(Hopefully, I'm asking question in the right place this time!)
I am trying to do the above exercise question (I don't have enough reputation to post inline picture, please click on "exercise 2") on Set Theory Axioms. I feel like I'm missing something obvious. I thought $T ∈ U ∈ V ∈ T$ leads to $T ∈ T$, but the answer says $T ∉ T$. Please enlighten me!
On
Remember that $a\in b\in c$ does not mean $a\in c$: consider e.g. $a=\emptyset, b=\{\emptyset\}, c=\{\{\emptyset\}\}$. (This is a key difference between "$\in$" and "$\subseteq,$" the latter of which is transitive.)
However, Foundation is strong enough to rule this "deeper self-elementhood" as well. The idea behind applying Foundation to rule out some configuration is to think of "$\in$" as meaning "smaller than," and whip up some set which is "circular." E.g. intuitively we have the following circle here: $$...\in U\in V\in T\in U\in V\in T\in ...$$ Note that already "$U\in V\in T\in U$" describes the whole circle. So our idea for a set with no "$\in$-least element" should just be
This circle itself: namely, $X=\{U, V, T,U\}$, or less redundantly just $X=\{U,V,T\}$.
And this would indeed be a counterexample to Foundation:
The set $X$ has three elements, namely $U,V,T$. And each of these has nonempty intersection with $X$: $T\in U\cap X$, $U\in V\cap X$, and $V\in T\cap X$.
The problem states that $T\in U\in V\in T$ and there are no other membership relations among them. $T\in T$ would be a membership relation among them in addition to $T\in U$, $U\in V$, and $V\in T$, and by hypothesis no such relation exists. For the same reason we know that $T\notin V$ and $V\notin V$ in the fourth part.