How can a continuous function $f:[0,1] \cap{\mathbb{Q}}\rightarrow \mathbb{R}$ not be uniformly continuous?

Question

If the domain $[0,1] \cap{\mathbb{Q}}$ of a continuous function $f$ consists of all rationals between zero and one, inclusive, how can the function not be uniformly continuous?

From my understanding, a continuous function can't be uniformly continuous if its image tends to an infinity, because then there can't exist a fixed delta such that the images are always within epsilon distance from each other.

I don't see how the "rationals only" aspect of the domain allows for the range to approach an infinity. Or is there some other condition that can make a continuous function not uniformly continuous?

Answer 1

Here is an example:$f(x)=1/(x-\sqrt\frac12)$. This is well-defined in $[0,1]\cap \Bbb Q$, and continuous at every rational point. But it is unbounded on $[0,1]\cap \Bbb Q$, so it can't be uniformly continuous.

Answer 2

There are other conditions besides a function approaching infinity that can cause it to fail to be uniformly continuous (e.g., increasingly rapid oscillations will prevent a function from being uniformly continuous). With your domain (which I think you mean to $[0,1]\cap\Bbb Q$), take $f(x)=\sin\left(\frac{1}{x-\frac{\sqrt{2}}{2}}\right)$ and you'll have a continuous-but-not-uniformly-continuous function.

This function fails to be uniformly continuous (even though it is bounded) due to the rapid oscillations around $\frac{\sqrt{2}}{2}$; if we were to remove an open interval around $\frac{\sqrt{2}}{2}$, though, the function would become uniformly continuous.

Answer 3

Just a note that adds a tiny bit to the other answers. Dealing with rational domain makes many tricks available. You can even think of a function $f: \mathbb Q \rightarrow \mathbb Q$ (unlike the examples given in the previous answers, and in your original question, where the function takes values in $\mathbb R$) that is continuous but not uniformly continuous. For example \begin{equation} f(x) = \frac{1}{x^2-2} \end{equation} or \begin{equation} f(x) = \begin{cases} 0 & \mbox{if}\ \ x^2<2\\ 1 & \mbox{if}\ \ x^2>2, \end{cases} \end{equation} the latter having also derivative equal to $0$ everywhere, despite of course not being constant at all.