# 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 $$s \in p$$

Exercise 1
2. If $$S$$ is a set and $$P(x)$$ is a predicate, then $$\{x \in S \mid P(x)\}$$ is a set.

Exercise 2
3. $$\forall x.[x \neq \emptyset$$ IMPLIES $$\exists y.y$$ is $$\in$$-minimal in $$x$$]

Exercise 3
4. $$\forall x [x \in y$$ IFF $$x \in z]$$ IFF $$\forall x[y \in x$$ IFF $$z \in x]$$

Exercise 4
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 $$\emptyset$$

Exercise 5
The empty set $$\emptyset$$ has no elements so it cannot have an $$\in$$-minimal element.
2. $$T$$

Exercise 6
$$T= \{V\}$$ 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. $$\{T\}$$

Exercise 7
$$T$$ is an $$\in$$-minimal element of $$\{T\}$$ because we know that $$T \notin T$$.
4. $$\{V, T\}$$

Exercise 8
$$\{V, T\}$$ has the $$\in$$-minimal element $$V$$, since we know that $$T \notin V$$ and $$V \notin V$$.
5. $$\{U, V, T\}$$

Exercise 9
None of the elements in $$\{U, V, T\}$$ is $$\in$$-minimal because each has one of the others as a member.
6. $$\{T, U, V\}$$

Exercise 10
$$\{T, U, V\} = \{U, V, T\}$$, so the answer is the same as above.
7. $$\{\emptyset, T, U, V\}$$

Exercise 11
Yes, $$\{\emptyset, T, U, V\}$$ has the $$\in$$-minimal element $$\emptyset$$.