Calulus on the set of natural numbers

Question

Can we think about Calculus on $\mathbf{N}$? How would work the notion of neighborhood, limit, continuity and differentiability and analyticity?

I need these to understand discrete dynamics over $\mathbf{N}$.

Any good reference is highly solicited.

Answer 1

Disclaimer: the answers are very boring. We assume that the distance metric on $\mathbb{N}$ is $|x-y|$.

1. An open ball of radius $r \in \mathbb{R}$ centered at $x$, $B_r(x)$ is the set of all points $y$ such that $|x-y| < r$. $B_r(x) = \{y \in \mathbb{N} : x - \lceil r\rceil + 1 \leq y \leq x + \lceil r\rceil - 1 \}$. In particular, if $r \leq 1$, $B_r(x) = \{x\}$.
2. A neighborhood of $x$ is any set which contains some $B_r(x)$. Since $B_{1/2}(x) = \{x\}$, any subset of $\mathbb{N}$ that contains $x$ is a neighborhood of $x$.
3. The sequence $\{x_n\}$ converges to a limit $L$ if for every $\delta > 0$ there is some $N \in \mathbb{N}$ we have that $n \geq N$ implies that $|x_n-L| < \delta$. A sequence has a limit if and only if that sequence is eventually constant, since $x,y \in \mathbb{N}$ implies that either $x = y$ or $|x-y| \geq 1$.
4. We say a function to have a limit $L$ at a point $a$ if for $\epsilon > 0$, we can choose a $\delta > 0$ such that $b \neq a$ and $|a-b| < \delta \Rightarrow |f(a)-L| < \epsilon$. Any natural number $L$ is the limit of any function $f:\mathbb{N} \rightarrow \mathbb{N}$. Why? Choose $\delta = 1/2$. Then $b \neq a$ and $|a-b| < 1/2$ is never true, so our statement is vacuously true.
5. Limits are not unique, thus the notion of continuity or differentiability are just nonsense. But using their definitions, we have that all functions are continuous, but the derivative (ignoring the fact you can't even use fractions in $\mathbb{N}$) of a function is not itself a function $f'(a) = L$ for any $L \in \mathbb{N}$.