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Suppose C1 and C2 are two cones; I should prove this statement:
$$ C1\subseteq C2\Rightarrow DualCone(C2)\subseteq DualCone(C1)$$ And this is the definition of DualCone in my textbook:$$DualCone(C) = \{y ∈ R^n: 〈y,x〉 ≥ 0, x ∈ C\}$$
My try:
By the hypothesis $C1\subseteq C2$, so we have $x ∈ C1\Rightarrow x∈C2$: $$DualCone(C1) = \{y ∈ R^n: 〈y,x〉 ≥ 0, x ∈ C1\} = \{y ∈ R^n: 〈y,x〉 ≥ 0, x ∈ C2\}$$
So with my conclusion: $$DualCone(C1) = DualCone(C2)$$

Am I wrong?

Asaf Karagila
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Arian Ghasemi
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  • Good! That's all there is to it. – geetha290krm Dec 23 '22 at 12:20
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    No, this isn't right. The equality$${y \in \Bbb{R}^n: \langle y,x \rangle \ge 0, x \in C_1} = {y \in \Bbb{R}^n: \langle y,x \rangle \ge 0, x \in C_2}$$needs to be justified, and is generally false. – Theo Bendit Dec 23 '22 at 12:32
  • @TheoBendit In my proof, this equality holds because $C1\subseteq C2$, so we have $x ∈ C1\Rightarrow x∈C2$. So if I add this would it be correct then? – Arian Ghasemi Dec 23 '22 at 12:36

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Suppose $y \in \operatorname{DualCone}(C_2)$. We wish to show $y \in \operatorname{DualCone}(C_1)$. That is, we wish to show, for all $x \in C_1$, $\langle x, y \rangle \ge 0$. Since $C_1 \subseteq C_2$, every such $x$ will also belong to $C_2$. And since $y \in \operatorname{DualCone}(C_2)$ and $x \in C_2$, we have $\langle x, y \rangle \ge 0$ as required. This is true for any $x \in C_1$, so $y \in \operatorname{DualCone}(C_1)$. And, since this is true for any $y \in \operatorname{DualCone}(C_2)$, we have $\operatorname{DualCone}(C_2) \subseteq \operatorname{DualCone}(C_1)$.

Theo Bendit
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