Antipodal generic points in a central relative ball

If $E=0$, then $u=0$, and there are no proper linear hyperplanes in $E$, so ${\mathcal{H}}^J$ is empty and $v=v^{\mathrm{\prime }}=u$ has the required properties. Suppose $\dim E>0$. By [F2], the relative hyperplanes meeting the ball form a finite set $\mathcal{K}$ and all contain $u$; by [F1], they are proper linear subspaces of $E$. Since $\mathbb{R}$ is infinite, [F3] supplies $h\in E\setminus {\bigcup }_{K\in \mathcal{K}}K$. If $\mathcal{K}$ is nonempty, this $h$ is nonzero because every $K$ contains $0$; if $\mathcal{K}$ is empty, choose any nonzero $h\in E$.

Choose $t>0$ with $t\mathrm{\parallel }h\mathrm{\parallel }<r$ and put $v=u+th$, $v^{\mathrm{\prime }}=u-th$, so both points lie in $B_E(u,r)$. For $K=\ker a\in \mathcal{K}$, [F1] gives a nonzero linear functional $a$, [F2] gives $a(u)=0$, and step 1.1 gives $a(h)\mathrm{\ne }0$. Thus $a(v)=ta(h)$ and $a(v^{\mathrm{\prime }})=-ta(h)$ are nonzero. No other relative hyperplane meets the ball, by the definition of $\mathcal{K}$, so neither point lies on any relative hyperplane. Also $v+v^{\mathrm{\prime }}=2u$ and $v-v^{\mathrm{\prime }}=2th\mathrm{\ne }0$. The segment parametrization in the Statement therefore gives $u=(1-1\mathrm{/}2)v+(1\mathrm{/}2)v^{\mathrm{\prime }}$ with $0<1\mathrm{/}2<1$, placing $u$ in its stated relative interior.