Property of closure of convex sets

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!

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 $$\overline{C}$$.
• 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? – dstivd Mar 1 at 16:55
• 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 $y_i$. – TheSilverDoe Mar 1 at 17:16