A Set is Affine iff it contains all of its Affine Combinations

Last updated Nov 1, 2022

Let $V$ be a Vector Space over $\mathbb{R}$ and $S \subset V$. $S$ is a Affine Set If and Only If $\forall n \in \mathbb{N}$, $x_{1}, \dots, x_{n} \in S$, all Affine Combinations (of $x_{1}, \cdots, x_{n}$) $\lambda_{1} x_{1} + \cdots + \lambda_{n} x_{n} \in S$.

$(\Leftarrow)$: Let $u, v \in S$. Note that for $\lambda \in \mathbb{R}$, $\lambda u + (1 - \lambda) v = w$ is a Affine Combination of $u,v$ and, thus, $w \in S$. Thus, by definition, $S$ is a Affine Set. $\checkmark$

$(\Rightarrow)$: We proceed by Induction. We skip the trivial case where $n=1$.

Base Case ($n = 2$): Let $x_{1}, x_{2} \in S$ and let $\lambda_{1}, \lambda_{2} \in \mathbb{R}$ so that $\lambda_{1} + \lambda_{2} = 1$. Then $\lambda_{2} = (1- \lambda_{1})$. Since $S$ is a Affine Set, we have that $$S \ni \lambda_{1}x_{1} + (1 - \lambda_{1})x_{2} = \lambda_{1}x_{1} + \lambda_{2}x_{2}$$ establishing the case when $n = 2$.

Inductive Step: Let $n \in \mathbb{N}$ so that $n > 2$, and assume the result holds true for $k \in [n-1]$. Let $\lambda_{1}, \dots, \lambda_{n} \in \mathbb{R}$ so that $\sum\limits_{i=1}^{n} \lambda_{i} = 1$. We must have $\exists j \in [n]$ so that $\lambda_{j} \neq 1$, otherwise $\sum\limits_{i=1}^{n} \lambda_{i} = n \neq 1$.Without Loss of Generality, let $j = 1$ (otherwise, we can just swap them). Then $\sum\limits_{i=2}^{n} \lambda_{i} = 1 - \lambda_{1}$ and $\sum\limits_{i=2}^{n} \frac{\lambda_{i}}{1- \lambda_{1}} = 1$. By Induction we know that $$\frac{\lambda_{2}}{1-\lambda_{1}}x_{2} + \cdots + \frac{\lambda_{n}}{1-\lambda_{1}}x_{n} \in S.$$ Because $S$ is a Affine Set $$\begin{align*} S &\ni \lambda_{1} x_{1} + (1 - \lambda_{1}) \left(\frac{\lambda_{2}}{1-\lambda_{1}}x_{2} + \cdots + \frac{\lambda_{n}}{1-\lambda_{1}}x_{n}\right)\\ &=\lambda_{1}x_{1} + \cdots + \lambda_{n} x_{n} \end{align*}$$ so $S$ is closed under Affine Combinations. $\checkmark \blacksquare$