# Thread: Axiom of Regularity Problem

1. ## Axiom of Regularity Problem

In conjunction with my other post on checking the regularity of a database, I am interested in solving Problem 1 on page 56 of Suppes' Axiomatic Set Theory:

Prove that for all sets $\displaystyle A$, $\displaystyle B$, and $\displaystyle C$ it is not the case that

$\displaystyle A\in B\land B\in C\land C\in A.$

Now, the proof of Theorem 2.106 on page 54 is bound to be similar. Here's the theorem statement:

$\displaystyle \neg(A\in B\land B\in A).$

Proof: (Quoting directly from Suppes) Suppose that $\displaystyle A\in B\land B\in A.$ Then

$\displaystyle (1)\quad A\in\{A,B\}\cap B\quad\text{and}\quad B\in\{A,B\}\cap A.$

By the axiom of regularity there is an $\displaystyle x$ in $\displaystyle \{A,B\}$ such that

$\displaystyle \{A,B\}\cap x=0$

and by Theorem 2.43 [Ackbeet: on page 31. It states that $\displaystyle z\in\{x,y\}\iff z=x\vee z=y.$]

$\displaystyle x=A\quad\text{or}\quad x=B.$

Hence

$\displaystyle \{A,B\}\cap A=0\quad\text{or}\quad\{A,B\}\cap B=0,$

which contradicts (1). QED.

So here's my attempt at a proof of the non-existence of the 3-cycle:

Suppose the assertion were false. Then the following hold:

$\displaystyle (1)\quad A\in\{A,B,C\}\cap B,$
$\displaystyle (2)\quad B\in\{A,B,C\}\cap C,$
$\displaystyle (3)\quad C\in\{A,B,C\}\cap A.$

By the axiom of regularity there is an $\displaystyle x\in\{A,B,C\}$ such that

$\displaystyle \{A,B,C\}\cap x=0.$

By the obvious extension of Theorem 43,

$\displaystyle x=A\quad\text{or}\quad x=B\quad\text{or}\quad x=C.$

Hence, at least one of $\displaystyle \{A,B,C\}\cap A$ or $\displaystyle \{A,B,C\}\cap B$ or $\displaystyle \{A,B,C\}\cap C$ is zero, contradicting (1), (2), or (3). QED.

Is this a valid proof?

2. Originally Posted by Ackbeet Is this a valid proof?
Took me a little while to decipher, but I say yes, it is valid. #### Search Tags

axiom, problem, regularity 