I encountered a problem that puzzles me a lot. Could anyone help me?
Let $f: \mathbb{N}^2 \to \mathbb{Z}$ be the function $f(x,y) = x-y$, where $\mathbb{N} = \{1,2,3,4,\ldots\}$ is the set of positive integers. Give a subset $S \subseteq\mathbb{N}^2$ such that $f: S \to \mathbb{Z}$ is a bijection.
I feel that the subset should be in the form $(x, a)$ where $x$ is still all the possible elements in $\mathbb{N}$ while $a$ becomes a fixed value like $1$ or $2$. However, $x-1$ will only lead to intergers in specific range, as $x$ can only be positive integers and the output should be all the possible integers.
Could anyone help me? Thanks a lot.
Draw a picture of $\mathbb{N}^2$ in the plane. Draw the level sets of the function $f(x,y)=x-y$, i.e. the sets of the form $f(x,y)=n$ for different values of $n$. For example, draw the set $f(x,y)=0$, the sets $f(x,y)=1$, $f(x,y)=2$, $f(x,y)=-1$, $f(x,y)=-2$, and so on. Observe how $\mathbb{N}^2$ intersects each of these level sets. Choose one element of $\mathbb{N}^2$ on each level set.