Problems 7 8


7) "Official" WFF for Empty Set Axiom:

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


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


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

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