Q18
Proof
Direct Proof of a Stated Identity or Equality
View
Let $T$ be a closed subset of $\mathcal{C}$, such that there exist $x, y \in T$ with $y \notin \{-x, x\}$. We assume that for all $a, b \in T$ with $b \notin \{-a, a\}$, we have that $\dfrac{b-a}{\|b-a\|_2}$ and $\dfrac{b+a}{\|b+a\|_2}$ belong to $T$. Show that $T = \mathcal{C}$.