Show that the following subset of $\mathbb{R}^2$ is open Question
I'm trying to show that the set $A=\{ (x,y) \in \mathbb{R}^2 : |x|+|y|<1\}$ is open in $\mathbb{R}^2$.
Attempt
Here is my try, but I'm not getting exactly what I would want:
Let $Q=(x_0,y_0) \in A$ then $|x_0|+|y_0| <1$, let $r=1-|x_0|-|y_0|>0$, let P=$(x,y) \in B(Q;r)$ then:

*

*$|x|-|x_0| \le |x-x_0| \le ||(x-x_0,y-y_0)|| < r = 1-|x_0|-|y_0| \Rightarrow |x| < 1-|y_0|$

*$|y|-|y_0| \le |y-y_0| \le ||(x-x_0,y-y_0)|| < r = 1-|x_0|-|y_0| \Rightarrow |y| < 1-|x_0|$
Adding the last two expressions: $|x|+|y| < 2-|y_0|-|x_0|$
If I get $|x|+|y| < 1-|y_0|-|x_0|$ then the proof is done, but how can I do this?
I'm also aware that the points of $A$ are inside a region bounded by $4$ lines, but I don't know how to approach it analytically to conclude the result.
 A: You are on the right track, you only have to choose a smaller disk: With $r = \frac 12 (1 - |x_0| - |y_0|) > 0$ and $P= (x,y) \in B(Q;r)$ is
$$
 |x| + |y| \le |x_0| + |x-x_0| + |y_0| + |y-y_0| \\
\le |x_0| + |y_0| + 2 \Vert P-Q \Vert < |x_0| + |y_0| +2r = 1
$$
so that $B(Q;r) \subset A$. This proves that $A$ is open.
A: Take $(x_0,y_0)\in A$ and let $r$ be the smallest of the numbers

*

*distance from $(x_0,y_0)$ to the line $x+y=1$ (which is equal to $\frac1{\sqrt2}|x_0+y_0-1|$);

*distance from $(x_0,y_0)$ to the line $x+y=-1$;

*distance from $(x_0,y_0)$ to the line $x-y=1$;

*distance from $(x_0,y_0)$ to the line $x-y=-1$.

Now, take $(x,y)\in B\bigl((x_0,y_0);r\bigr)$. Since $(x,y)$ is closer to $(x_0,y_0)$ than the distance from $(x_0,y_0)$ to the line $x+y=1$, and since $(x_0,y_0)$ is below that line, $(x,y)$ is also below that line. By the same argument, $(x,y)$ is below the line $x-y=-1$ and above the lines $x+y=-1$ and $x-y=1$. In other words, $(x,y)\in A$. This proves that $B\bigl((x_0,y_0);r\bigr)\subset A$. So, $A$ is an open set.
