What Would It Look Like?

Calculus Level 3

Does there exist a function that is continuous at all irrational points but not continuous at all rational points?

Definition (for this question) : A function $f$ is said to be continuous at a point $n$ if for any positive $\epsilon$ there exists an $x$ such that for all real $|l|<x, \space |f(n+l)-f(n)|<\epsilon$

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Wen Z
Aug 19, 2016

Relevant wiki: Continuous Functions

We'll define a function first and then we'll prove it works.

$f(x)=0$ if x is irrational If $x$ is rational, let $x=\frac{p}{q}$ where $p,\space q \in \mathbb{z}$ and $q>0$ then we let $f(x)=\frac{1}{q}$

I would recommend trying to visualise this function, or at least drawing a rough sketch of it before carrying on so you have a 'feel' of the function.

Now at any irrational point, say $r$ , it is continuous because no matter how small epsilon gets (in the definition above), there are finitely many $x$ with $f(x)>\epsilon$ where $|x-r|<1$ . This is because $q$ (above) can only take finitely many values.

At any rational point, say $r$ , there exist irrational points arbitrarily close to it as the irrationals are dense in the reals as required.

Geoff Pilling
Sep 3, 2016

How about f(0)=1. f(x)=0 otherwise?

What Would It Look Like?

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