Let $S = \{1,2,3,4\}. X, Y \in \mathcal{P}(S)$ and R be a relation $R(X,Y): |X \cap Y| = 1$. Is the relation R transitive? I'm reading up on Relations and Functions with the book Discrete Mathematics by G. Chartrand, P. Zhang and one of the exercises there proposes that the relation R over $\mathcal{P}(S)$ where $R(X,Y): |X \cap Y| = 1$ is symmetric only — which I can make sense with since my mental model with this is that $(a, b) \in R$ never occurs because $|X \cap Y| = 1$ is only true when $|X|$ or $|Y|$ is 1, therefore only $(a,a) \in R$, thus the implication for symmetry is vacuously true.
My concern is that with my thinking of how the relation R is symmetric, then this can also mean that it's transitive as well, since $(a,b) \in R \land (b,c) \in R$ never occurs and therefore is vacuously true also.
Is my thinking correct? How is relation R symmetric only and why can't it be transitive as well?
Edit
Okay. I had $(a,a) \in R(X \cap Y, X \cap Y)$ in mind for whatever reason when I was thinking $(a, a) \in R$.
 A: I have absolutely no idea why you think $|X\cap Y|$ only if $|X|$ or $|Y|$ is $1$.
Simple case $X=\{1,2\}$ and $Y=\{2,3\}$ then $X\cap Y = \{2\}|$ and $|X\cap Y| = 1$ and $R(X,Y)$.
Further more if $Z = \{3,4\}$ then $R(Y,Z)$ but $X\cap Z =\emptyset$ so $\not R(X,Z)$ even though $R(X,Y)$ and $R(Y,Z)$.  So not transitive.
....
Any way  $X \cap X = X$ so $R(X,X)$ only if $|X| = 1$.  So if $|X| \ne 1$ we do not have $R(X,X)$ so it is not reflexive.
$X\cap Y=Y\cap X$ so $|X\cap Y| = |Y\cap X|$ so $R(X,Y) \iff R(Y,X)$ so symmetry holds.
Transitivity:  There no way to determine $X\cap Z$ from $X\cap Y$ and $X\cap Z$.  It is certainly possible for $X$ and $Y$ to have only one thing in common and $Y$ and $Z$ to have one thing in common but $X$ and $Z$ to have nothing or many things in common.
Let $X = $ the prime natural numbers.  Let $Y = $ the even natural numbers.  And let $Z=$ then numbers between $11$ through $13$ inclusively.
$X \cap Y= \{2\}$.  $Y\cap Z=\{12\}$ and $X\cap Z = \{11,13\}$.  Not transitive.
