# How to prove that if $A\subseteq A\times A$, then $A=\varnothing$?

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?

• 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 ($\in$). – BrianO Nov 10 '15 at 17:37
• – Nex Nov 10 '15 at 18:29
• I do not buy his argument that the set A does not contain any empty sets. This means that ∅∉ A. – Eric Brown Nov 11 '15 at 0:50

## 1 Answer

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 $A$.