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Sep 20, 2021 at 22:16 comment added Frode Alfson Bjørdal @ZuhairAl-Johar The construction will work without extensionality as well, if the goal is to achieve a transitive closure.
Sep 20, 2021 at 22:15 comment added Frode Alfson Bjørdal @ZuhairAl-Johar I hope the following interests you: mathoverflow.net/questions/404362/…
Sep 17, 2021 at 4:29 comment added Frode Alfson Bjørdal @ZuhairAl-Johar Yes, I was looking for at least one transitive closure for each set. If you may help me state an easy solution, without using the recursion theorem, I would be grateful.
Sep 15, 2021 at 18:04 comment added Zuhair Al-Johar what do you exactly mean by the transitive closure? As long as Extensionality is lost, there is no quarantee that there is a unique transitive closure for a set, for each set there can be many transitive closures and all of them are of course co-extensional. If you meant to prove existence of a transitive closure for each set, then this is easy, its moreorless the same traditional proof with some modification to follow co-extensionality in replacing from $\omega$ to the $n$-unions instead of identity as usual.
Sep 14, 2021 at 19:57 history edited Frode Alfson Bjørdal CC BY-SA 4.0
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Sep 14, 2021 at 19:27 history asked Frode Alfson Bjørdal CC BY-SA 4.0