Confused about elements of a set Here is an exercise from this book.

Suppose that $$A \in B$$ and $$B \in C$$ . Does this mean that $$A \in C$$ ?
Give an example to prove that this does NOT always happen (and explain
why your example works). You should be able to give an example where
$$|A| = |B| = |C| = 2$$ .

I could not find an example of 3 sets proving this statement. With my beginner understanding of sets, I think this cannot happen. If the set $b$ is an element of the set $c$ , then the elements of $b$ can never be related to $c$ because $b$ is not a subset of $c$. So $a$ cannot possibly be an element of $c$.
However, from the wording of the question (prove this does NOT always happen), this implies that the statement is usually true but there are cases where this is false. Can anyone help me think of an example where this statement is true or false?
 A: *

*Consider $B=\{A\}$ and $C=\{B\}=\{ \{A\}\}$, then $A\in B, B\in C$ but $A\notin C$.

*Consider $B=\{A\}$ and $C=\{B, A\}=\{A, \{A\}\}$, then $A\in B, B\in C$ and $A\in C$.

A: It can happen that $A\in B$, $B\in C$ and $A\in C$.
Take $A=\emptyset$. In order to have $A\in B$, the minimum requirement is $B=\{\emptyset\}$. In order to have $A\in C$ and $B\in C$, the minimum requirement is that $C=\{\emptyset,\{\emptyset\}\}$.
On the other hand, if you take $A=\emptyset$, $B=\{\emptyset\}$ and $C=\{\{\emptyset\}\}$ you have $A\in B$, $B\in C$, but $A\notin C$, because the only element of $C$ is not the empty set.
Now try out an example with $|A|=|B|=|C|=2$. Hint: take for $A$ your favorite two-element set; then you want $B=\{A,u\}$ where $u\ne A$ and $C=\{B,v\}$. Can you finish?
A: If you want $|A| = |B| = |C| = 2$ then, using $D,E,F,G$ unrelated to each other, you can have $A \in C$ or not as you wish

*

*Let $A=\{D,E\}$ and $B=\{A,F\}$ and $C=\{A,B\}$.
Then $A \in B$ and $B \in C$ and $A \in C$

*Let $A=\{D,E\}$ and $B=\{A,F\}$ and $C=\{B,G\}$.
Then $A \in B$ and $B \in C$ but $A\not \in C$, since $A \not= B$ and $A\not =G$
I would say in a handwaving way that the second case might be seen as more usual than the first
