Property of closure of convex sets

Question

Let $C \subseteq \mathbb{R}^n$ be a convex set. I would like to show that if $x \in \overline{C} \setminus C$ then there exists a sequence $(x_k) \notin \overline{C}$ converging to $x$.

How can I show this "from scratch"?

Thanks!

Answer 1

Let's first prove that $x \notin \mathrm{Int}(\overline{C})$, where $\mathrm{Int}$ denotes the interior. To do this, suppose that $x \in \mathrm{Int}(\overline{C})$ : then there exists a ball $B(x, \varepsilon)$, of center $x$ and radius $\varepsilon > 0$ such that $B(x, \varepsilon) \subset \overline{C}$. Let's choose a certain numbers of points $y_1, ..., y_p$ in this ball, such that $x$ is contained in the interior of the convex hull of the $y_i$. The $y_i$ are all in $\overline{C}$, so you can choose for each $i$ a point $y'_i \in C$, sufficiently near $y_i$, such that $x$ is contained in the convex hull of the $y'_i$. Of course this would imply, by convexity of $C$, that $x \in C$. So this is absurd.

So you have $x \in \overline{C} \setminus \mathrm{Int}(\overline{C})$, i.e. $x$ belongs to the boundary of $\overline{C}$. It is easy now to check that you can approach $x$ by a sequence of points that do not belong to $\overline{C}$.