Let $f:\mathbb{R}→\mathbb{R}$ is function such that,
$$ f(x) = \begin{cases} n &\text{if $x=n, x\in\Bbb N$}\\ 0,&\text{otherwise}\end{cases}$$
and $T= \mathbb{N}∪\{n+\frac{1}{n} | n ∈ \mathbb{N}\}$
Then under usual metric on $\mathbb{R}$, $f$ is uniformly continuous on
a) $\mathbb{N}$ but not on $T$
b) $T$ but not on $\mathbb{N}$
c) both $\mathbb{N}$ and $T$
d) neither $\mathbb{N}$ nor on $T$
I get stuck on this question from hours, but didn't able to solve it.
I know function is always continuous on isolated points of domain if it is defined there, but point of $\mathbb{N}$ are not isolated point of domain $\mathbb{R}$. Further, $$\lim_{x\to n+} f(x)=\lim_{x\to n-} f(x)= 0≠ f(n)$$ so how can be function is continuous on points of $\mathbb{N}$? I may be wrong. Please help me.
All function on $\mathbb N$ is uniformly continuous (due to the definition: Take $\eta = 1/2$ for any $\epsilon$).
$f$ is not uniformly continuous on $T$ since $|f(n+1/n)-f(n)| \to \infty$ while $|n+1/n - n| \to 0$.