1.11 Infinite Sets

Set Theory Axioms [Optional]

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

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

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

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

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

  2. 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.

    1. The empty set \(\emptyset\)

      The empty set \(\emptyset\) has no elements so it cannot have an \(\in\)-minimal element.

    2. \(T\)

      \(T= \{V\}\) 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.

    3. \(\{T\}\)

      \(T\) is an \(\in\)-minimal element of \(\{T\}\) because we know that \(T \notin T\).

    4. \(\{V, T\}\)

      \(\{V, T\}\) has the \(\in\)-minimal element \(V\), since we know that \(T \notin V\) and \(V \notin V\).

    5. \(\{U, V, T\}\)

      None of the elements in \(\{U, V, T\}\) is \(\in\)-minimal because each has one of the others as a member.

    6. \(\{T, U, V\}\)

      \(\{T, U, V\} = \{U, V, T\}\), so the answer is the same as above.

    7. \(\{\emptyset, T, U, V\}\)

      Yes, \(\{\emptyset, T, U, V\}\) has the \(\in\)-minimal element \(\emptyset\).