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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.

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    $\begingroup$ Affine sets are certainly not elements of an affine space. They are often defined as certain subsets of an affine space. The question is not meaningful without reference to a specific definition of "affine set", though. $\endgroup$ – Eric Wofsey May 12 '18 at 21:16
  • $\begingroup$ @EricWofsey My apologies, I meant to write subsets instead of elements. By affine set, I mean a set of points where a line through any two distinct points in the set is also in the set. $\endgroup$ – The Pointer May 12 '18 at 21:20
  • $\begingroup$ But what do you mean by "point" and "line"? Where do your affine sets live? Or is your question what the full definition of "affine set" is? $\endgroup$ – Eric Wofsey May 12 '18 at 21:52
  • $\begingroup$ @EricWofsey See Page 21, 2.1.2 of Convex Optimisation by Boyd and Vandenberghe. That is the text that I am currently studying. $\endgroup$ – The Pointer May 12 '18 at 21:59
  • $\begingroup$ For future questions remember that the reader cannot understand concepts which you do not properly explain (unless they are really standard concepts, which is not the case for affine sets). $\endgroup$ – Paul Frost Aug 7 '18 at 12:47
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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 $w_1 + w_2 \in W$.

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