Supposse $\Omega \subset \mathbb{R^{n}}$ is an isolated point set, $\textit{i.e}$, $\forall x \in \Omega, \exists \delta \gt 0: \forall \epsilon \gt 0, B_{\epsilon}(x) \backslash \{x\} = \emptyset$. We know for a fact that a function $f$ is continuous on every isolated point, and since the $\Omega$ has only isolated points, $f$ is continuous on $\Omega$. My question is: is $f$ uniformly continuous in $\Omega$?
I thought of something but I almost sure that it is wrong:
Let $D = \{||x_{k+1}-x_{k}||, x_{k} \in \Omega\}$ be the set of distance between two consecutive points of $\Omega$. Since every point of $\Omega$ is an isolated one, zero is a lower bound of $D$, hence, the infimum of $D$ is well defined.
Let $\delta_{0} = \inf D$. Then $\forall x,y \in \Omega, ||x-x|| \lt \delta_{0} \implies 0=||f(x)-f(x)|| \lt \epsilon, \forall \epsilon \gt 0$, hence $f$ is uniformly continuous.
But then $\delta_{0}$ is dependent of an arbitrary point of $\Omega$, which contradicts the definition of uniformly continuity.
What do you think? Can $f$ be uniformly continuous in this set?
What you did is wrong. Why should we have $\delta_0>0$?
And, no, $f$ doesn't have to be uniformly continuous. Take, for instance $D=\left\{\frac1n\,\middle|\,n\in\mathbb N\right\}$ and $f\colon D\longrightarrow\mathbb R$ defined by $f\left(\frac1n\right)=n$. Assuming that we are dealing with the usual distance in $\mathbb R$ (and in $D$), $f$ is not uniformly continuous.