# Euclidean space: $k$ points in $\mathbb{R}^n$

Consider $$k$$ points $$p_1,\dots,p_k$$ in $$\mathbb{R}^n$$. Then, $$\forall i,j=1,\dots,k: p_i + \operatorname{span}(p_1-p_i,\dots,p_k-p_i) = p_j + \operatorname{span}(p_1-p_j,\dots,p_k-p_j).$$ Assume that $$D$$ is an affine space with $$p_1,\dots,p_k \in D$$, then $$p_1 + \operatorname{span}(p_2-p_1,\dots,p_k-p_1) \subseteq D$$.

Proof: Since $$p_l - p_j = (p_l - p_i)-(p_j-p_i)$$ for all $$l,i,j=1,\dots,k$$, we have that $$\underline{\operatorname{span}(p_1-p_i,\dots,p_k-p_i) = \operatorname{span}(p_1-p_j,\dots,p_k-p_j)}$$.

From $$\underline{p_i-p_j \in \operatorname{span}(p_1-p_j,\dots,p_k-p_j)}$$ follows that $$\underline{p_i + \operatorname{span}(p_1-p_i,\dots,p_k-p_i) = p_j + \operatorname{span}(p_1-p_j,\dots,p_k-p_j)}$$.

Now assume $$p_1,\dots,p_k \in D$$, then $$D = p_1 + D_0$$. This implies that $$p_2-p_1,\dots,p_k-p_1 \in D_0$$, and therefor $$\operatorname{span}(p_2-p_1,\dots,p_k-p_1) \subseteq D_0$$. This completes the proof.

(1) By the observation in the first line ("Since …"): $$p_1-p_j=(p_1-p_i)-(p_j-p_i)\in\operatorname{span}(p_1-p_i,\ldots,p_k-p_i),$$ and similarly for $$p_2-p_j,\ldots,p_k-p_j$$. Since all these vectors turn out to lie in $$\operatorname{span}(p_1-p_i,\ldots,p_k-p_i)$$, we have demonstrated that $$\operatorname{span}(p_1-p_j,\ldots,p_k-p_j)\subseteq\operatorname{span}(p_1-p_i,\ldots,p_k-p_i).$$ By reversing their roles, we can show that the opposite inclusion is also true. Therefore, the two subspaces are equal.
(2) $$p_i-p_j\in\operatorname{span}(p_1-p_j,\dots,p_k-p_j)$$ by definition, since it's one of these vectors whose span we're taking. From the previous part, we already know that $$\operatorname{span}(p_1-p_j,\ldots,p_k-p_j)=\operatorname{span}(p_1-p_i,\ldots,p_k-p_i)=A,$$ where $$A$$ is just a name I want to give to this subspace for brevity. Then the second property simply states that $$p_i+A=p_j+A$$, which is true precisely because $$p_i-p_j\in A$$. In a bit more detail: $$p_i-p_j\in A \implies (p_i-p_j)+A=A \implies p_i+A=p_j+A.$$