Proof of Small Cases, Reduction, and Membership for Convex Combinations

Sums of real numbers are the finite sums of the field $\mathbb{R}$, and coordinates of convex combinations are as in Convex Combination of Finitely Many Points of $\mathbb{R}^n$. Write $S$ for the successor map of Natural Numbers; by Arithmetic of Addition on the Natural Numbers we have $p+1=S(p)$ for every natural number $p$, so claim 1 of Properties of Finite Sums may be applied in the form $\sum_{k=1}^{p+1}a_k=\sum_{k=1}^{p}a_k+a_{p+1}$. Record that $0\,t=0$ for every $t\in\mathbb{R}$, since $0\,t=(0+0)\,t=0\,t+0\,t$ by distributivity and claim 2 of Additive Cancellation and Elementary Additive Identities in a Field applies.

Claim 1. Let $t$ be a system of convex weights of length $1$. By claim 1 of Properties of Finite Sums, $\sum_{k=1}^{1}t_k=t_1$, so $t_1=1$; and for $j\in[n]$ the same claim gives $\bigl(\sum_{k=1}^{1}t_kx_k\bigr)_j=t_1\,(x_1)_j=(x_1)_j$. Hence $\sum_{k=1}^{1}t_kx_k=x_1$.

Let $u$ be a system of convex weights of length $2$. By claim 1 of Properties of Finite Sums, $1=\sum_{k=1}^{2}u_k=u_1+u_2$, so, by commutativity and associativity of addition together with claims 3 and 4 of Additive Cancellation and Elementary Additive Identities in a Field, $u_2=(u_1+u_2)-u_1=1-u_1$. For $j\in[n]$ the same claim gives

which by Sum of Points of $\mathbb{R}^n$ and Scalar Multiple of a Point of $\mathbb{R}^n$ is the $j$th coordinate of $u_1y_1+u_2y_2$. Hence $\sum_{k=1}^{2}u_ky_k=u_1y_1+(1-u_1)y_2$.

Claim 2. By claim 1 of Properties of Finite Sums, in its restriction part and its recursion part,

so $s=1-t_{N+1}$ as in the computation of $u_2$ above. Since $0\le t_k$ for every $k\in[N+1]$, claim 5 of Properties of Finite Sums gives $0\le s$, and claim 6 of that lemma gives $t_{N+1}\le1$; since $1-s=t_{N+1}$ is nonnegative, claim 3 of Elementary Arithmetic in an Ordered Field gives $s\le1$.

For $j\in[n]$, claim 1 of Properties of Finite Sums also gives

(a) Suppose $t_{N+1}=1$. Then $s=1-1=0$, so claim 5 of Properties of Finite Sums forces $t'_k=0$ for every $k\in[N]$. Every summand of $\sum_{k=1}^{N}t'_k(x'_k)_j$ is then $0\,(x'_k)_j=0$, so that sum equals $0$ by claim 7 of Properties of Finite Sums. By $(*)$, the $j$th coordinate of $\sum_{k=1}^{N+1}t_kx_k$ is $(x_{N+1})_j$ for every $j\in[n]$, that is, $\sum_{k=1}^{N+1}t_kx_k=x_{N+1}$.

(b) Suppose $t_{N+1}\ne1$. With $t_{N+1}\le1$ this gives $t_{N+1}<1$, so $0<1-t_{N+1}=s$ by claim 1 of Elementary Order Arithmetic in an Ordered Field. By claim 7 of Elementary Order Arithmetic in an Ordered Field, $s^{-1}$ exists and $0<s^{-1}$, so $\tau_k=s^{-1}t'_k$ satisfies $0\le\tau_k$ by claim 5 of Elementary Arithmetic in an Ordered Field, and $s\,\tau_k=s\,(s^{-1}t'_k)=t'_k$ by associativity. By claim 3 of Properties of Finite Sums,

so $\tau$ is a system of convex weights of length $N$ and the convex combination $y=\sum_{k=1}^{N}\tau_kx'_k$ is defined. Using $t'_k=s\,\tau_k$ and claim 3 of Properties of Finite Sums again,

Substituting into $(*)$ and using $t_{N+1}=1-s$ shows that the $j$th coordinate of $\sum_{k=1}^{N+1}t_kx_k$ equals $s\,y_j+(1-s)\,(x_{N+1})_j$, which by Sum of Points of $\mathbb{R}^n$ and Scalar Multiple of a Point of $\mathbb{R}^n$ is the $j$th coordinate of $s\,y+(1-s)\,x_{N+1}$. This proves (b).

Claim 3. We use the induction principle for the natural numbers. Let $\mathcal{P}$ be the set of natural numbers $N$ with $1\le N$ such that, for every convex $C\subseteq\mathbb{R}^n$, every $x:[N]\to\mathbb{R}^n$ taking all of its values in $C$ and every system of convex weights $t$ of length $N$, the combination $\sum_{k=1}^{N}t_kx_k$ lies in $C$.

By claim 1 the combination for $N=1$ is $x_1$, which lies in $C$; so $1\in\mathcal{P}$.

Suppose $N\in\mathcal{P}$, and let $C$, a map $x:[N+1]\to\mathbb{R}^n$ with values in $C$, and a system $t$ of convex weights of length $N+1$ be given; adopt the notation of claim 2. If $t_{N+1}=1$, then by claim 2(a) the combination equals $x_{N+1}$, which lies in $C$. Otherwise claim 2(b) applies: $\tau$ is a system of convex weights of length $N$ and the restriction $x'$ takes all of its values in $C$, so $y\in C$ by the induction hypothesis, and

with $0\le s$ and $s\le1$. Since $C$ is convex, this point lies in $C$. Hence $N+1\in\mathcal{P}$, and by induction $\mathcal{P}$ contains every natural number $N$ with $1\le N$.