# Prove the relations are an equivalence relation

Let $A$ be the be the standard plane in $\mathbb{R}^2$ and $x$ and $y$ be points.

1. Now in $A$ we define $R$ by $xRy$ if and only if $x$ and $y$ are co-linear. Is $R$ and equivalence relation on $A$.

2. Also let $B$ be the the set of all lines in $A$. Define $R$ in $B$ by $aRb$ if and only if $a$ and $b$ do not intersect.

My attempt for the first part. Letting $x$ and $y$ be points in $A$ since all points lie on the same line as themselves then $xRx$ so the relation is reflexive. Now Suppose $xRy$ then $x$ and $y$ lie on the same line. Since two points determine a unique line then $yRx$ so the relation is symmetric. Now suppose $xRy$ and $yRz$ such that $z$ is also a point. Then $x$ and $y$ lie on the same line and $y$ and $z$ lie on the same line. Therefore $x$ and $y$ determine a unique line say $M$ and $y$ and $z$ also determine a unique line say $N$. Now the intersection of $M$ and $N$ contain $y$ so they are not the same line. Therefore $x$ is not related to $z$ so the relation is not transitive

Let $a,b,c$ be distinct lines in $B$. Since $a$ contains at least two distinct points it is a unique then $aRa$ so the relation is reflexive. Now assume $aRb$ therefore $a \cap b$ is empty. So it follows that $bRa$ therefore the relation is symmetric. Now suppose $aRb$ and $bRc$ then $a\cap b$ and $b \cap c$ is empty implies that $a \cap c$ is empty so $aRc$ so the relation is transitive. Therefore this relation is an equivalence relation.

I'm unsure if this is sufficient I was least sure on the transitivity portion for both relations.

• If $a$ is a line, then $a\cap a\neq\varnothing$, so $(a,a)\notin R$ and thus $R$ cannot be an equivalence relation. – Aweygan Jan 25 '17 at 7:10
• Line $a$ must certainly does intersect with itself! $a \cap a = a$. So the relation is not reflexive. – fleablood Jan 25 '17 at 7:17
• "Since a contains at least two distinct points it is a unique then aRa so the relation is reflexive." I can not parse any meaning out of this. " then a∩b and b∩c is empty implies that a∩c is empty" How on earth does that follow. Let A = {a,b,c} and B= {d,e,f} and C = {g,a,b}. How would $A \cap B= \emptyset$ and $B\cap C = \emptyset$ imply $A\cap C = \emptyset$? More to the point a, b, and c are parallel. So a is either equal to c or parallel to c. It is not transitive as $a = c$ is an option and the relationship is absolutely NOT reflexive. – fleablood Jan 25 '17 at 7:24
• 1) is very good though. – fleablood Jan 25 '17 at 7:24
• A bit confusing that you're using $R$ to signify the given relation in both problems, especially when comments don't specify which one they're referring to... Anyways, in 1., any two points are colinear, which makes this into the trivial equivalence relation, where each point is related to all other points (remember that $x$ and $z$ also define a unique line $L$, independent of $y$). That being said, perhaps they meant "$x$ and $y$ are colinear with the origin"? – Arthur Jan 25 '17 at 7:28