of $\Bbb R^2$(for example, $S_1$ is open...)

A: Hint:  Notice if $(x,y) \in A$ then $x\le \max(x,y) < 2$ and $y\le \max(x,y) < 2$.
Hint 2:  Notice if $||(a,b),(x,y)|| < \epsilon$ then $|b-y| < \epsilon$ and $|a-x| < \epsilon$.  (why?)
So if you let $\epsilon\le  2-\max(x,y) >0$ then if $(a,b) = B_{\epsilon}$ then $|a-x| < \epsilon$ and $|b-y|<\epsilon$.  Can you conclude that $\max(a,b) < 2$?
And if we let $\epsilon \le \max(x,y) -(-1)> 0$ then if $\max(x,y)= x$ then $|a-x| < \epsilon$ and if $\max(x,y) = y$ then $|b-y| < \epsilon$.  If $|a-x| < \epsilon$ can you conclude $a > -1$.  If $|b-y| < \epsilon$ can you conclude $b > -1$.  And as $a \le \max(a,b)$ and $b \le \max(a,b)$ then we must have $\max (a,b) > -1$.
Now can you see how if we make $\epsilon \le \min(2-\max(x,y), \max(x,y)+1)$ that that will take care of both conditions?
Can you see how we would now be done?
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Let $(x,y) \in C$.  So $-1 < \max(x,y) < 2$.  Let $\epsilon = \min(2-\max(x,y), \max(x,y)-(-1)$.
Claim:  $B_{\epsilon}(x,y) \subset C$.
Pf:  Let $(a,b) \in B_{\epsilon}(x,y)$.  Then $||(a,b),(x,y)|| = \sqrt{(x-a)^2 + (y-b)^2} < \epsilon$.
Either $\max(a,b) = a$ or $\max(a,b) = b$.
Case 1: $\max(a,b) = a$.
$|x-a| = \sqrt{(x-a)^2}\le \sqrt{(x-a)^2 +(y-b)^2} < \epsilon \le 2-\max(x,y)\le 2 - x$
So $x-2 < a - x < 2-x$ so $2x-2 < a < 2$.  SO $a < 2$.
Case 2: $\max(a,b) = b$.  Then by the exact same argument.
$|y-b| = \sqrt{(y-b)^2}\le \sqrt{(x-a)^2 +(y-b)^2} < \epsilon \le 2-\max(x,y)\le 2 - y$
So $y-2 < b - y < 2-y$ so $b < 2$.
So $\max(a,b) < 2$.
Now either $\max(x,y) = x$ or $\max(x,y) =y$.
Case 1: $\max(x,y)=x$
$|x-a| = \sqrt{(x-a)^2} \le \sqrt{(x-a)^2 + (x-b)^2} < \epsilon \le \max(x,y)+1=x+1$.
As $x = \max(x,y) > -1$ we have $x+1 > 0$.  And so $|x-a|< x+1$ so $x-a \le |x-a| < x+1$ so $a > -1$.  So $\max(a,b) \ge a > -1$.
Case 2.  $\max(x,y) = y$
By exact same argument $|y-b| < \epsilon \le y+ 1$ so $y-b \le |y-b| < y+1$ so $\max(a,b) \ge b \ge -1$.
So either way $\max(a,b) > -1$ and $-1 < \max(a,b) < 2$ so $(a,b) \in C$ so $B_{\epsilon}(a,b) \subset C$ and so $C$ is open.
