Define a graph $\phi$ as;
$\phi(x)=|x|, \forall x\in [-1,1]$ and $\phi(x+2)=\phi(x)$.
I'm trying to prove that $\phi$ is a function, but it seems something is wrong in my argument.
It's not hard to see that "$\forall x\in \mathbb{R}, \exists y\in[0,1]$ such that $(x,y)\in \phi$"
I'm trying to prove that $x_1=x_2 \Rightarrow \phi(x_1)=\phi(x_2)$ and here's my argument below.
============= Suppose $x_1=x_2$.
Then there exists a unique $n\in\mathbb{Z}$ such that $n_i≦\frac{x_i + 1}{2}
Thus $\phi(x_1)=|x_1 - 2n_1|=\phi(x_2)$. Q.E.D
===============
I'm sure something's wrong in my argument, but don't know what it is..
(This argument works even when $\phi(1)≠1$, so something's wrong here..)