Continuity of function defined on natural numbers

Question

I was wondering how one may define continuity if a function takes input only from the set of natural numbers, not real numbers. When we define continuity, we use limit notation, i.e., h tends to zero. But now since h can only be zero or some natural number, tending or belonging between 0 and 1 is nonsense.

Answer 1

Asking whether a function is continuous is by nature a topological question. Whenever you check that a real function $f$ of one real variable is continuous at a point $x_0$ you claim that for each $\varepsilon > 0$ there exists a $\delta > 0$ such that $\lvert x-x_0\rvert < \delta$ implies $\lvert f(x)-f(x_0)\rvert < \varepsilon$. This is to say that the preimage of an open subset of $\mathbb{R}$, in this case $(f(x_0)-\varepsilon,f(x_0)+\varepsilon)$, is an open subset of $\mathbb{R}$. Notice that here open means open in the standard topology of $\mathbb{R}$, where open sets are by definitions unions of open intervals.

You can generalize as follows: if $X$ and $Y$ are two topological spaces, a function $f\colon X \to Y$ is continuous if preimages of open sets are open.

When $X = \mathbb{N}$ with the discrete topology, every subset of $\mathbb{N}$ is open. Therefore no matter what preimage you are taking, it will be open by definition, so every function defined on the natural numbers is continuous.

Answer 2

Using the open set formulation of continuity, every function is continuous as $\mathbb N$ has the discrete topology as a subspace of $\mathbb R$.

Answer 3

This won't be rigorous, but might give you some intuition.

If a function is defined on set of natural numbers, the absolute value of a difference between any two arguments is a natural number. Let's call this number $n$. Clearly exists number $\delta=n+1\in\mathbb{N}$.

And this very existence of the number $\delta$ is what you need in the delta-epsilon definition.