I think that the answer is yes. If $f(x) = x^{2}$ were uniformly continuous on $\mathbb{N}$, then for every pair of sequences $\{u_{n}\}$ and $\{v_{n}\}$ satisfying

$$\lim_{n\to\infty} \left( u_{n} - v_{n}\right) = 0,$$

we must also have

$$\lim_{n\to\infty} \left(f(u_{n}) - f(v_{n})\right) = 0. $$

But, I cannot come up with two such sequences in $\mathbb{N}$ such that the difference of them equals $0$. So, I think there is nothing to check, and the function is uniformly continuous.

  • 2
    $\begingroup$ wait, you want to prove uniformly continuous and you start by saying "if $f$ were uniformly continuous, then ..." and try to verify ... $\endgroup$ – mathworker21 Nov 16 '18 at 7:39

In $\mathbb N$ $|x-y| <1$ implies $x=y$. So ANY function is uniformly continuous.


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