Approximating points in the boundary of a convex set

Question

Let $y$ be a point in the boundary of a convex set $K\subseteq \mathbb{R}^n$. I would like to show that all neighbourhoods of $y$ contain a point in $\text{Ext}(K) = \mathbb{R}^n\setminus \overline{K}$ where $\overline{K}$ denotes the closure of $K$.

Equivalent problem: Prove that if $K$ is convex and $U$ is open then $U\subseteq \overline{K}$ implies $U\subseteq K$.

It follows from the definition of boundary that all neighbourhoods of $y$ contain a point in the complement of $K$. If we drop the convexity assumption then a counterexample exists: take a non-trivial open dense set in $\mathbb{R}$.

I am taking a first course in convex optimization, so I'm looking for an elementary proof.

All neighbourhoods indeed contain a point in the complement. What is the union of the complement and Ext of $K$? — Eemil Wallin, Jan 25 '24 at 12:02
@Eemil Isn't the exterior contained in the complement? I don't see how this helps. — Hernán Ibarra Mejia, Jan 25 '24 at 14:12
I meant it as a food for thought. Another hint: dense sets are not connected, but convex sets are. — Eemil Wallin, Jan 25 '24 at 19:44

score 1 · Answer 1 · answered Jan 26 '24 at 14:47

If $K$ contains an interior point $x$ then consider the line from $x$ through $y$ parametrized as $\phi(t):= x + t(y-x)$. It is easy to see that $\phi(t)$ is an interior point for all $t\in (0,1)$. In addition, $\phi(t) \not \in \bar K$ for $t>1$, as otherwise $\phi(1)=y $ would be an interior point, see this answer

If $K$ has no interior points then it is contained in a $(n-1)$-dimensional hyperplane. Take a vector $e$ that is orthogonal to the hyperplane, then $y+te\not \in \bar K$ for all $t\ne 0$, see this answer

Approximating points in the boundary of a convex set

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