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Let $K$ be a non-empty closed convex set in a real normed space $X$. Let $x_0$ be a boundary point of $K$ i.e. $x_0 \in \partial K$. Is there a convex set $Y$ disjoint from $K$ such that $x_0$ is also a boundary point of $Y$? Can you ensure $Y$ has an interior point?

Background: I was trying to prove supporting hyperplane theorem for normed spaces and my strategy was to use Hahn-Banach separation theorem on two convex sets ($K$ and $Y$) to obtain a continuous functional separating the sets. Then by the fact $x_0$ lies on the boundary of both sets and by continuity, it must be a maximum or minimum for the functional, giving us the result. Ideally $Y$ would also have an interior point, so I may actually apply the theorem.

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  • $\begingroup$ I suspect this may be easily proven by assuming supporting hyperplane theorem. Guess I really am just looking for a proof of supporting hyperplane theorem. $\endgroup$ – E.Lim Jan 8 '17 at 2:02
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As a simple example suppose $\mathbb{R}$ as a normed space.Let $K=[0,1] $ which is convex and closed and $1\in \partial K$. Now let $Y=(1,2)$ we have also $1\in \partial Y.$

Certainly we can consider $\mathbb{R}^2$ and other important spaces.

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  • $\begingroup$ I mean $K$ as a given closed convex set in a real normed space. $\endgroup$ – E.Lim Jan 7 '17 at 15:15

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