How do I prove that $A\subseteq (A \times A)\rightarrow A = \varnothing$ in axiomatic set theory?

This is theorem 107 of Patrick Suppes "Axiomatic Set Theory". I do not buy his argument that the set $A$ does not contain any empty sets because, according to him, the set $A$ is a subset of $A\times A$. Are there any other arguments that prove this theorem? Why does the set $A$ not have the empty set as a member?

  • $\begingroup$ First, "does not contain ANY empty sets": there is only one empty set (extensionality). Second, what does $A\subseteq A\times A$ have to do with it? How could this possibly imply that $\emptyset \in A$? Every set includes the empty set ($\emptyset \subseteq A$), but that does not mean contains ($\in$). $\endgroup$ – BrianO Nov 10 '15 at 17:37
  • $\begingroup$ See math.stackexchange.com/questions/1516538/… $\endgroup$ – Nex Nov 10 '15 at 18:29
  • $\begingroup$ I do not buy his argument that the set A does not contain any empty sets. This means that ∅∉ A. $\endgroup$ – Eric Brown Nov 11 '15 at 0:50

Suppose towards a contradiction that $A\subseteq A\times A$ and $A$ is nonempty. Pick an element $a$ of $A$ of minimal rank. Then $a=(b, c)$ for some $b, c\in A$, since $A\subseteq A\times A$. But then $b$ and $c$ each have rank strictly less than that of $a$, which is a contradiction.

Note that this depends on the precise definition of the pairing function; for some pairing functions we may in fact have $A\subseteq A\times A$ for nonempty $A$.


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