A function can be defined by specifying a set of tuples. If I write the definition of a function $f = \lbrace(0, f) \rbrace$, is this function sound? Will this lead to a paradox? The domain of this function is $\lbrace 0\rbrace$, but what is its range?

Related: Can a function be applied to itself? , How can a set contain itself?

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    $\begingroup$ The definition proposed is not a ZF set construction. $\endgroup$ Jul 8, 2015 at 21:31

1 Answer 1


If I write the sentence "The table police the dog, while drinking", can I write it? Does it have any meaning?

Not everything that we can write, syntactically, must have meaning mathematically. Just like we can write utter nonsense (just look at the career Monty Python got out of it!), we can also write mathematical nonsense.

You can write $f=\{(0,f)\}$. Sure. But does it have any meaning?

Assuming $\sf ZF$, the answer is no. You can prove that such $f$ does not exist. Namely, there does not exist a set $f$ such that the unique element of $f$ is an ordered pair $(x,y)$ such that $x=0$ and $y=f$ itself. The reason is that any reasonable method for encoding ordered pairs will eventually create a cycle in the $\in$ relation.

Specifically, in the Kuratowski definition, $(0,f)=\{\{0\},\{0,f\}\}$, and then $f\in\{0,f\}\in(0,f)\in f$.

On the other hand, if you reject the axiom of regularity (also known as Foundation axiom), then it is a possibility for $x$ to be an element of an element of an element of $x$ itself. So it is possible to obtain such $f$. In that case, $f$ is indeed a function, and its range is $\{f\}$, of course. What else would it be?

  • $\begingroup$ You know that man the Israelis have, who can bend his legs over his head on each step? That's not me. I'm a retired wi-window cleaner and pacifist, without doing warcrimes! I was head of Gestapo for 10 years! $\endgroup$
    – Asaf Karagila
    Jul 8, 2015 at 22:26
  • $\begingroup$ Tsch tsch tsch! I was not head of Gestapo at all. I make joke! $\endgroup$
    – Asaf Karagila
    Jul 8, 2015 at 22:26
  • $\begingroup$ Now I see the mistake that I made in my previous comment: I forgot that $f$ is not $(0,f)$ but $\{ (0,f) \}$. $\endgroup$
    – Ian
    Jul 8, 2015 at 23:18
  • $\begingroup$ Assuming $\mathsf{NF(U)}$ as your set theory, you can find a function that takes itself as both an input and a value in a very silly way: the identity function on the universe exists. $\endgroup$ Jul 10, 2015 at 16:28
  • $\begingroup$ @Malice: Right. Working with any of the anti-foundation axioms works as well, since they literally give you a way to prove the existence of (or sometimes even prove the uniqueness of) $f=\{(0,f)\}$ or $f=\{(f,0)\}$ or so on. $\endgroup$
    – Asaf Karagila
    Jul 10, 2015 at 16:33

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