3

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?

Community
  • 1
miselico
  • 275

1 Answers1

13

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?

Asaf Karagila
  • 405,794
  • 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! – Asaf Karagila Jul 08 '15 at 22:26
  • Tsch tsch tsch! I was not head of Gestapo at all. I make joke! – Asaf Karagila Jul 08 '15 at 22:26
  • Now I see the mistake that I made in my previous comment: I forgot that $f$ is not $(0,f)$ but ${ (0,f) }$. – Ian Jul 08 '15 at 23:18
  • 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. – Malice Vidrine Jul 10 '15 at 16:28
  • @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. – Asaf Karagila Jul 10 '15 at 16:33
  • @AsafKaragila: Actually, that's interesting... I can't think of a way off the top of my head to prove the existence of one of those in NF. Probably I'm being daft, it certainly seems like it should be lurking in there somewhere... – Malice Vidrine Jul 10 '15 at 17:21
  • @Malice: That's an interesting question, can $\sf NF$ prove some sort of anti-foundation axiom. – Asaf Karagila Jul 10 '15 at 17:42
  • @AsafKaragila: Well, it obviously disproves foundation with $V\in V$, or that the set of all ordinals is a member of a member of the set of all ordinals; but it doesn't prove the existence of Quine atoms and such which have more the flavor of $f={(0,f)}$. I'm sure you can squeeze such a thing out of a suitable permutation model, in any case. – Malice Vidrine Jul 10 '15 at 17:47
  • @Malice: Yes, of course it disproves foundation. But AFA say more, they say essentially that any type of failure happens, in some meaningful way which I am too lazy to explain right now. – Asaf Karagila Jul 10 '15 at 18:02
  • @Asaf: That's one of those things perpetually on my "read up on it" list. Maybe this is the week... – Malice Vidrine Jul 10 '15 at 18:19