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There are some related problems:
1. dual cone is closed
2. Why is any subspace a convex cone?

Consider a cone $\mathcal{C}(A)$:

$$\mathcal{C}(A) = \{Ax: x\geq 0\}$$

This is a cone generated by the columns of $A$. Since $x\geq 0$, the construction meets the definition of being a convex cone. Therefore $\mathcal{C}(A)$ is a convex cone.

Is it closed?

I think it is without further proof.

However, in the following lecture from Stanford:

http://web.stanford.edu/class/msande310/lecture03.pdf (p.9 and p.11)

In the proof of Farkas lemma, the author cannot guarantee that $\mathcal{C}$ is closed convex before further proof. So the author utilize convergence sequence to prove that.

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sleeve chen
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  • Why do you think it is? – Eric Wofsey Jun 09 '16 at 06:09
  • You should find out what closed means in this context. E.g. are linear combinations of vectors from the cone part of the cone. – mvw Jun 09 '16 at 06:09
  • Why I think it is....this is from the following lecture from Upenn: http://www.cis.upenn.edu/~cis610/convex1-09.pdf Definition 3.2.4 – sleeve chen Jun 09 '16 at 06:11
  • $C$ is a convex cone iff $C$ is closed under positive linear combination – sleeve chen Jun 09 '16 at 06:12
  • This is what confuses me. If it is from definition, we do not have to prove it further. – sleeve chen Jun 09 '16 at 06:14
  • Just a fussy detail: What type of object is $A$? (I presume that is a linear map from a finite dimensional vector space to a finite dimensional vector space. However, if "linear", "finite", and "finite" in that sentence are not correct then we might need to be careful with "closed".) – Eric Towers Jun 09 '16 at 06:43
  • $A\in \mathbb{R}^{m\times n}$ – sleeve chen Jun 09 '16 at 08:09
  • In this context "closed" is used in the topological sense: a subset of a topological space is closed if its complement is an open subset. So there is indeed something to be proved. – Lee Mosher Jun 09 '16 at 12:47

1 Answers1

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The closedness of $\{A\,x \mid x \ge 0\}$ is proved on p. 11 of http://web.stanford.edu/class/msande310/lecture03.pdf.

However, there are convex cones, that are not closed, e.g. $C = (0,\infty)^2 \cup \{(0,0)\}$ is a convex cone in $\mathbb{R}^2$ which is not closed.

gerw
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