Suppose $$f:=\begin{cases}n,\ \ \text{if}\ x\in\mathbb{N}\\0\ \ \text{otherwise}\end{cases}$$. Then, is $f$ uniform continuous over $\mathbb{N}$ or the set $A=\{n+\frac1{n}:n\in\mathbb{N}\}$? Or both?
My reasoning is that since the stance $|x-y|\le\delta$ implies that $x=y$ in the set $\mathbb{N}$, therefore $f$ is uniformly continuous in $\mathbb{N}$. Whereas, in the set $A$, for example, if we take $x=3$ and $y=3+1/d\,,d>\frac1{\delta}$, then for all $\delta$, we have $\epsilon<n$ such that $|x-y|<\delta\not\implies|f(x)-f(y)|=n<\epsilon$. Is my rasoning right? Thanks beforehand.
The restriction of $f$ to $A$ is not uniformly continuous simply because it is not continuous there: you have $2\in A$ (since $2=1+\frac11$), $f(2)=2$, and $f(a)=0$ for every $a\in(1,3)\setminus\{2\}$. But you cannot take $x=3$, since $3\notin A$.
And $f|_{\Bbb N}$ is uniformly continuous by the reason that you mentioned.