$A$ is a convex set iff $\overline{A}$ is a convex set

Question

I need to show this claim for a convex set $A \subset \mathbb{R}^n$ ($\overline{A}$ is the closure of $A$). Do you have any idea on how to do this? I think we need to use the fact that $x \in \overline{A}\Leftrightarrow \exists \{x_n\}$ with $x_n \in A$, $\forall n \in \mathbb{N}$ that converges to $x$, but I do not understand how to do it.

Thanks in advance.

Answer 1

• Assume that $A$ is convex. Let $x,y\in \overline{A}$, there are sequences $x_n,y_n$ converging, respectively, to $x,y$. Because $x_n,y_n\in A$, for each $\lambda\in [0,1]$, $$ (1-\lambda)x_n + \lambda y_n = z_n \in A. $$ Clearly $z_n\to (1-\lambda)x + \lambda y$, so that, by closure, $$ (1-\lambda)x + \lambda y \in \overline{A}, \quad \forall \lambda \in [0,1],\ x,y\in \overline{A}. $$
• The opposite implication is false! For istance, $[0,1]$ is convex but $[0,1]\setminus \{1/2\}$ is not, whereas $\overline{[0,1]\setminus \{1/2\}} = [0,1]$.

Answer 2

If $x, y\in\overline{A}$ then show that $(1-t)x+ty\in\overline{A}$ for all $0\leq t\leq 1$.

Since $x, y\in \overline{A}$, there exists sequences $x_n$ and $y_n$ in $A$ such that $x_n\rightarrow x$ and $y_n\rightarrow y$. Consider the sequence $(1-t)x_n+ty_n$. This is a sequence in $A$ and it converges to $(1-t)x+ty$ so it is in $\overline{A}$.