# Affine Subspace Confusion

I'm having some trouble deciphering the wording of a problem.

I'm given $V$ a vector space over a field $\mathbb{F}$. Letting $v_1$ and $v_2$ be distinct elements of $V$, define the set $L\subseteq V$: $L=\{rv_1+sv_2 | r,s\in \mathbb{F}, r+s=1\}$.

It's the next part where I can't figure out what they mean.

"Let $X$ be a non-empty subset of $V$ which contains all lines through two distinct elements of $X$."

No idea what this set $X$ is. Once I figure that out, I'm supposed to show that it's a coset of some subspace of $V$. I'm hoping this part will become clearer once I know what $X$ is...

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For example, $V=\Bbb R^2$ and $X=\{(x,y)|y=1\}$ satisfy this condition. But many other choices of $X$ work also. –  Andrew Sep 23 '12 at 17:54
@Andrew: Still confused! Sorry... –  AsinglePANCAKE Sep 23 '12 at 17:59
A line is a subset of $V$ of the form given by $\{rv_1+sv_2|r+s=1\}$. A line goes "through" a collection of points $S$ if every point is on the line, i.e. $S\subseteq L$. For any two distinct points $x,y\in V$, there is a unique line containing both points. (Set $v_1=x,v_2=y$ in the set-builder notation I gave.) The set $X$ is presumed to be any arbitrary subset of $V$ with the property that for any distinct elements $x,y\in X$, $X$ also contains the unique line $L(x,y)$ through the points $x,y$, i.e. $$\forall x,y\in X, x\ne y,\{rx+sy:r,s\in\Bbb F,r+s=1\}\subseteq X .$$ –  anon Sep 23 '12 at 18:03
@anon, so for any two points it contains it contains the line between them? So if there's a point z, not on the line spanned by x and y, X also contains the lines L(x,z), L(y,z) and all the lines connecting points of L(x,y) to points of L(x,z) and so on and so on? –  AsinglePANCAKE Sep 23 '12 at 18:10
If z is not on the line L(x,y), X does not necessarily contain L(x,z) or L(y,z): it does though if X contains z. –  anon Sep 23 '12 at 18:17

By definition the set $L$ in your question consists of all the points on a line. So you may think of $L$ as a line (or the line that passes through the two points $v_1$ and $v_2$).
Hence if you are considering the two points ($v_1$, $v_2$) giving you the line $L$, then a subset $X$ containing all lines (the one line) through the two points, is a subset $X$ containing $L$: $L \subseteq X$.
Note: There might be a bit confusion here since by saying "$X$ contains all lines..." you might be understood as saying that the elements of $X$ are lines. But that would mean that $X$ is not a subset of $V$, so I assumed that you by "$X$ contains all lines..." mean that $X$ contains all the points on the lines (all the points that make up the lines).