Exercise involving the Regularity Axiom

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!

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, $$C\notin A.$$

• 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! – HalfAFoot Mar 16 at 21:04
• 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! – Cameron Buie Mar 16 at 21:09
• 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! – HalfAFoot Mar 16 at 21:34
• Nicely done! I've edited my answer to lead you in that direction, and to pose an alternative to proof by contradiction. – Cameron Buie Mar 16 at 21:35
• We can but try! – Cameron Buie Mar 16 at 21:42