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For what set theoretic reasons can a function not be included in its own domain?

Thanks

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    @MarianoSuárez-Alvarez well, it is the primary interpretation I know for Lambda calculus. Even the set theoretic definition of function is just a formality that we interpret as something like what we think of as a "function." As soon as I saw a question about "functions which are in their own domain," I thought of Lambda calculus.2012-12-01

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Corrected: A non-empty function can be a subset of its own domain. Let $x_0=0$, for each $n\in\omega$ let $x_{n+1}=\langle x_n,0\rangle$, and let $X=\{x_n:n\in\omega\}$; then

$f\triangleq\big\{\langle x_n,0\rangle:n\in\omega\big\}=\{x_{n+1}:n\in\omega\}\subseteq X=\operatorname{dom}f\;.$

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    Very nice answer. Another variation would be to take $f$ defined on $V_\omega$ to be $f(x)=x\cup\{x\}$, then $f\subseteq V_\omega$ as well.2012-12-03
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One could use such a function to construct a set that violates the Axiom of Regularity.

Note: This answer assumes that "included" means "is an element of."

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    Presumably, if $f\in\mathrm{Dom}(f)$, that'd be the descending sequence $f\ni \big(f,f(f)\big)=\lbrace f,\lbrace f,f(f)\rbrace\rbrace\ni f\ni \big(f,f(f)\big)=\lbrace f,\lbrace f,f(f)\rbrace\rbrace\ni f\ni\cdots$2012-12-01