# Show that a set is an affine subset

Let $V$ be a vector space over $\mathbb{F}$ and $S \subseteq V$ a nonempty set.

Theorem: $S$ is an affine subset of $V$ if and only if $$\forall u,v \in S, \forall \lambda \in \mathbb{F}, \lambda u + (1-\lambda) v \in S$$

The only if direction is easy, but the converse is not. Let $v_0 \in S$ and define $W = \{w \in V\mid w + v_0 \in S\}$. We wish to show that $W$ is a subspace of $V$. Trivially $0 \in W$. If $u',v' \in W,\lambda \in \mathbb{F}$, there exist $u,v \in S\mid u' = u - v_0, v' = v - v_0$, and \begin{align*} u' + \lambda v' &= u - v_0 + \lambda v - \lambda v_0 \\ &= \lambda u - v_0 + \lambda v + (1 - \lambda) v_0 - v_0 - \lambda u + u \\ &= \underbrace{\lambda u + (1 - \lambda)v_0}_{\text{in } S} - v_0 - v_0 + \lambda v + u - \lambda u \\ &= \underbrace{\lambda u + (1 - \lambda)v_0}_{\text{in } S} - v_0 - v_0 + \underbrace{\lambda v + (1 - \lambda) u}_{\text{in } S} \end{align*} How do I show that this is in the form of $w' - v_0, w' \in S$?

We have a claim that $W$ is a subspace That is for $u',\ v'\in W$ we must show that $u'+cv'\in W$
(1) Hence $$v_0\in S, \ v_0+v'\in S$$
Hence $$cv' +v_0=c(v_0+v') + (1-c) v_0 \in S$$ so that $$cv'\in W$$
(2) Consider $$u'+v_0,\ cv'+v_0 \in S$$
That is $$(1-d)(u'+v_0) +d(cv' + v_0) \in S$$ If $d=1/2$ then $$1/2 (u' +cv') + v_0 \in S \Rightarrow 1/2(u'+cv')\in W$$
• What if $1/2 \not\in \mathbb{F}$? – Henricus V. Mar 1 '16 at 23:58
• $(1-d)u' + dcv' \in W$ so that $u' + \frac{dc}{1-d}v'\in W$ If we redefine $c$ to be $\frac{dc}{1-d}$, then we proved. – HK Lee Mar 27 '16 at 8:25