2
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Any function whose domain if $\mathbb N$ is continuous. $\endgroup$ Feb 8, 2019 at 10:18

3 Answers 3

8
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ Any discrete interval is considered Open? So if f: {2,3,5} -> {6,2,9} is considered continuous? $\endgroup$ May 10, 2020 at 3:36
  • $\begingroup$ 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. $\endgroup$
    – Gibbs
    May 10, 2020 at 13:22
  • $\begingroup$ 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. $\endgroup$ May 10, 2020 at 13:48
  • 1
    $\begingroup$ 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. $\endgroup$
    – Gibbs
    May 10, 2020 at 14:17
  • 1
    $\begingroup$ 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. $\endgroup$
    – Gibbs
    May 10, 2020 at 14:52
1
$\begingroup$

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

$\endgroup$
4
  • $\begingroup$ How can we prove that $\endgroup$
    – pde
    Feb 8, 2019 at 10:20
  • $\begingroup$ can you give me some reference material or so , because I have never read topology , thanks for help $\endgroup$
    – pde
    Feb 8, 2019 at 10:24
  • $\begingroup$ Alternatively, every $x \in \mathbb N$ is an isolated point. Thus every function with domain $\mathbb N$ is continuous by the $\epsilon-\delta$ formulation. $\endgroup$ Feb 8, 2019 at 10:27
  • 1
    $\begingroup$ @KeshavSharma Munkre's Topology is the standard reference. $\endgroup$ Feb 8, 2019 at 10:28
1
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .