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Is there any non empty set a such that for every $x∈a$ , $\{x\}∈a$ ? If it is true, how do you deduce the statement from ZFC ( heavy use of the Axiom of Infinity I suspect) & if it false, where does the contradiction appear from ? And, if it is true, does it work as a weaker version of Infinity ? It seems to be the one originally used by Zermelo.

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    $\begingroup$ An example would be $V_\omega$. If $a\in V_\omega$ then $a\in V_k$ for some k, so $\{a\}\subseteq V_k$ so $\{a\}\in V_{k+1}\subseteq V_\omega$. $\endgroup$ – Wojowu Oct 4 '15 at 7:06
  • $\begingroup$ What is Vω ? An ordinal ? $\endgroup$ – user175081 Oct 4 '15 at 7:10
  • $\begingroup$ en.wikipedia.org/wiki/Cumulative_hierarchy $\endgroup$ – Wojowu Oct 4 '15 at 7:10
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Of course (as Wojowu points out in the comments) $V_\omega$ is an example of such a set (as indeed is any $V_\lambda$ for $\lambda$ a limit ordinal), but we can also prove in ZFC that the set $$\{\{\}, \{\{\}\}, \{\{\{\}\}\}, . . .\}$$ of finite Zermelo ordinals exists, which I think is what you were specifically looking for. We can prove this set exists using mainly Infinity and Replacement (it might be possible to avoid Replacement, but I'm tired and lazy right now :P) - one particularly silly way to do it is to define it as the set of elements of $V_\omega$, every element of whose transitive closure has exactly one element. (Replacement is being used here to talk about the transitive closure.)


By the way, you're quite right that Infinity is crucial: $V_\omega$ is a model of all the $ZFC$ axioms except Infinity, and no set as you describe exists in $V_\omega$.

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