Proving if a relation is or is not transitive. Let S be a set that contains at least two different elements. Let R be the
relation on $P(S)$, the set of all subsets of S, defined by $(X, Y ) \in R$ if and only if
$X \cap Y = \emptyset$. 
I am trying to prove if this is transitive or not.  I from what I am understanding about transitivity I don't think it is.
Here is my answer right now:
$R$ is not transitive because if $X=\{1,2\}$ and $Y=\{3,4\}$ and $Z=\{2, 6\}$ then $X \cap Y = \emptyset$ and $Y\cap Z = \emptyset$ but $X \cap Z = \{2\}$.
I am not sure if this works as a counter example.
 A: Yes that works, the idea being that just because $X,Y$ are disjoint, and $Y,Z$ are disjoint does not always imply that $X,Z$ are disjoint. 
A: You can find a very simple counterexample using the facts that the relation is symmetric, there are some $X,Y$ such that $X R Y$, and for all $X\ne \emptyset$, not $XRX$. All that's required is for $S\ne \emptyset$, it doesn't need to have two or more elements. Any subset of $S$ bears $R$ to $\emptyset\subseteq S$, in particular $S$ itself does, and vice versa:
$$\begin{align}
S R \emptyset &\quad\text{because $S\cap \emptyset = \emptyset$, and} \\
\emptyset R S &\quad\text{for the same reason};
\end{align}$$ 
However, it's not the case that $S R S$, as $S = S\cap S$ is nonempty.

Your counterexample works, but you should come up with a similar proof using just the data you're given — a set with $\ge 2$ elements (yours has at least $6$).

Perhaps the question(er) meant to disqualify the empty set from the range of the relation. In that case, it is essential to have at least two elements in $S$. Say $a,b\in S$, $a\ne b$. Let $X=Z = \{a\}, Y=\{b\}$. Then $X R Y$ and $Y R Z$, but not $X R Z$.
A: You have the right idea, but introduce extra entities.   You have used six specific elements.
$S$ is given as a set with at least 2 elements.   Any counter example we use to prove a general result about $S$ should come from the smallest case. 
That is: don't use more than two elements; and don't use any particular elements.   Be generic.

We know $S$ contains at least two distinct elements.   Let us label them $a, b$.   As such, $\mathcal P(S)=\{\emptyset,\{a\},\{b\},\{a,b\},\ldots,S\}$
Now $\{a\}\cap \{b\}=\emptyset$ and $\{b\}\cap \{a\}=\emptyset$ but $\{a\}\cap \{a\}\neq \emptyset$.   So the relation on any such $S$ is not transitive.
