Let $\mathbb{R}^m_+ = \lbrace x \in \mathbb{R}^m : x \geqslant 0 \rbrace.$ What is the boundary set of the set $ \mathbb{R}^m_+$?

Question

Let $ x \in \mathbb{R}^m,$ $ x = [x_i], i = 1,2, \dotsm , m.$ Define $\mathbb{R}^m_+ = \lbrace x \in \mathbb{R}^m : x_i \geqslant 0, 1 \leqslant i \leqslant m \rbrace.$ What the boundary set $ \partial \mathbb{R}^m_+$ of the set $ \mathbb{R}^m_+$? Is not it the set $ \lbrace{ 0\rbrace} \in \mathbb{R}^m_+?$

Answer 1

The definition of boundary I'm assuming is $\partial X = \bar X \smallsetminus X^{\circ}$, where $X^{\circ}$ is the (topological) interior of $X$.

It's not too hard to check that the interior of $\mathbb{R}^m_+$ is the set $$ \{x = (x_1, \dots, x_m) \in \mathbb{R}^m\, :\, x_i > 0\ \text{for all}\ 1 \le i \le m \}$$

This leaves the set of points at least one of whose coordinates is zero as the boundary: any open set around such a point must contain a point with a strictly negative coordinate, which does not lie in $\mathbb{R}^m_+$.

Intuitively, you have an infinite cube one of whose vertices is the origin and the rest of whose vertices are 'at infinity' in the positive directions.