I came across this brief statement in my lecture notes that I'm unable to demonstrate to myself:
"In $ZF^{--} +$ Infinity, if $r\subseteq V^2$ is a set-like relation, then for every set $x$ there is a minimal(with respect to inclusion) r-closed set $y$ that contains $x$. Further, this does not hold in $ZF$ without Infinity, and it generally does not hold if $r$ is not set-like."
Note:
(1) $ZF^{--}$ stands for the set of axioms: Existence, Pairing, Union, Extensionality, Separation and Replacement axioms.
(2) a relation is set-like if the collection of all $r$-predecessors of a set is a set.
I am also not certain what "with respect to inclusion" implies:
Does it mean that
(a) $\forall a, a\in y \rightarrow$ "a is not r-closed"?
Or
(b) $\forall a, a\subseteq y \rightarrow$ "a is not r-closed"?
or something else?
The way I understand your question, a set $y$ is $r$-closed if whenever $a\in y$ and $b\ r\ a$ holds then $b\in y$. For a set $a$, let $p(a)=\{b\vert b\ r\ a\}$ deonte the $r$-predecessors of $a$. As $r$ is set-like, $p(a)$ is a set for any $a$. Start with a set $x$. We want to find $x\subseteq y$ such that $y$ is $r$-closed and whenever $x\subseteq z\subsetneq y$, $z$ is not $r$-closed (this is what I understand to be the ''inclusion minimal'' part). Just take $y=\bigcup_{n<\omega} \bigcup p^n[x]$ where $p^0=id$ and $p^{n+1}(a)=p(p^n(a))$. It is not hard to check that $y$ is as desired in $ZF^{--}$.
For the second part: $V_\omega$ is a model of $ZF-Inf$. Consider the relation $r\subseteq\omega^2$ where $m\ r\ n$ iff $m=n+1$. The inculsion least $r$-closed $\{0\}\subseteq y$ would be $\omega$, but $\omega\notin V_\omega$.