Interior, Exterior and Boundary proofs in $\mathbb{R}^2$ What's the general form for proving Interior, Exterior and Boundary points in two dimensions? Using the open ball $B(x,r)$.
E.g. for the set $\{(x, y) ∈ \mathbb{R}^2 : |x| ≤ 1, y ≥ 0\}$?
I know it's a "generalisation" of the $\mathbb{R}$ version to $\mathbb{R}^2$, but the cross products are causing some difficulties.
 A: A neighbourhood of $x$ in $\mathbb R^2$ is an open ball centered at $x$ with radius $r$, denoted by $B(x,r)$.
$x_1$ is said to be an interior point of $A$ if $x_1$ has a neighbourhood that is contained in $A$.
In other words, if you want to prove that $x_1$ is an interior point, then you should guess for a radius $r_1$. After that, you should prove that the open ball centered at $x_1$ with radius $r_1$ is contained in $A$.
$x_2$ is an exterior point of $A$, if $x_2$ is an interior point of $A^c$.
$x_3$ is a boundary point if every neighbourhood of $x_3$ has elements both from $A$ and $A^c$.
For your example: For example $(1,3)$ is a boundary point, because whatever $r$ you choose $B\big( (1,3),r \big)$ contains elements from both $A$ and $A^c$.
$(1- \frac r 2 , 3) \in A$ and also this point is in $B\big( (1,3),r \big)$.
However, $(1+ \frac r 2 , 3) \in A^c$ and also this point is in $B\big( (1,3),r \big)$.
So, you can see that whatever $r$ we choose, the open ball around $(1,3)$ with radius $r$ contains elements from both $A$ and $A^c$. Hence $(1,3)$ is a boundary point.
A: $\{(x, y) ∈ \mathbb{R}^2 : |x| < 1, y > 0\}$ are interior point.
$\{(x, y) ∈ \mathbb{R}^2 : |x| = 1, y > 0\}\cup \{(x, y) ∈ \mathbb{R}^2 : |x| \leq 1, y = 0\}$  are boundary point.
exteror points are $\{(x, y) ∈ \mathbb{R}^2 : |x| > 1, y > 0\}\cup \{(x, y) ∈ \mathbb{R}^2 : y<0\}$
