A finite positive-spanning poll set has a positive uniform angle

Given: Positive spanning and directional score Positive-spanning poll sets, Euclidean geometry The Euclidean inner product $⟨x,y⟩={\sum }_{k<n}{x}_k{y}_k$ on ${\mathbb{R}}^n$, compactness and extrema (Heine-Borel in ${\mathbb{R}}^n$: with the Euclidean metric a subset of ${\mathbb{R}}^n$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line, A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value).

For every unit vector $u$, positive spanning writes $-u={\sum }_i{\lambda }_i{d}_i$ with ${\lambda }_i\ge 0$ (Positive-spanning poll sets).

Cauchy–Schwarz gives $\mathrm{\mid }⟨u,d_i⟩\mathrm{\mid }\le \mathrm{\parallel }u{\mathrm{\parallel }}_2\mathrm{\parallel }{d}_i{\mathrm{\parallel }}_2$ (Cauchy-Schwarz $\mid ⟨x,y⟩\mid \le \parallel x{\parallel }_2\parallel y{\parallel }_2$ with its equality case, the triangle inequality for $\parallel \cdot {\parallel }_2$, the parallelogram law and polarisation).

By the coordinatewise vector-sum convention, $⟨u,{\sum }_i{\lambda }_i{d}_i⟩={\sum }_{k<n}{u}_k{\sum }_{j<m}{\lambda }_{j+1}(d_{j+1}{)}_k$. Finite distributivity and reordering the two finite sums, using finite-sum additivity/scaling and induction on their lengths, gives ${\sum }_i{\lambda }_i⟨u,d_i⟩$. A finite sum of nonnegative real terms is nonnegative (Positive-spanning poll sets, The Euclidean inner product $⟨x,y⟩={\sum }_{k<n}{x}_k{y}_k$ on ${\mathbb{R}}^n$, Laws of finite sums and finite products).

If $M={\max }_i\mathrm{\mid }{a}_i-b_i\mathrm{\mid }$ for two nonempty finite lists, then $a_i\le b_i+M\le {\max }_j{b}_j+M$ for each $i$; taking maxima and reversing the lists gives $\mathrm{\mid }{\max }_i{a}_i-{\max }_i{b}_i\mathrm{\mid }\le M$ (Every nonempty finite set of reals has a maximum and a minimum, Maximum and minimum of a set).

The unit sphere is nonempty, closed, and bounded. For $n\ge 1$, the standard vector $e_0$ has Euclidean norm $1$ (The standard list $e:n\to F^n$ with $e_i(i)=1_F$ and $e_i(j)=0_F$ for $j\mathrm{\ne }i$ is an ordered basis of $F^n$; hence ${\dim }_F{F}^n=n$, and $F^0$ is the zero space with basis $\mathrm{\varnothing }$ and dimension $0$, The Euclidean inner product $⟨x,y⟩={\sum }_{k<n}{x}_k{y}_k$ on ${\mathbb{R}}^n$). If $\mathrm{\parallel }x{\mathrm{\parallel }}_2\mathrm{\ne }1$, choose $\epsilon =\mathrm{\mid }\mathrm{\parallel }x{\mathrm{\parallel }}_2-1\mathrm{\mid }\mathrm{/}2$; the reverse triangle inequality shows every $y$ with $d_2(x,y)<\epsilon$ also has $\mathrm{\parallel }y{\mathrm{\parallel }}_2\mathrm{\ne }1$, so the complement is open. Every unit vector lies in $B_2(0,2)$, so the sphere is bounded. Hence it is compact by Heine–Borel (Cauchy-Schwarz $\mid ⟨x,y⟩\mid \le \parallel x{\parallel }_2\parallel y{\parallel }_2$ with its equality case, the triangle inequality for $\parallel \cdot {\parallel }_2$, the parallelogram law and polarisation, Metric space: $d(x,y)=0$ iff $x=y$, symmetry, and the triangle inequality; pseudometric and ultrametric, The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement, Open ball, closed ball and sphere in a metric space, Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space, Heine-Borel in ${\mathbb{R}}^n$: with the Euclidean metric a subset of ${\mathbb{R}}^n$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line).

The reals form a totally ordered field makes $\mathbb{R}$ a totally ordered field, in particular the product of two strictly positive reals is positive; Multiplication by zero: $0\cdot a=0$ gives multiplication by zero. Hence if $a,b\ge 0$, either one is zero and $ab=0$ (using field commutativity if needed), or both are strictly positive and $ab>0$. Thus $ab\ge 0$.