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First some definitions.

Definition: A subset $A \subseteq \mathbb{R}^n$ is called an affine set if for every distinct pair of points $x, y \in A$, the line determined by $x$ and $y$ is contained in $A$.

This definition (along with any other's listed here) is taken from Rotman's Introduction to Algebraic Topology

The above definition I assume that "the line determined by $x$ and $y$" means that for any $x, y \in A$ it follows that $tx + (1-t)y \in A$ for all $t \in \mathbb{R}$

Definition: An ordered set of points $\{p_0, p_1, \dots, p_m\} \subseteq \mathbb{R}^n$ is affine independent if $\{p_1 - p_0, p_2 - p_0, \dots p_m - p_0\}$ is a linearly independent subset of $\mathbb{R}^n$.

Now onto my question:

Question: Does every affine set have an affine independent subset?

Now there's a theorem that states any affine set $A = x + S$ for some $x \in \mathbb{R}^n$ and some vector subspace $S \subseteq \mathbb{R}^n$. If $-x \in A$ then $A = x+S = 0+S = S$ in which case $A$ is a vector subspace of $\mathbb{R}^n$.

In this case (when the affine set $A$ is a vector space) I can answer the above question. Since $A = S$ we have that $\dim(A) = \dim(S) = k \leq n$ and thus $A$ contains a linearly independent subset $\{s_1, \dots, s_k\}$, then since $0 \in A$, the set $\{0, s_1, \dots, s_k\}$ is an affinely independent subset of $A$.

But in the case when the affine set in question isn't necessarily a vector space, I'm not sure if there would exist an affine independent subset.

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  • $\begingroup$ You've got the notion of "line" wrong; the $x, y \in A$, it means that $tx + (1-t)y \in A$ for all $t \in \Bbb R$. $\endgroup$ – John Hughes Sep 9 '18 at 16:42
  • $\begingroup$ Ohh whoops @JohnHughes, I'll change that now $\endgroup$ – Perturbative Sep 9 '18 at 16:42
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Every set, affine or not, contains an affine independent set - the empty set. But I am assuming you refer to a maximal affine independent set. Translation by vector is an affine automorphism, therefore if $A$ is independent, so does $x+A:=\{a+x: a \in A\}$. If $S$ is a vector space and $A$ is a maximal affine independent set in $S$, then so does $x+A$. This result is a relatively basic one.

BTW: Proving that every vector space has a basis requires Axiom of Choice.

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