If $(V, ||\cdot||)$ is a normed vector space and $A \subseteq C$ is compact (and therefore closed since the topology on $V$ induced by $||\cdot||$ is Hausdorff) and convex, and $x \in V \backslash A$, then does there exist an open, convex subset $U$ of $V$ such that $A \subseteq U$ but $x \notin U$?
I'm thinking that there is such a set $U$, if we define $ \varepsilon := \inf_{a \in A} ||a – x||$ (is this infimum necessarily finite?), then does $U := A + \frac 1 2 B(0, \varepsilon)$ satisfy these criteria?
Best Answer
Yes. We can even use (one of) the Hahn-Banach separation theorem(s) to deduce that there is a hyperplane strictly separating these sets. This means we can even do it with weakly open sets.
Yes again. Note that $\|a - x\|$ is bounded below by $0$, hence the infimum is not $-\infty$. So long as $A \neq \emptyset$ (which, presumably, it is), the infimum is not $+\infty$. So, the infimum should be finite.
Yes once more. Every point in $u \in U$ satisfies $\inf_{a \in A} \|u - a\| < \varepsilon / 2$. Indeed, any $u \in U$ can be expressed in the form $a' + \frac{1}{2}b$, where $b \in B(0, \varepsilon)$. In particular, this means, $$\frac{\varepsilon}{2} > \frac{1}{2}\|b\| = \|u - a'\| \ge \inf_{a \in A} \|u - a\|.$$ Note that this does not hold for $x$.