TITLE: Membership via transitive closure
QUESTION [5 upvotes]: I would like to find a condition which is both necessary and sufficient to $x\in y$, using only $=$, $\cup$, $\cap$, $\emptyset$, and the transitive closure operator.
(A little bit of context may help: I am trying to give a categorical interpretation to the membership relation, given that I have managed to interpret all other operations above.)
My question is: is this even possible? I suspect it might not be, but I have no clue as to how to find a proof of this.
Edit. There is no power-object in said category, which makes it a whole lot harder.

REPLY [4 votes]: It is not possible to define $x \in y$ using only $=$, $\cap$, $\cup$, $\operatorname{tcl}$, and $\varnothing$.
Indeed, if it were possible, then it would certainly be possible to do so when $x = \varnothing$ and $y$ is a non-transitive set. But $\{ \varnothing, y, \operatorname{tcl} y \}$ is already closed under the operations $\cap$, $\cup$, $\operatorname{tcl}$, and $\varnothing$, and since the only binary relation we have available is $=$, any binary relation definable in this language cannot depend on what $y$ actually is. But there are non-transitive sets $y$ where $\varnothing \in y$ and also non-transitive sets $y$ where  $\varnothing \not\in y$.