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I am trying to understand the notion of convex cones. So, here are my questions.

I can understand that the non-negative orthant, $\mathbb{R^n_+}$, defined as $\left\{ (x_1, \ldots, x_n) \in \mathbb{R}^n | x_i \geq 0, i = 1, \ldots,n \right\}$ is a convex cone. But I do not understand what is special about this.

This raises the following questions:

  1. In $\mathbb{R}^2$, is every quadrant a convex cone?

  2. What can we say about the octants in $\mathbb{R}^n$ in general? For example, the non-positive octant in $\mathbb{R}^3$?

I have split the question into two cases to avoid any special cases that might only be valid in $\mathbb{R}^2$.

TShiong
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Gokulakrishnan CANDASSAMY
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    It will be great if people who are downvoting the question (even before any response) could explain why they have done so. I am still looking for an explanation to my questions above – Gokulakrishnan CANDASSAMY May 23 '23 at 15:08
  • The downvotes generally occur before response (if any). I did not, but please have a look at Quick beginner guide for asking a well-received question and avoid "no clue" questions – Anne Bauval May 23 '23 at 15:19
  • If you explicit how you "can understand that the non-negative orthant, $\mathbb{R^n_+}$ [...] is a convex cone", you will be able to answer your two questions. – Anne Bauval May 23 '23 at 16:17
  • @AnneBauval Prof, let me try to answer the two questions: my understanding of a convex cone is that a set that is both convex (for any two points in the set, the line segment between the points is included in the set) and a cone (for every point in the set, non-negative scalar multiples are included in the set). Continuing this reasoning, I am forced to believed that every orthant in $\mathbb{R}^n$ is a convex cone indeed. Am I right in this reasoning? – Gokulakrishnan CANDASSAMY May 23 '23 at 18:19
  • The result is true but where is the reasoning? You only recalled the definitions. – Anne Bauval May 23 '23 at 20:35
  • I wonder why the post has been reopened without having been enriched with your attempts. Never mind, the 2 following posts give (one possible version of) the missing reasoning: https://math.stackexchange.com/questions/2158543 and https://math.stackexchange.com/questions/3512063 – Anne Bauval May 24 '23 at 08:35

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Every orthant in $\mathbb{R}^n$ can be expressed as the intersection of a finite number of closed halfspaces containing the origin. i.e. an orthant in $\mathbb{R}^n$ can be described as the set $\mathcal{S} = \lbrace x \in \mathbb{R}^n \vert a_i^Tx \leq 0, i = 1, \cdots, p\rbrace$.

Clearly, from the above definition, we can see that $\forall \alpha \in \mathbb{R}_+, \alpha x \in \mathcal{S}$. Thus, every orthant is a closed convex cone.

Gokulakrishnan CANDASSAMY
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