Prove the next equivalence:

(1) Axiom or Regularity

(2) If $X$ is a non-empty set, $X$ has an $\epsilon_x$-minimal element. This means, There exists a $z \in x$ such that $\lnot (x \in_x z )$ for all $x \in X.$

I got (1) $\rightarrow$ (2) wrong in an exam.

Because I didn't prove that "the $z$ existed." I am not being able to understand this observation, and so, I don't know how to correct this proof, or taking it into the right direction. So here's my attempt:

The Axiom of Regularity says: Let $A$ be a set, with $A \neq \emptyset $. There exists an $a \in X$ such that, if $b \in a$, then $b$ is not in $A$.

Proof. As $X$ is not empty, there exists a $z\in X$. If $z\in X, $ and for some $x$, $x\in X$ then, by the Axiom of Regularity, $x \notin_x z.$ So, $\lnot(x \in_x z).$

  • $\begingroup$ What does "$\in_x$" mean? $\endgroup$ – Noah Schweber Jun 3 '17 at 19:17
  • $\begingroup$ It means $\in$ restricted to the set $x$ $\endgroup$ – Trux Jun 3 '17 at 19:25
  • $\begingroup$ Then you should write "$\in_X$" instead of "$\in_x$" (and usually we'd just write "$\in$," since we don't treat different membership relations simultaneously). $\endgroup$ – Noah Schweber Jun 3 '17 at 19:29

Your second-to-last sentence is a non-sequitur. If I understand it correctly, you're trying to say "If $x\in X$, then by regularity $x\not\in z$," where your assumption on $z$ is just "$z$ is in $X$."

But this isn't enough to conclude what you want! Consider $X=\{1, \{1\}\}$ and take $z=\{1\},x=1$. Then $z\in X, x\in X$, but $x\in z$.

  • $\begingroup$ If I exchange the choices of $x$, $z$for, "If $x\in X, $ and for some $z$, $x\in X$ then, by the Axiom of Regularity, $x \notin_x z.$ So, $\lnot(x \in_x z),$" there would be still something missing? $\endgroup$ – Trux Jun 3 '17 at 19:45
  • $\begingroup$ @Trux "by the Axiom of Regularity, $x\not\in z$." How are you arguing this? Right now $x$ and $z$ are just arbitrary elements of $X$; there's no reason for $x$ to not be an element of $z$. The axiom of regularity tells us that $X$ has some element $z$ with a certain property; you need to use the axiom of regularity, not just assert that it solves the problem. (Also what does "for some $z$, x\in X$" mean? We already know $x\in X$ by the previous phrase.) $\endgroup$ – Noah Schweber Jun 3 '17 at 19:50
  • $\begingroup$ By contradiction. Because of, Axiom of regularity, there exists an $x$, such that, if $x \in_x z$ then $x$ is not in $X.$ But $x \in X$, so $x \notin_x z$. $\endgroup$ – Trux Jun 3 '17 at 20:05
  • $\begingroup$ You have a typo in your statement of the axiom - that should be "there exists a $z$." More to the point, note that the way you use regularity is to introduce an element - e.g. "By regularity, there is some $z$ such that ..." You didn't do that at all in your original answer. $\endgroup$ – Noah Schweber Jun 3 '17 at 20:15

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