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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).

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    $\begingroup$ What does $u'x$ mean? $\endgroup$ Mar 17 '19 at 9:39
  • $\begingroup$ probably inter product with $u'$ the tranpose $\endgroup$
    – dmtri
    Mar 17 '19 at 9:41
  • $\begingroup$ @JoséCarlosSantos Usual product in $\mathbb{R^n}$. Edited. $\endgroup$
    – Lecter
    Mar 17 '19 at 9:43
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    $\begingroup$ why the intersection of closed sets is not a close set? $\endgroup$
    – dmtri
    Mar 17 '19 at 9:45
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    $\begingroup$ @dmtri It's done. $\endgroup$ Mar 17 '19 at 9:49
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Actually, your argument works: $X^0$ is closed because it can be expressed as an intersection of closed sets.

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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).

Yes you can. One of the axioms of topology is that any union of open sets is open. (This is easy to show directly for the standard topology on a metric space).

Taking complements, you get that any intersection of closed sets is closed.

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