What is the relationship between affine sets and affine spaces? Specifically, are all affine sets subsets of an affine space?
Rahul's comment in this question seems to imply that this is the case, but I'd appreciate a more concrete, clear, and definitive answer, if possible.
Thank you.
You should first understand the concept of an affine space. See for example
https://en.wikipedia.org/wiki/Affine_space .
To be formal, an affine space is a triple $A = (X,V,\alpha)$ consisting of a set $X$, a (real) vector space $V$ and a function $\alpha : X \times V \to X$ written as $\alpha(x,v) = x + v$ such that various axioms are satisfied. Moreover, there is a fairly obvious notion of an affine subspace of an affine space $X$.
Each vector space $V$ can be regarded as an affine space $A(V) = (V,V,a)$, where $a(x,v)$ is the sum of $x$ and $v$ in $V$. The affine subspaces of $A(V)$ are precisely the sets $x + W = \{x + w \mid w \in W \}$, where $x \in V$ and $W$ is linear subspace of $V$.
Your book defines the concept of an affine set $C \subset \mathbb{R}^n$. Now we can show that $C \subset \mathbb{R}^n$ is an affine set if and only if it is an affine subspace of $\mathbb{R}^n$.
First consider an affine subspace $S = x + W$. Given $x_i = x + w_i \in S$, we see that $\theta x_1 + (1 -\theta) x_2 = \theta x + \theta w_1 + (1 -\theta) x + (1 -\theta) w_2 = x + \theta w_1 + (1 -\theta) w_2 \in S$ since $\theta w_1 + (1 -\theta) w_2 \in W$.
Next consider an affine set $C$. Pick any $x \in C$ and define $W = -x + C$.
Let us first check that $W$ is an affine set. Given $w_i = -x + c_i \in W$, we see that $\theta w_1 + (1 - \theta) w_2 = -\theta x + \theta c_1 - (1 -\theta) x + (1 - \theta) c_2 = -x + \theta c_1 + (1 -\theta) c_2 \in W$ since $\theta c_1 + (1 - \theta) c_2 \in C$.
We now show that $W$ is a linear subspace of $\mathbb{R}^n$; this implies that $C = x + W$ is an affine subspace.
(1) $0 = -x + x \in -x + C = W$.
(2) For all $\theta \in \mathbb{R}$ and all $w \in W$ we have $\theta w = \theta + (1 - \theta)0 \in W$ since $w, 0 \in W$.
(3) For all $w_1, w_2 \in W$ we have $\frac{1}{2}(w_1 + w_2) = \frac{1}{2}w_1 + (1 - \frac{1}{2})w_2 \in W$ so that by (2) also $w_1 + w_2 \in W$.