Prove dual cone is closed

Question

I have difficulties understanding this proof:

Dual cone of $C$ is defined to be

$$C^*=\left\{ y \in X^*: \langle y,x\rangle \geq 0 , \forall x \in C > \right\}$$

For each $x$, $\left\{y :\langle y,x \rangle \geq 0\right\}$ is a closed half space. Hence $C^*$ is an intersection of closed half-spaces.

My problem is with this part:

Hence $C^*$ is an intersection of closed half-spaces.

How can we say "for each x, $\left\{y:\langle y,x \rangle \geq 0\right\}$" is an intersection?
I'm new to convex analysis and will appreciate any explanations for this part .

Your problem seems to be with basic set theory rather than convex analysis. — geetha290krm, Dec 23 '22 at 09:58
@geetha290krm Maybe you are right, but please note that I've never seen this conversion before. You would call it a lack of experience and I'm here to correct my mistakes — Arian Ghasemi, Dec 23 '22 at 10:03
@geetha290krm and I never accept something until I fully understand it — Arian Ghasemi, Dec 23 '22 at 10:11

Anne Bauval · Accepted Answer · 2022-12-23T09:57:45.067

1

$$C^*=\bigcap_{x\in C}\left\{y\in X^*:\langle y,x \rangle \geq 0\right\}.$$

More generally, $$\left\{y\in A:\forall x\in B\quad P(x,y)\right\}=\bigcap_{x\in B}\left\{y\in A:P(x,y)\right\}.$$

This is because an element $y$ of $A$ belongs to the LHS iff $\forall x\in B\quad P(x,y)$, and to the RHS iff... same condition!

edited Dec 23 '22 at 09:57

answered Dec 23 '22 at 09:55

Anne Bauval

34,650

1

Finally, thanks! – Arian Ghasemi Dec 23 '22 at 10:07

Prove dual cone is closed

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