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I am not sure how to proceed with the following question and would appreciate some help:

Let $A,B$ and $C$ be sets.Further suppose that $A\in B$ and $B \in C$.Using the Regularity Axiom show that $C \notin A$.

My attempt:

Since $A \in B$ and $B \in C$ therefore $C:=${{$A$}} or equivalently {$A$}$\in C$. By the regularity axiom, since $C \neq \emptyset$, then there exists an element $x$ such that $x \in C$ and $x \cap C= \emptyset$. With {$A$} being the only element in $C$,it follows that {$A$}$\cap C= \emptyset$. To prove that $C \notin A$, suppose instead that $C \in A$ and derive a contradiction.

With $C \in A$ and {$A$} $\in C$ then by the pairing axiom the set {$A,C$} exists and hence $A \cap${$A,C$}$=C$ and $C \cap${$A,C$}$=\emptyset$....i do not really know how to proceed...looking at other sets and pairing them up does not yield any contradiction for me...

Help!

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There are a few issues to address.

Since $A \in B$ and $B \in C$ therefore $C:=${{$A$}} or equivalently {$A$}$\in C$.

You seem to be assuming that $A$ is the only element of $B$ and that $B$ is the only element of $C.$ However, this has not been given.

By the regularity axiom, since $C \neq \emptyset$, then there exists an element $x$ such that $x \in C$ and $x \cap C= \emptyset$.

Very true.

With {$A$} being the only element in $C$,it follows that {$A$}$\cap C= \emptyset$.

Well, again, we don't actually know for sure that $\{A\}$ is an element of $C,$ though it is a subset of an element of $C.$

To prove that $C \notin A$, suppose instead that $C \in A$ and derive a contradiction.

Ah! Now that's a good idea!

With $C \in A$ and {$A$} $\in C$ then by the pairing axiom the set {$A,C$} exists and hence $A \cap${$A,C$}$=C$ and $C \cap${$A,C$}$=\emptyset$....i do not really know how to proceed...looking at other sets and pairing them up does not yield any contradiction for me...

Try showing that $\{A,B,C\}$ is a set.

Once that's done, from here, we can use Regularity to get our contradiction. By Regularity, one of $A,B,C$ must have empty intersection with $\{A,B,C\},$ but $C\in A,$ $A\in B,$ and $B\in C,$ so this is not possible.

We could proceed even more easily with a direct proof. We first prove that $\{A,B,C\}$ is a set. Next, since $A\in B$ and $B\in C,$ we must have that $A\cap\{A,B,C\}=\emptyset$ by Regularity, so in particular, $C\notin A.$

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  • $\begingroup$ Thank you for your nice response ! Would the following be correct : Assuming that $C \in A$ and with {$A,B,C$}$\neq \emptyset $ then $A \cap ${A,B,C}$=C$.But by the regularity axiom, there exists an $x \in ${$A,B,C$} with $x \cap ${$A,B,C$}$= \emptyset$...is this right so far? if it is ..what allows me to choose x=A and thereby conclude a contradiction? Thank you! $\endgroup$ – HalfAFoot Mar 16 at 21:04
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    $\begingroup$ You're close. Rather, we can say that $C\in A\cap\{A,B,C\}.$ However, we're not done, yet! We still need to justify that $B\cap\{A,B,C\}$ and $C\cap\{A,B,C\}$ are non-empty, and that $\{A,B,C\}$ is even a set! $\endgroup$ – Cameron Buie Mar 16 at 21:09
  • $\begingroup$ Thank you very much! Applying the pairing axiom on (i) $A$ with itself and (ii) $B$ and $C$ yield the following sets {$A,A$} and {$B,C$}.Collecting these two sets into a family F and applying the union axiom suggests the existence of a set that consists of all elements belonging to at least one set in F hence {$A,B,C$} is assured.Now the following hold : (i)$C \in A \cap$ {$A,B,C$} ; (ii) $A \in B \cap$ {$A,B,C$} ; (iii) $B \in C \cap$ {$A,B,C$}...By the regularity axiom one of A,B,C must be disjoint from the set {$A,B,C$}...but (i)-(iii) seem to be non empty! $\endgroup$ – HalfAFoot Mar 16 at 21:34
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    $\begingroup$ Nicely done! I've edited my answer to lead you in that direction, and to pose an alternative to proof by contradiction. $\endgroup$ – Cameron Buie Mar 16 at 21:35
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    $\begingroup$ We can but try! $\endgroup$ – Cameron Buie Mar 16 at 21:42

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