I am working on a problem that states the following:
When is it possible to define the irreflexive closure of a relation $R$, that is, a relation that contains $R$, is irreflexive, and is contained in every irreflexive relation that contains $R$?
Attempt: First note that a relation $R$ defined on a set $A$ is irreflexive if $(a,a) \notin R$ for all $a \in A$.
Suppose $R$ is a relation defined in a set $A$. Let $S$ be an irreflexive relation on $A$ that contains $R$. Then $(a,a)\notin S$ for all $a \in A$ and $R \subseteq S$ which implies that $(a,a) \notin R$ for all $a \in A$. Therefore, $R$ is irreflexive on $A$. Hence, unless $R$ is irreflexive, we can't define irreflexive closure of $R$.
Finally, note that when $R$ is irreflexive, the irreflexive closure of $R$ is $R$ itself.
Question: is my approach correct?
Your answer is perfectly correct! After all, since the closure $S$ must be irreflexive, then $(a,a) \not \in S$ $\forall a \in A$, but that in turn enforces the same restrictions on subsets of $S$, namely the relation we're trying to close. Thus the closure only exists for relations which themselves are already irreflexive.
Mostly just posting this to get this out of the unanswered queue. Posting as Community Wiki in particular since I have nothing further to add.