Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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...

share|improve this question
    
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
1  
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

1 Answer 1

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).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.