Continuity of function defined on natural numbers

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.

• Any function whose domain if $\mathbb N$ is continuous. Feb 8, 2019 at 10:18

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.

• Any discrete interval is considered Open? So if f: {2,3,5} -> {6,2,9} is considered continuous? May 10, 2020 at 3:36
• You cannot say "continuous" without specifying a topology. If your sets are equipped with the natural topology induced by $\mathbb{N}$, then yes your function is continuous. May 10, 2020 at 13:22
• If you were to tell a high schooler whether f(x) = 2x f:{2,3,4} -> N is continuous in its domain, what would you say? My level of Mathematics is not even close to comprehending what you said in the comment above. May 10, 2020 at 13:48
• Do you accept the $\varepsilon$-$\delta$ definition as in the first part of my answer? If so it is enough to restrict to natural numbers as follows. The set of the natural numbers $x$ such that $|x-x_0| < \delta$ is an interval centered at $x_0 \in \mathbb{N}$ with radius $\delta$, so if the radius is less than $1$ the interval contains only $x_0$ and coincides with $\{x_0\}$. Now choose any $\varepsilon > 0$ and take $\delta < 1$. Then we have just seen that any $x$ such that $|x-x_0| < \delta$ must be $x_0$, so $f(x) = f(x_0)$, and the definition is always true. Thus $f$ is continuous. May 10, 2020 at 14:17
• No, the only element $x$ in a $\delta$-neighbourhood (or interval centered at $x_0$ or radius $\delta$) is $x_0$, thus $f(x)=f(x_0)$, which makes the definition of continuity trivial. May 10, 2020 at 14:52

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

• How can we prove that
– pde
Feb 8, 2019 at 10:20
• can you give me some reference material or so , because I have never read topology , thanks for help
– pde
Feb 8, 2019 at 10:24
• Alternatively, every $x \in \mathbb N$ is an isolated point. Thus every function with domain $\mathbb N$ is continuous by the $\epsilon-\delta$ formulation. Feb 8, 2019 at 10:27
• @KeshavSharma Munkre's Topology is the standard reference. Feb 8, 2019 at 10:28

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.