# Calulus on the set of natural numbers

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.

• The notion of neighborhood is trivial, as every set is open, and thus every function is continuous. Limits don't happen. – JHance Oct 16 '14 at 6:02
• Define continuity over $\mathbf{N}$. Please give me proof of this... – Fukuzita Oct 16 '14 at 6:04
• how do you define open set unless you have a concept of nbd? – Fukuzita Oct 16 '14 at 6:27
• Well if you are just considering it with the same topology as it has a subset of $\mathbb{R}$ then every point is isolated, which means every point is open in the subspace topology, and means that there are no sequences converging to the point except the constant sequence. – JHance Oct 16 '14 at 6:31
• @Fractas For simplicity, use the subspace topology inherited from $\mathbb R$, that make $\}mathbb N$ a discrete space. – Hagen von Eitzen Oct 16 '14 at 6:31

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 $L \in \mathbb{N}$.