Can a relation with less than 3 elements be considered transitive?
The general rule for a relation $\;\sim\;$ to be a transitive relation on a set $S$, we must have that for all $a, b, c \in S$, with $a, b, c\;$ not necessarily distinct.
- IF $\;a \sim b\;$ AND IF $\;b \sim c,$
- THEN we MUST have that $a \sim c$
If there happens to be less than three elements, then provided reflexivity holds for all $a \in S$, and symmetry holds for all $a, b \in S$, then transitivity follows.
Say we have the relation $R$ denoted by $\sim$ on the set $\{a, b\}$.$\,\,\,$
Then provided $a \sim a$, $b \sim b$, $a \sim b$ AND $b \sim a$, so that $R =\{(a, a),(b, b), (a, b), (b, a)\}$, then $R$ is reflexive, and symmetric, and must therefore be transitive, given there are only two elements.
If $S = \{a\}$, then any relation that is reflexive, i.e. any relation for which $R = \{(a, a)\}$ happens also to be (trivially) symmetric and transitive.
The only time transitivity fails is when there exists a, b, c such that
$a \sim b$ and $b \sim c$, BUT NOT $a \sim c$.
Sometimes it's easier to understand that a relation is transitive, UNLESS there exists a counterexample as described immediately above.
Yes, and yes, respectively. For example, the relation $\{(a,b),(b,a)\}$ is not transitive; its transitive closure is $\{(a,a),(a,b),(b,a),(b,b)\}$. So the useful information you get from knowing that you have a transitive relation containing $(a,b)$ and $(b,a)$ is that you must also have $(a,a)$ and $(b,b)$.