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In Tao's Analysis I, the following exercise is posed:

Use the axiom of regularity (and the singleton set axiom) to show that if $A$ is a set, then $A \notin A$. Furthermore, show that if $A$ and $B$ are two sets, then either $A\notin B$ or $B\notin A$ (or both).

I was able to demonstrate the first portion of this (i.e. that $A \notin A$), but I am having trouble formulating the proof necessary to demonstrate the second claim. Here is what I have so far.

Firstly, if $A = B$, then we know by $A\notin A$ that the statement $A\notin B \lor B\notin A$ is true, so let's consider the case where $A \neq B$.

Define $A=\{...\}$ and $B=\{...'\}$. (just my shorthand notation for denoting that they contain different elements and are therefore not equal).

Consider the case where $A \in B$. Then we have $B = \{A,...' \}$. Suppose also, by contradiction, that $B\in A$. Then we could rewrite $B=\{A,...'\}$ as $B=\{\{B,...\},...'\}$. But because $A\in B$, we could then rewrite $B$ as $B=\{\{\{A,...'\},...\},...'\}$. This process would continue infinitely.

I believe there is a contradiction here because I remember reading something about infinitely nested sets not being allowed in set theory where the axiom of regularity is asserted...so I think I am close.

What I see so far (in the little nested examples that I presented), is that $A=\{\{A,...'\},...\}$. I feel like this somehow contradicts the idea that $A \notin A$, but I am not certain why.

Any help is greatly appreciated!

Andrés E. Caicedo
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S.C.
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1 Answers1

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Hint. By regularity,
{A,B} has an element that is disjoint from {A,B}.

William Elliot
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  • $(A \cap {A,B} = \emptyset) \lor (B \cap {A,B}=\emptyset)$. This results in the statement: $(A \notin A \land B \notin A) \lor (B \notin B \land A\notin B)$ . Is that what you're getting at? – S.C. Feb 09 '20 at 03:05
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    Yes, you can take a hint. – William Elliot Feb 09 '20 at 05:15