Partial orders: Prove transitive and irreflective is equivalent to asymmetric

Question

Given set $S$ and binary relation $R \subseteq S \times S$, let:

• $R$ is irreflexive iff $\{(a,a)|a\in S\} \cap R$ is empty
• $R$ is asymmetric iff $(a,b) \in R \implies (b,a) \notin R$
• $R$ is transitive iff $(a,b) \in R \land (b,c) \in R \implies (a,c) \in R$

For any transitive $R$, prove irreflexive is equivalent to asymmetric.

This paper, at Lemma 1.1(iv), says the proof is "obvious." However, it seems trivial to construct a counterexample, e.g., a complete digraph. Or, let:

$S \equiv \{ a,b,c \}$ and $R \equiv \{(a,b),(b,a),(b,c),(c,b),(a,c),(c,a)\}$

Here $R$ appears to me to be transitive, irreflexive but yet symmetric. What am I missing?

Answer 1

Your $R$ is not transitive, because $(a,b)\in R$ and $(b,a)\in R$ but $(a,a)\notin R.$