Exercise 3.2.1

Question

Show that the universal specification axiom, Axiom 3.8, if assumed to be true, would imply Axioms 3.2, 3.3, 3.4, 3.5, and 3.6. (If we assume that all natural numbers are objects, we also obtain Axiom 3.7.) Thus, this axiom, if permitted, would simplify the foundations of set theory tremendously (and can be viewed as one basis for an intuitive model of set theory known as “naive set theory”). Unfortunately, as we have seen, Axiom 3.8 is “too good to be true”!

Khagan Gulusoy · Accepted Answer

\begin{proof}
\begin{enumerate}
    \item (Axiom 3.8 $\implies$ Axiom 3.2) Let $P(x)$ be a property of $x$ such that $x
eq x$, and consider the set specified by this property, i.e., 
    $$\left\{ x : x 
eq x ight\}.$$
    Then the above set is identical with the empty set $\emptyset$ from Axiom 3.2 since no object $y$ can satisfy $P(y)$. 
    
    \item (Axiom 3.8 $\implies$ Axiom 3.3) Let $P(x)$ be a property of $x$ such that $x=a$, and consider the set specified by this property, i.e., 
    $$\left\{ x : x = a ight\}.$$
    Then the above set is identical with the singleton set $\{a\}$ from Axiom 3.3 since the only object $y$ contained in this set is $y = a$.
    
    \item (Axiom 3.8 $\implies$ Axiom 3.4) Let $P(x)$ be a property of $x$ such that $x \in A$ or $x \in B$, and consider the set specified by this property, i.e., 
    $$\left\{ x : x \in A 	ext{ or } x \in B ight\}.$$
    Then the above set is identical with the pairwise union set $A \cup B$ from Axiom 3.4.
    
    \item (Axiom 3.8 $\implies$ Axiom 3.5) Let $Q(x)$ be a property of $x$ such that $x \in A$ and $P(x)$ is true, and consider the set specified by this property, i.e., 
    $$\left\{ x : x \in A 	ext{ and } P(x) 	ext{ is true} ight\}.$$
    Then the above set is identical with the specification set $\{x \in A : P(x) 	ext{ is true}\}$ from Axiom 3.5.
    
    \item (Axiom 3.8 $\implies$ Axiom 3.6) Let $Q(y)$ be a property of $y$ such that $P(x,y)$ is true for some $x \in A$ is true, and consider the set specified by this property, i.e., 
    $$\left\{ y : P(x,y) 	ext{ is true for some $x \in A$} ight\}.$$
    Then the above set is identical with the replacement set $\{y : P(x,y) 	ext{ is true for some } x \in A\}$ from Axiom 3.6.
    
    %\item (Axiom 3.8 $\implies$ Axiom 3.6) Let $P(x):= Q(x,z)$ for some $z\in A$ and $P(y):=Q(y,z)$ for some $y\in A$. Then, we get
    %$$\{y \in \{x : P(x) 	ext{ is true} \} \iff P(y) 	ext{ is true} \}\implies \{y \in \{x : Q(x,z) 	ext{ is true } \} \iff Q(y,z) 	ext{ is true} \}, $$
    %which means that Axiom 3.8 $\implies$ Axiom 3.6.
\end{enumerate}
\end{proof}

Exercise 3.2.1

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Exercise 3.2.1

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