$f:A=\mathbb{Q}\cap (0,7)\to\mathbb{R}$ be an uniformly continuous function
can anyone tell me which of the following are correct?
$1. f \text{ is bounded}$
$2. f$ must be constant
$3. f$ is differentiable at all rational points
$4. f$ is differentiable in $(0,7)$
as uniformly continuous function is send bounded set to bounded set so $1$ is true, I have no counter example for $2$, as the domain is totally disconnected so $3,4$ are false. am I right?
I don't think $f$ has to be constant. $f(x) = x$ seems to be uniformly continuous on $A$. In fact, if $g: [0, 7] \to \mathbb R$ is continuous, its restriction to $A$ is uniformly continuous. (To see this, suppose $\epsilon > 0$ is given, then there exists $\delta > 0$ such that $|x - y| < \delta$ implies $|g(x) - g(y)| < \epsilon$, where $x, y \in [0, 7]$. Obviously the part of the sentence about $x, y$ is true for $x, y \in A \subseteq [0, 7]$ also, so $f$ is uniformly continuous.)
That $f$ is not necessarily differentiable at rational points is not a result of the domain being disconnected.
You can just pick $f(x) = \max\{1, x\}$. $f$ is uniformly continuous but not differentiable at $1$ because the left limit and the right limit of $\frac{f(x - 1)}{x - 1}$ as $x \to 0$ are different.
