6
$\begingroup$

Can functions with domain in the Natural Numbers be continuous? In the high school, it is teached an intuitive notion of continuous functions: functions which will always appear as an "unbroken curve", no matter how close one zooms in the graph. This would be the opposite of discrete functions, which would appear as "dots" if one zooms close enough.

In Analysis, this intuitive notion is abandoned to make way for formalism. Rudin, in "Principles of Mathematical Analysis" defines continuous functions as it follows:

Which, for me, can be written in formal language as:

$\forall\varepsilon>0\quad\exists\delta>0 \quad\forall x \in E\quad ((d_{x}(x,p)<\delta)\implies(d_{y}(f(x),f(y))<\epsilon))$

Functions with domain in the natural numbers will appear as dots in a graph, instead of an unbroken curve.

I am trying to formally prove that no function with domain on the Naturals is continuous, can somebody help me out?

PS: What is the domain of $\varepsilon$ and $\delta $?

$\endgroup$

1 Answer 1

11
$\begingroup$

The "unbroken curve" idea works for functions defined on an interval, but not in general.

In fact, every function $\Bbb N\to Y$ (for some space $Y$) is continuous! The kicker is to make $0<\delta\leq1.$ But why does this work? Well, take any $x\in\Bbb N,$ and suppose $y\in\Bbb N$ is such that $\lvert x-y\rvert<\delta.$ The only way this is possible is if $\lvert x-y\rvert=0$ (why?), meaning that $x=y.$ Thus, regardless of the function $f:\Bbb N\to Y,$ we have $f(x)=f(y)$ whenever $\lvert x-y\rvert<\delta,$ so our choice of $\epsilon>0$ is completely irrelevant.

Spaces (like $\Bbb N$) with this property (that all functions on it are continuous) are known as discrete spaces--more often, they are defined to be spaces in which every subset is open, or (equivalently) in which every one-point subset is open.

As for $\epsilon$ and $\delta,$ the only general restriction on them is that they be positive real.

$\endgroup$
2
  • 1
    $\begingroup$ As for the intuition behind the more formal definition given by Rudin, you can see here. For some discussion of why Rudin uses that definition instead of the "unbroken curve" heuristic, see this answer and the commentary below it. For a related question, see here. $\endgroup$ Aug 25, 2015 at 20:21
  • 1
    $\begingroup$ One has to forget completely what one has learnt at school... Thanks for the answer! $\endgroup$ Aug 2, 2018 at 7:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.