PRELIMINARY

7) "Official" WFF for Empty Set Axiom:

$\exists y \forall x \neg (x \in y)$

8)

a. "Unofficial" Definition of y, the intersection of x, $\cap x$, with all sets z belonging to all elements w of x.

$\forall{z}\forall{w}(z \in (w \in x)) \rightarrow (z \in y)$

b. This definition still does make sense in the case that x is the null set $\emptyset$ because, in that case, the zs that are satisfy $z \in w \in x$ are the null set, and the intersection of the null set is the null set.

c. Lemma: For the set of natural numbers, $\mathbb{N} = \{\emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\}, \{\emptyset, \{\emptyset,\{\emptyset\}\}\},...\}$, $\cap{\mathbb{N}} = \emptyset$.

We will prove by induction by considering that in a generic infinite set $S$, $\{S_{1}, S_{2}, ..., S_{n-1}\} \cap \{S_{n}\} = \cap S$. Consider base case $n = 2$. $\mathbb{N}_{1} \cap \mathbb{N}_{2} = \emptyset \cap \{\emptyset\} = \emptyset$. Suppose this holds for $n+1$. We know that $\emptyset \in \mathbb{N}_{n+1}$ holds for all $n$ because each element of $\mathbb{N}$ is the union of the element before it and the set containing the element before it. Thus, for all $n$, $\emptyset \cap \mathbb{N}_{n} = \emptyset$. Therefore, we can be sure that, by induction, $\cap \mathbb{N} = \emptyset$. QED

FINAL:

7) "Official" WFF for Empty Set Axiom:

$\exists y \forall x \neg (x \in y)$

8)

A. "Unofficial" WFF Definition of the Intersection of x, with all sets z belonging to all elements w of x.

$\forall{z}((z \in y) \leftrightarrow \lnot (x = \emptyset)\wedge\forall{w}( (w \in x) \rightarrow (z \in w)))$

B. This definition does make sense for the null set, $\emptyset$. We know intuitively that the intersection of all elements of the set with no elements is nothing. This definition is clearly false for all z if x is the empty set, meaning it defines $\cap\emptyset$ to be $\emptyset$. Without the condition that x not be the empty set, if x were the empty set, by vacuous truth all sets z would be in the $\cap{x}$, thus making $\cap{x}$ the set of all sets, which clearly cannot exist.

C. Lemma: For the set of natural numbers, $\mathbb{N} = \{\emptyset, \{\emptyset\}, \{\emptyset, \{\emptyset\}\}, \{\emptyset, \{\emptyset,\{\emptyset\}\}\},...\}$, $\cap{\mathbb{N}} = \emptyset$.

By the definition, any element in the intersection of a set must be in all elements of that set. Consider the first two elements of $\mathbb{N}$, $\emptyset$ and $\{\emptyset\}$. There are no elements of the first element of $\mathbb{N}$, thus there are no elements of the second element of $\mathbb{N}$ that are in any elements of the first element of $\mathbb{N}$. Thus, the intersection of $\mathbb{N}$ must be $\emptyset$.

Comments - Andrew Furash

Nice work here. You addressed and solved the issue 8b was getting at.