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"?



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 $\overline{C}$.

  • $\begingroup$ Thanks for your answer! I follow everything except the bit about slightly perturbing the $y_i$ while keeping $x$ in the convex hull. It's intuitively clear, but how can one "cleanly" fill in the details for that? $\endgroup$ – dstivd Mar 1 at 16:55
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    $\begingroup$ The fact that $x$ is in the interior of the convex hull of the $y_i$ means that there exists $\lambda_1, ..., \lambda_p$, such that $\sum \lambda_i y_i = x$ and $0 < \lambda_i < 1$, and $\sum \lambda_i = 1$. Then the $\lambda_i$ are solution of a linear system (plus the condition $0 < \lambda_i < 1$), and probably you can show that the solution is continuous with respect to the $y_i$. $\endgroup$ – TheSilverDoe Mar 1 at 17:16

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