Link to similar post: Set theory relation: irreflexive and transitive
My question is a little bit different from the one above. My question is why we say that transitivity property is satisfied "by default" or "vacuously" for a relation $R$ in a set $X$ when we are not able to find any single pair of elements in $R$ of the forms $(a,b)$ and $(b,c)$ where $a,b,c \in X$ so that we can check for transitivity.
I understand that if we said that $R$ is not transitive, then we reach the conclusion from it that $R$ contains elements such that $(a,b), (b,c) \in R$ but $(a,c)\notin R$ which is, of course, not true since it does not contain any pair of elements in $R$ of the forms $(a,b)$ and $(b,c)$ where $a,b,c \in X$.
But still, doesn't saying that $R$ is transitive imply that it satisfies the condition of transitivity, that is
if $(a,b), (b,c) \in R$
$\implies (a,c)\in R$
$\forall$ $ a,b,c \in X$ ?
Yes the condition must be satisfied, and also the condition is satisfied in this particular case.
Read the condition as:
"for every pair $(a,b),(b,c)\in R$ we must have $(a,c)\in R$"
Is there a pair $(a,b),(b,c)\in R$ for which this is not the case here?
No!
Actually even stronger: we do not even have any pair $(a,b),(b,c)\in R$.
(so there is definitely not a pair $(a,b),(b,c)\in R$ for which it is not true that $(a,c)\in R$)