Characterization of affine space by straight lines

Question

Let $\mathcal{A}$ be an affine space with $\overrightarrow{\mathcal A}$ its vector space.

Show that $\mathcal{F}\subset\mathcal{A}$ is an affine subspace iff $\forall A,B\in \mathcal{A}$, such that $A \neq B$, we have $(AB)\subset \mathcal{A}$

I think the right proposition is equivalent to $\forall C \in (AB)~\exists k\in\mathbb{R}, C=A+k \overrightarrow{AB}$ and thus $C\in \mathcal{A}$. So for the left to right implication we may need to use this latter relation to show that between two point $C$ and $A$ there always exist a straight line.

Answer 1

Here is a brief hint. First, in my notation: $\mathbf{K}$ is an arbitrary field, $\mathbf{V}$ a left $\mathbf{K}$-vector space and $\mathscr{A}$ an affine space whose translation space (I shall use the syntagm "director" space instead) is $\mathrm{Dir}\mathscr{A}=\mathbf{V}$. Given arbitrary vector $x \in \mathbf{V}$ and point $X \in \mathscr{A}$ I will write ${}^{x}X$ for the $x$-translate of $X$. Similarly, given two points $X, Y \in \mathscr{A}$ I will write $\overrightarrow{XY}$ for the vector with origin at $X$ and vertex at $Y$ (i.e. the unique $x \in \mathbf{V}$ such that ${}^{x}X=Y$). I will also employ the notation: $${}^{\mathbf{X}}\mathscr{M}\colon=\{{}^{x}X\}_{\substack{x \in \mathbf{X}\\X \in \mathscr{M}}}$$ for any subsets $\mathbf{X} \subseteq \mathbf{V}$ and $\mathscr{M} \subseteq \mathscr{A}$. For any pair of points $X, Y \in \mathscr{A}$ let $XY$ denote the affine subspace generated by the set $\{X, Y\}$. This subspace is an affine line when $X \neq Y$ -- such that $\mathrm{Dir}(XY)=\mathbf{K}(\overrightarrow{XY})$ -- and is equal to the $0$-dimensional affine subspace $\{X\}$ when $X=Y$. The description: $$XY=\left\{\lambda X+(1_{\mathbf{K}}-\lambda)Y\right\}_{\lambda \in \mathbf{K}}$$ is valid in both the general ($X \neq Y$) as well as the singular ($X=Y$) case.

1. According to my style of definition, a subset $\mathscr{S} \subseteq \mathscr{A}$ is called an affine subspace provided that there exists a vector subspace $\mathbf{U} \leqslant_{\mathbf{K}} \mathbf{V}$ such that ${}^{\mathbf{U}}\mathscr{S} \subseteq \mathscr{S}$ -- in other words $\mathscr{S}$ is stable with respect to the translations given by $\mathbf{U}$ -- and such that the action of $\mathbf{U}$ induced on $\mathscr{S}$ (by restricting the one of $\mathbf{V}$ on $\mathscr{A}$) is regular (i.e. transitive with all elements free, of trivial stabilisers). This is equivalent to asserting that there exists a point $O \in \mathscr{S}$ and a subspace $\mathbf{U} \leqslant_{\mathbf{K}} \mathbf{V}$ such that $\mathscr{S}={}^{\mathbf{U}}O$ and entails the fact that $\mathrm{Dir}\mathscr{S}=\mathbf{U}$ (the director space of $\mathscr{S}$ is precisely $\mathbf{U}$, which exists uniquely satisfying the forementioned conditions).
2. An elementary equivalent characterisation of the assertion ""$\mathscr{S} \subseteq \mathscr{A}$ is an affine subspace" is the conjunction between $\mathscr{S} \neq \varnothing$ and the claim that for any family of scalars $\lambda \in \mathbf{K}^{\left(\mathscr{S}\right)}$ of finite support (which means that the subset of indices $P \in \mathscr{S}$ for which $\lambda_P \neq 0_{\mathbf{K}}$ is finite) and such that $\displaystyle\sum_{P \in \mathscr{S}}\lambda_P=1_{\mathbf{K}}$, the affine combination $\displaystyle\sum_{P \in \mathscr{S}}\lambda_PP \in \mathscr{S}$ is also an element of $\mathscr{S}$. In other words, the affine subspaces are the nonempty subsets closed with respect to affine combinations.
3. Given that $\mathscr{S} \subseteq \mathscr{A}$ is a nonempty set (you don't seem to have that hypothesis) such that $XY \subseteq \mathscr{S}$ whenever $X, Y \in \mathscr{S}$, you can draw the conclusion that $\lambda X+(1_{\mathbf{K}}-\lambda)Y \in \mathscr{S}$ for any $\lambda \in \mathbf{K}$ and $X, Y \in \mathscr{S}$. In general, if the field $\mathbf{K}$ has at least $3$ elements, one can extend this by induction on the cardinality of the support of the family of coefficients to the claim that any affine combination of points of $\mathscr{S}$ remains in $\mathscr{S}$. The converse implication that if $\mathscr{S}$ is an affine subspace then along with every two points it contains the subspace generated by them is trivial, since by definition the affine subspace generated by a nonempty subset $\mathscr{X}$ is the intersection of all the affine subspaces which include $\mathscr{X}$.

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Final remark: while it is true that ultimately every author will adopt his own personal style, I would say that the bracketed notation $(XY)$ is not to be chosen to refer to the line through $\{X, Y\}$, because this would leave you without the very natural and intuitive notation for the open segment of extremities $X$ and $Y$, given an ordered field of scalars $\mathbf{K}$. Similarly for sets of three points or more: $XYZ$ suffices to denote the at most $2$-dimensional subspace generated by $\{X, Y, Z\}$ while $(XYZ)$ should be reserved to refer to the open triangular surface delimited by the triangle $(X, Y, Z)$, given that these points are indeed affinely independent in an affine plane over an ordered field of scalars (in which case the rigorous definition would be: $$(XYZ)=(XY, Z \cap (YX, Z \cap (ZX, Y,$$ where for any line $\mathscr{D}$ and point $P \notin \mathscr{D}$ the syntax "$(\mathscr{D}, P$" denotes the open half-plane given by all the points of the ambient plane $\mathscr{P}$ which are on the same side of line $\mathscr{D}$ as point $P$).