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Assume (i). An antipodal map $g:S^n\to S^{n-1}$, followed by the inclusion $S^{n-1}\subset {\mathbb{R}}^n$, would be an antipodal map with no zero. Hence (i) implies (ii). Conversely, if an antipodal $f:S^n\to {\mathbb{R}}^n$ had no zero, thenwould be an antipodal map to $S^{n-1}$. Thus (ii) implies (i).

If $F:B^n\to S^{n-1}$ is antipodal on the boundary, regard $S^n$ as two copies of $B^n$ glued along their boundary. Use $F(x)$ on the upper copy and $-F(-x)$ on the lower copy. The boundary condition makes these definitions agree on the seam, and the resulting map $S^n\to S^{n-1}$ is antipodal. Thus (ii) implies (iii).

Conversely, an antipodal map $g:S^n\to S^{n-1}$ restricted to a closed hemisphere, identified with $B^n$, is antipodal on its equatorial boundary. Hence (iii) implies (ii). The three assertions are equivalent; they are standard forms of the Borsuk-Ulam theorem.

For $v\in S^{n-1}$, orthogonally project every member of ${\mathcal{C}}_i$ onto the oriented line ${\ell }_v$. These projections are compact intervals. They are pairwise intersecting because the original sets are, and pairwise intersecting intervals have a common intersection. Let $m_i(v)$ be the midpoint of that common interval and let $f_i(v)$ be its signed coordinate on ${\ell }_v$. Compactness makes $f_i$ continuous, and reversing the orientation gives

Define the continuous antipodal mapThe Borsuk-Ulam theorem gives $v$ with $F(v)=0$. Write the common value as $t$. The hyperplanemeets every set in every ${\mathcal{C}}_i$, because $t=f_i(v)$ lies in the projection interval of each such set. Thus $H$ is the required common hyperplane transversal.