Why this function is not uniformly continuous? Let a function $f: \mathbb R \rightarrow \mathbb R$ be such that $f(n)=n^2$ for $n \in \mathbb N$. 
Why $f$ is not uniformly continuous?
Thanks

Suppose it is. i.e. for every $\varepsilon>0$ there is some $\delta>0$ such that if $|x-y| \leq \delta$ then $|f(x)-f(y)| \leq \varepsilon$. Any interval $[n,n+1]$ can be broken into at most $\lceil \frac{1}{\delta} \rceil$ intervals with length$<\delta$ using some partition $x_0=n<x_1<\ldots<n+1=x_k$. Using the triangle inequality, $|f(n)-f(n+1)| \leq |f(n)-f(x_1)|+|f(x_1)-f(x_2)|+\ldots+|f(x_{k-1})-f(n+1)| \leq \lceil \frac{1}{\delta} \rceil \varepsilon$. But this can't hold for all $n$ because $|f(n)-f(n+1)|=1+2n$ grows larger than any bound.
