Let $S=\{(x,y)\in\mathbb{R}^2:|x|<1 , |y|<1$ Prove S is a open set Good morning, i'm stuck with this exercise.
Let $S=\{(x,y)\in\mathbb{R}^2:|x|<1 , |y|<1$ Prove S is a open set
Definition: $A$ set is open if for all $a∈S$ exists $r>0$ such that $B(a,r)⊂S$
My work:
Suppose $(x,y)\in S$.
Let $r=?$
We need prove $B((x,y),r)\subset S$
Let $(a,b)\in B((x,y),r)$ then $d((a,b),(x,y))<r$
In this step i'm stuck. Can someone help me?
 A: It helps if you draw a picture. The set $S$ is a square, with side length $2$, whose centre is at the origin. You just need $r$ to be the smallest distance from the point $(x, y)$ to a side of the square: either $x = 1$, $y = 1$, $x = -1$, or $y = -1$. So, your $r$ will be the minimum of four possibilities:
$$r = \min \lbrace 1 - y, y + 1, 1 - x, x + 1 \rbrace.$$
You need to convince youself that these are indeed the right formulas for the distance of $(x, y)$ from the four sides. Or, really, you just need to show this particular $r$ works.
A: HINT: You can check that $S=(-1,1)\times (-1,1)$. For a hint on this, think about what $\lvert x\rvert<1$ means on the real axis. 
Now, take $p\in S$. Let $p=(p_1,p_2)$ for $\lvert p_1\rvert<1$ and $\lvert p_2\rvert<1.$ Take $m=\max(\lvert p_1\rvert, \lvert p_2\rvert)$. Set $r=\frac{1-m}{2}$. Finally, verify that $B_r(p)\subset S$. To see this, fix $q\in B_r(p)$ written as $q=(q_1,q_2)$ and see how large the $\lvert q_1\rvert$ and $\lvert q_2\rvert$ can be. This should show that $S $ is open.
