Show that an "open square" is an open set: Show that $\{(x,y) \in \mathbb{R}^2$ such that $-1How do I prove that an "open" square, centered in the origin is in fact an open set? I've already have this geometrical argument:
Let $S$ denote the square.
Suppose $(x,y) \in S$. Let $\delta = \min \{1 - |x|, 1 - |y|\}$. 
Then, geometrically it is clear that $B_\delta(x,y) \subseteq S$. Hence $S$ is open. 
However, How can I write this down and prove it in a formal matter?
 A: Let $S$ denote the square you wrote in the comments. 
Suppose $(x,y) \in S$. Let $\delta = \min \{1 - |x|, 1 - |y|\}$. 
Then $B_\delta(x,y) \subseteq S$. Hence $S$ is open. 
A: It is enough to show that every point in the open square is the center of some open ball that is included entirely within the open square.  Find the distance from the point in question to the nearest point on the boundary.  Then the ball with that radius will serve.
PS in response to comments: Suppose the distance from $(x,y)$ to the nearest side of the square is $\delta=1-x$.  If the distance from $(u,v)$ to $(x,y)$ is less than $\delta$ then $(u-x)^2+(v-y)^2<\delta^2$.  From that it follows that $(u-x)^2<\delta^2$ so $|u-x|<\delta$, and from that we get $x-\delta<u<x+\delta=x+(1-x)=1$.  Since $u>1$, the first component of the pair $(u,v)$ is such that the point must be strictly to the left of the right boundary of the square.  And the other boundaries are farther away.  Similar arguments handle the cases where one of the three other boundaries is the nearest one.
A: It depends on what an open set is. If you mean that every point inside the square is contained in an open ball that is entirely contained within the square, then you can explicitly give the open ball $B_r(x)$ of radius $r$ centered around $x$ that is contained within the square.
