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We have set $A \subseteq \mathbb{R}^n $ and define $A^*$ as $$ A^* := \{s \in \mathbb{R}^n : x^Ts \geq 0, \forall x \in A \}$$

Suppose $A^*$ is nonempty. What are the conditions needed for $A = (A^*)^*$?

(Question actually asks to prove (or disprove) that $A$ is a closed convex cone.)

sedrick
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  • Sketch: The dual of a set is a cone, and the dual of a cone is a closed convex cone. So A** is a closed convex cone, and if A** = A then A is a closed convex cone. Conversely, if A is a closed convex cone, then since A** is the closure of A when A is a cone, and A is closed, we have A** = A. – Ragib Zaman Dec 01 '19 at 05:59
  • @RagibZaman you mentioned that "A** is the closure of A when A is a cone". Why is this so? – sedrick Dec 01 '19 at 06:04
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    That fact should be developed in the text that you are learning from, but you can also find it here: https://math.stackexchange.com/q/28727 – Ragib Zaman Dec 01 '19 at 06:20
  • Got it. Thanks! – sedrick Dec 01 '19 at 06:25

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