I'm studying for an exam of mathematical logic. This question envolves the Peano axioms, I think.
Prove that, for all $ n \in \omega$, $ n \notin n$.
It's kind of obvious that it's true but I don't know how to prove it. I tried to assume that $ n \in n$ and the first element $0 \in 0$ is false but honestly it doesn't seem right to do this. If I go with $ n \notin n$, a set is never an element of itself so it seems irrelevant to go further.
Any ideas?
Edit:
Ok, so from what I've seen from Von Neumann ordinals I know the following:
- $ 0 \in \omega$ and thus $0 \notin 0$ because $ 0 = \{ \}$ and $ 0 \notin \{ \}$
- $n \in \omega$ and thus $n^+ \in w$ with $ n^+ = n$ $\cup$ $\{n \}$
Now, if $n \notin n$ then $n^+ \notin n^+$ right?
If I assume that $ n^+ \in n^+$ is true, can I say that, since $ n^+ = n$ $\cup$ $\{n \}$, then $ n^+ = n$ or $ n^+ \in \{n\}$? And then claim this is absurd?
Does even Peano arithmetics have $\in$ relation? I think it doesn't. It is a general statement that $x\notin x$ for all $x$. In ZFC its proof would look like this: Set $S=\{y\in x\colon y=x\}$, if it's nonempty then by regularity axiom there is $z\in S$ such that $z\cap S=\emptyset$ but $z\cap S=x\cap S\supset S$, contradiction. If $S$ is empty then we are also done.