Dense and irreflexive binary relation

Question

Let $R \subseteq A \times A$ be a irreflexlive relation($\forall a \in A: (a,a) \notin R)$. I want to proof that if $R$ is dense, then $A$ cannot be finite. A relation $R$ is dense if $\forall (a,b) \in R$: there is $(a,c) \in R$ and $(c,b) \in R$.
I am not exactly sure where to start on this proof. I tried to go through an example:
Let $(a,b) \in R$, then there have to be $(a,c)$ and $(c,b)$ in $R$ with $a\neq b\neq c$. Because $(a,c)$ is in $R$ we also need to have $(a,d)$ and $(d,c)$ in $R$.
If I continue like this there will be infintely many elements that are in relation with $a$. This will be only be possible if $A$ is not finite. I know this is not a formal proof and probably not even correct.
I just have no idea where to start on this proof and would appreciate any help.

Answer 1

Your idea on how to prove it is correct, though, as you said, it is a bit informal. There are many ways to make if formal. For example you can show using more or less your argument that there exists an injective function from $\mathbb{N}$ to $R[a]=\{b\in A\mid (a,b)\in R\}$. Another idea is the following: Show via strong induction on the size of $A$ that if $A$ is finite then any irreflexive relation on $A$ is not dense.

By the way, notice that the way you give the definition creates a slight problem: The empty relation on any set $A$ is irreflexive and dense. You need to make the extra assumption that $R$ is not empty, i.e. there is some $(a,b)\in R$.