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Let $E$ be the class of ordered pairs such that $a\in B$:

$$E=\{(a,b):a\in b\} . $$

I want to prove that $E$ is not a set. Kelley (General Topology Theorem 104 p.267) does the next:

He supposes $E$ is a set, so that $\{E\}$ is also a set. And because $E\in\{E\}$, the pair $(E,\{E\})$ is in $E$, by definition, so we have the relation

$$ E\in \{E\}\in(E,\{E\})= \{E, \{E,\{E\}\}\, \}\in E. $$

Now he considers the set

$$ \{x:x=E \mbox{ or } x=\{E\} \mbox{ or } x=\{E,\{E,\{E\}\}\}\, \} $$

and says it contradicts the Axiom of Regularity. My question is how it contradicts the axiom.

I mean, if I have sets $$ a\in b \in c \in a, \tag{$\ast$} $$ why the set

$$S=\{a,b,c\}$$

contradicts the axiom?

My attempt:

I want to find an element $x\in S$ such that

$$x\cap S = \emptyset.$$

Suppose $x=a$:

$$a\cap S = \{z:z\in a \mbox{ and } (z=a,b\mbox{ or } c\}.$$

Now, what happens if z=a? I mean, $a\in b\in \cdots$, but for me $\in$ is not transitive, so $a\in a$ not follows from ($\ast$). I suppose my mistake is that, but I cannot understand why.

Thanks

PD: Here is a picture with the proof from Kelley's book.

Dog_69
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1 Answers1

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$S$ is not disjoint from any of its elements $a,b,c$. Namely: $c\in S\cap a$ (because $c\in S$ and $c\in a$), and, similarly, $a\in S\cap b$ and $b\in S\cap c$.

  • Ooohhh. Yes yes. Thank you for showing me the intersection relations. It is an easier way to proof that, instead of check them one-by-one. – Dog_69 Feb 06 '18 at 23:28