Proof of Finite Sums of the Coordinates of a Point of ℝⁿ

Throughout, a set $u$ lies in $\mathbb{R}^{n}$ exactly when it is a map $u:[n]\to\mathbb{R}$, and then $u_{k}=u(k)$ for $k\in[n]$ (Euclidean Space $\mathbb{R}^n$: n-Tuples of Real Numbers with Their Componentwise Operations (ii)); $\mathbb{R}$, with its addition, multiplication and order, is an ordered field (The Real Numbers Form an Ordered Field in Which Every Nonempty Set Bounded Above Has a Supremum (i)), in particular a commutative ring; and since its addition is associative and commutative, as the addition of a commutative ring, for a map $a:[n]\to\mathbb{R}$ the sum $\sum_{k=1}^{n}a_{k}$ of Finite Sums in a Commutative Ring and in an Ordered Field: Distributivity, Differences, Telescoping, Constant Terms, Comparison and the Triangle Inequality, read over the interval $\{1,\dots,n\}$, equals the finite sum of the statement, both being the sum over the finite set $[n]$ (Natural Numbers as Indices: ℕ ⊆ ℕ₀, 1 ∈ ℕ₀, and the Two Readings of an Operation from 1 to n (v)).

Restriction. Let $t\in\mathbb{R}^{N+1}$, so $t:[N+1]\to\mathbb{R}$. Now $[N]\subseteq[N+1]$ by Intervals of Natural Numbers: Initial Segments, Adding One Element, Splitting and Shifting (v), as $N\le N+1$. So $t|_{[N]}:[N]\to\mathbb{R}$ is a map (a set, being contained in $t$) with $(t|_{[N]})(k)=t(k)$ for every $k\in[N]$, and hence lies in $\mathbb{R}^{N}$, and its components are $(t|_{[N]})_{k}=t_{k}$ for $k\in[N]$. This proves (i).

The first step. Let $t\in\mathbb{R}^{1}$, so $t:[1]\to\mathbb{R}$. The first identity of Iterated Operations: Recursion, Splitting, Reordering, Termwise Combination and Homomorphisms (i), applied to the addition of $\mathbb{R}$ with $n=1$ and $a=t$, gives $\sum_{k=1}^{1}t_{k}=t_{1}$, which is (ii).

The successor step. Let $t\in\mathbb{R}^{N+1}$, so $t:[N+1]\to\mathbb{R}$. The second identity of Iterated Operations: Recursion, Splitting, Reordering, Termwise Combination and Homomorphisms (i), applied to the addition of $\mathbb{R}$ with $n=N$ and $a=t$, gives

$$\sum_{k=1}^{N+1}t_{k}=\Big(\sum_{k=1}^{N}t_{k}\Big)+t_{N+1},$$

where, by Iterated Operations: Finite Sums and Finite Products (ii), the inner sum is the finite sum of $t|_{[N]}$. By (i), $t|_{[N]}\in\mathbb{R}^{N}$ with components $(t|_{[N]})_{k}$, so the inner sum is $\sum_{k=1}^{N}(t|_{[N]})_{k}$, and (iii) follows.

Membership. Let $t\in\mathbb{R}^{N}$, so $t:[N]\to\mathbb{R}$. By Iterated Operations and Operations over Finite Sets Take Values in the Underlying Set (i), applied to the addition of $\mathbb{R}$, $\sum_{k=1}^{N}t_{k}\in\mathbb{R}$; this is (iv).

Scalar multiples. Let $\tau\in\mathbb{R}^{N}$ and $c\in\mathbb{R}$. By Finite Sums in a Commutative Ring and in an Ordered Field: Distributivity, Differences, Telescoping, Constant Terms, Comparison and the Triangle Inequality (i), applied to the commutative ring $\mathbb{R}$ with $a=\tau$,

$$c\sum_{k=1}^{N}\tau_{k}=\sum_{k=1}^{N}c\,\tau_{k}.$$

By Euclidean Space $\mathbb{R}^n$: n-Tuples of Real Numbers with Their Componentwise Operations (iv), $c\tau\in\mathbb{R}^{N}$ and $(c\tau)_{k}=c\,\tau_{k}$ for every $k\in[N]$. The right-hand sum is, by Iterated Operations: Finite Sums and Finite Products (i), the finite sum of the map $k\mapsto c\,\tau_{k}$ on $[N]$ (which is $c\tau$, since both have values $c\,\tau_{k}$), so the right-hand sum is $\sum_{k=1}^{N}(c\tau)_{k}$. This proves (v).

Nonnegative terms. Let $t\in\mathbb{R}^{N}$ with $0\le t_{k}$ for every $k\in[N]$. By Finite Sums in a Commutative Ring and in an Ordered Field: Distributivity, Differences, Telescoping, Constant Terms, Comparison and the Triangle Inequality (vi), applied to the ordered field $\mathbb{R}$ with $a=t$, we get $t_{j}\le\sum_{k=1}^{N}t_{k}$ for every $j\in[N]$, and $0\le\sum_{k=1}^{N}t_{k}$. This proves (vi).