I'm trying to show that given $C$, a closed set:
C is convex $\Leftrightarrow \forall \{x,y\} \subseteq C: \frac{1}{2}x + \frac{1}{2}y \in C$
But i'm stuck at the $\leftarrow$ part. I'm trying to prove by contradiction:
$C$ is closed, $\forall \{x,y\} \subseteq C: \frac{1}{2}x + \frac{1}{2}y \in C$ and $C$ is not convex.
If $C$ is not convex, $\exists \lambda \in [0,1], x,y \in C: \lambda x+(1-\lambda)y \notin C$
But can't see how to proceed. An alternative approach that i thought was saying that if i take the average point between $x$ and $y$, infinitely(since $\forall \{x,y\} \subseteq C: \frac{1}{2}x + \frac{1}{2}y \in C$) then the whole segment between $x$ and $y$ belongs to $C$, hence $C$ is convex. But i'm not sure this is true(i think that maybe there are gaps between $x$ and $y$).
What's the best approach? Thanks in advance.
Hint: Let $(x,y) \in C^2$, $\tau \in [0,1]$. We want to show that $x + \tau(y-x) \in C$. Let $\epsilon > 0$. Given that $x + j\cdot 2^{-n}(y-x) \in C$ for all $n$ and $j \in \{0,\dots, 2^n\}$ (why ?), we're hoping to find $j_\epsilon$ and $n_\epsilon$ such that:
\begin{align*} || x + \tau(y-x) - (x + j_\epsilon \cdot 2^{-n_\epsilon}(y-x))|| = ||(\tau - j_\epsilon \cdot 2^{-n_\epsilon})(y - x) || < \epsilon. \end{align*}
Such $j_\epsilon$ and $n_\epsilon$ exist as the bisection method reminds us, therefore given any $\epsilon >0$,$ B(x + \tau(y-x), \epsilon) \bigcap C \neq \emptyset$, yielding that $x + \tau(y-x) \in \text{cl}(C) = C$