Show that an open rectangle is an open set

Question

Show that the open rectangle $\square=(a,b)\times(c,d)$ is open in $\Bbb R^2$.

I can see that if we took any point $(x,y) \in \square$ with $\epsilon < \min\{|x-a|,|x-b|,|y-c|,|y-d|\}$ then the ball $B_\epsilon (x,y) \subset \square$. But I can't quite prove it

Answer 1

Let $(z_1, z_2) \in B_\epsilon (x,y)$ which means

$$\sqrt{(z_1-x)^2+(z_2-y)^2} < \epsilon$$

Let me show that $z_1 > a$.

$$|z_1-x|\leq\sqrt{(z_1-x)^2+(z_2-y)^2} < \epsilon \leq x-a$$

$$a-x<z_1-x < x-a$$

$$\color{blue}{a<z_1}<2x-a$$

I will leave the checking of other boundaries as an exercise.