# Is it true that $f(x) = x^{2}$ is uniformly continuous on $\mathbb{N}$?

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.

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

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