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On page 51, the authors applied the separating hyperplane theorem to the sets ${x_0}$ and the interior of $C$ to prove the supporting hyperplanes theorem (assuming the interior of $C$ is nonempty). However, the original claim of the theorem is applied to the set $C$ instead of its interior. I can envision the use of following limit argument: for each point $x$ in $C$ (and its boundary), there exists a sequence ${x_k}$ in its interior that converges to $x$. Then the proof follows readily by taking such limit.

Is there a quick and rigorous way to show the above argument? It seems to me that convexity has to play some role. Otherwise, I can easily construct a non-convex set $C$ such that the above argument fails.

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The interior of a convex set is either empty or dense in the set. See Closure of interior and interior of closure in a topological vector space.

In our case, the interior of $C$ is dense in $C$. Therefore, if a continuous function $f:\mathbb R^n\to \mathbb R$ satisfies $f\le 0$ on the interior of $C$, this inequality also holds on $C$. In particular, this applies to linear functions.

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