# Why is a polar cone a closed set?

Let $$X \subset \mathbb{R}^n$$. We define the polar cone as

$$Xº:=\{x\in\mathbb{R}^n\,|\,\langle u,x\rangle\leq 0,\forall u\in X\}$$

How can I show that this set is closed?

If I fix some $$u\in X$$ then I have that $$\{x\in\mathbb{R}^n\,|\,\langle u,x\rangle\leq 0\}$$ is a closed halfspace; but if $$X$$ is infinite we can't conclude that the intersection of closed sets is also a closed set (as far as we are talking in terms of usual topology).

• What does $u'x$ mean? Mar 17 '19 at 9:39
• probably inter product with $u'$ the tranpose Mar 17 '19 at 9:41
• @JoséCarlosSantos Usual product in $\mathbb{R^n}$. Edited. Mar 17 '19 at 9:43
• why the intersection of closed sets is not a close set? Mar 17 '19 at 9:45
• @dmtri It's done. Mar 17 '19 at 9:49

Actually, your argument works: $$X^0$$ is closed because it can be expressed as an intersection of closed sets.