Is the non-negative orthant a closed convex set?

Question

I know that non-negative orthant is convex. How can we show that it is a closed convex set or not. I read the definition given for closed convex sets on Wikipedia but it is hard to put that definition into a mathematical formula to show this.

Answer 1

HINT:

You know $\Bbb R^n_+$ is convex, so for $\forall x,y \in \Bbb R^n_+$ we have $\lambda x + (1-\lambda)y \in \Bbb R^n_+,\forall \lambda \in [0,1]$. Therefore, any $x,y \in \Bbb R^n_+$ is also a limit point defined by a sequence of $\lambda_i$.

More specifically, take any component $i \leq n$ of $z \in \Bbb R^n_+$, then $z_i \in [0,\infty)$. Note that $0$ is part of $\Bbb R^n_+$ and is the lower boundary for $[0,\infty)$. Similarly to above, we see that $0$ is a limit point of a decreasing sequence in $[0,\infty)$. Therefore, the boundary point $z_b=(z_1,z_2,...,z_{i-1},0,z_{i+1},...)$ is a limit point and also inside $\Bbb R^n_+$. By symmetry, we can do this for any component of $z$ and for any $z \in \Bbb R^n_+$.

Answer 2

The non-negative orthant can be defined by the following linear matrix inequality (LMI)

$$ \begin{bmatrix} x_1 & 0 & 0 & \dots & 0\\ 0 & x_2 & 0 & \dots & 0\\ 0 & 0 & x_3 & \dots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & 0 & \dots & x_n\end{bmatrix} \succeq {\bf O}_n$$

and, thus, it is a (convex) spectrahedron.

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_{Related: Every convex polyhedron is a spectrahedron}

Answer 3

The non-negative orthant can be defined by a conjunction of (strict) linear inequalities of the form $ {\bf A} {\bf x} \leq {\bf b}$, where ${\bf A} = - {\bf I}$ and ${\bf b} = {\bf 0}$, and, thus, is a (convex) $\mathcal H$-polytope.