Show that an open rectangle is an open set
Matthew Barrera 
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
$\endgroup$1 Answer
$\begingroup$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.
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