> [!theorem] > > Let $x \in \mathbb{S}^n$, and let $r > 0$. There exists $r' > 0$ such that the open cone > $ > C = \bigcup_{\alpha \in \real \setminus \bracs{0}}\alpha B(x, r') > $ > is contained in > $ > P = \bigcup_{\alpha \in \real \setminus \bracs{0}}\alpha [B(x, r) \cap \mathbb{S}^n] > $ > *Proof*. Let $r'$ such that $1/(1 - r')$ > > Let $y = x + z \in B(x, r')$ with $\norm{z} < r'$, then $\norm{y} \in (1 - r', 1 + r')$. Let > $ > \hat y = \frac{y}{\norm{y}} = \frac{x}{\norm{y}} + \frac{z}{\norm{y}} > $ > then if $r' < 1$, > $ > \norm{\hat y - x} < \abs{1 - \frac{1}{\norm{y}}} + \frac{r'}{\norm{y}} < \abs{1 - \frac{1}{\norm{y}}} + \frac{1}{\norm{y}} > $ > If $r'$ is chosen such that > $ > \abs{1 - \frac{1}{1 \pm r'}} + \frac{1}{1 \pm r'} < r > $ > then $\hat y \in B(x, r) \cap \mathbb{S}^n$ and $y \in P$.