Set Theory Axioms [Optional]

Match the name of the set theory axiom to its corresponding statement. Try not to look back at the slides!

\(\forall x \exists p\forall s.s \subseteq x\) IFF \(s \in p\)

If \(S\) is a set and \(P(x)\) is a predicate, then \(\{x \in S \mid P(x)\}\) is a set.

\(\forall x.[x \neq \emptyset\) IMPLIES \(\exists y.y\) is \(\in\)minimal in \(x\)]

\(\forall x [x \in y\) IFF \(x \in z]\) IFF \(\forall x[y \in x\) IFF \(z \in x]\)


Suppose the Foundation Axiom did not hold and there are sets \(T, U, V\) such that \(T \in U \in V \in T\), and there are no other membership relations among them.
For each of the following sets, indicate whether it has an \(\in\)minimal element.
The empty set \(\emptyset\)
The empty set \(\emptyset\) has no elements so it cannot have an \(\in\)minimal element. 
\(T\)
\(T= \{V\}\) is possible, and since \(V \notin V\), the element \(V\) is \(\in\)minimal in \(T\) in this case. But if \(T= \{V,S\}\), where \(V \in S\) and \(S \in V\) for some set \(S\), then \(T\) has no \(\in\)minimal element. 
\(\{T\}\)
\(T\) is an \(\in\)minimal element of \(\{T\}\) because we know that \(T \notin T\). 
\(\{V, T\}\)
\(\{V, T\}\) has the \(\in\)minimal element \(V\), since we know that \(T \notin V\) and \(V \notin V\). 
\(\{U, V, T\}\)
None of the elements in \(\{U, V, T\}\) is \(\in\)minimal because each has one of the others as a member. 
\(\{T, U, V\}\)
\(\{T, U, V\} = \{U, V, T\}\), so the answer is the same as above. 
\(\{\emptyset, T, U, V\}\)
Yes, \(\{\emptyset, T, U, V\}\) has the \(\in\)minimal element \(\emptyset\).
