Let $A= \{(x,y) \in \mathbb{R}^2 : |x|+|y|<2\}$. Show that $\{(x,y) \in \mathbb{R}^2 : |x|+|y|=2\} \subset \partial A$.
As all norm are equivalents in $\mathbb{R}^2$, it is reasonable the use the norm $\| \cdot \|_1$ on our open balls, i.e. $(\mathbb{R}^2,\| \cdot \|_1)$.
So let $(x_0,y_0) \in \mathbb{R}^2$ such that $|x_0|+|y_0|=2.$
We want to show that for each $r>0$, $B((x_0,y_0),r) \cap A \not= \emptyset$ and $B((x_0,y_0),r) \cap \mathbb{R}^2-A \not= \emptyset.$
$r>\|(x,y)-(x_0, y_0)\| = \|(x-x_0, y-y_0)\| = |x-x_0|+ |y-y_0| \geq |x|-|x_0|+|y|-|y_0| = |x|+|y|-(|x_0|+|y_0|) = |x|+|y| - 2.$
I blocked on this problem for a while. Is anyone could help me to solve this problem?
Write $B=\{(x,y)\in\mathbb{R}^2:|x|+|y|=2\}$. We want to show that for all $(x,y)\in B$ all open balls around $(x,y)$ intersect both $A$ and $\mathbb{R}^2\setminus A$.
Given $(x,y)$ with $|x|+|y|=2$, set $(x_n,y_n)=(1-1/n)(x,y)$ for natural numbers $n$. It's pretty obvious that $(x_n,y_n)\to (x,y)$. We have $|x_n|=(1-1/n)|x|$ and $|y_n|=(1-1/n)|y|$ so that $|x_n|+|y_n|<|x|+|y|=2$ and so $(x_n,y_n)\in A$. Every open ball around $(x,y)$ must contain some elements of $(x_n,y_n)$ since they converge to $(x,y)$, and so every such ball intersects $A$.
For $\mathbb{R}^2$, set $(x_n',y_n')=(1+1/n)(x,y)$. They lie in $\mathbb{R}^2\setminus A$ and also converge to $(x,y)$, so every open ball around $(x,y)\in B$ contains some $(x_n', y_n')$, and so intersects $\mathbb{R}^2\setminus A$.