Clarification needed, show it is not an equivalence relation: $a \sim b \Leftrightarrow (a>b \wedge b>a)$ for $a, b \in \mathbb{R}$.

Question

A question from HW:

$a \sim b \Leftrightarrow (a>b \wedge b>a)$ for $a, b \in \mathbb{R}$.

Show it is not an equivalence relation.

My problem - For instance, how can I even check for transitivity? I mean, let $a,b,c \in \mathbb{R}$, then $a \sim b \Leftrightarrow (a>b \wedge b>a)$ and $b \sim c \Leftrightarrow (b>c \wedge c>b)$. Both are empty sets in $\mathbb R$, so how does this imply that $(a>c \wedge c>a)$?

Thanks.

Answer 1

Equivalence relations must satisfy $a \sim a$.

Answer 2

$\sim\, = \emptyset$.

That is, the relation is the empty relation: there are no $a,b\in \mathbb{R}$ such that $a\sim b$. So it can't be reflexive, as a relation on $\mathbb{R}$. (It is vacuously transitive and symmetric!)