Prove tht $f(x) = x^{2}$ is uniformly continuous on $\bigcup_{n = 1}^{\infty} [n, n + 1/n^{3}]$. 
How can I show that the function $f(x) = x^{2}$ is uniformly continuous on $\bigcup_{n = 1}^{\infty} [n, n + \frac{1}{n^{3}}]$? 

As $n \to \infty$, I know that we get only all of the integers; however, the result doesn't make sense to me. I'm aware of both the sequential and $\epsilon-\delta$ definitions of continuity, but I didn't get anywhere with them.
 A: Hint
Let $\epsilon>0$. You need to choose $\delta$ so $|x-y|<\delta$ implies $|x^2-y^2|<\epsilon$, as $x,y$ range over your set. You can always make $\delta$ smaller; choose $\delta$ small enough so that $|x-y|<\delta$ implies they are in the same interval $[n,n+n^{-3}]$.
The maximum value of $|x^2-y^2|$ for $x, y\in [n,n+n^{-3}]$ is $(n+n^{-3})^2-n^2=2n^{-2}+n^{-6}$. We can choose $N$ large enough so $n>N$ implies this difference is less than $\epsilon$. This allows us to restrict our attention to the intervals $[n,n+n^{-3}]$ where $n<N$. On those intervals, $f$ has a bounded derivative.
A: Note that for $x,y\in [m,m+{1\over m^3}]$ for some $m$ we have$$|x^2-y^2|\le \left(m+{1\over m^3}\right)^2-m^2={2\over m^2}+{1\over m^6}\le {3\over m^2}$$then for some $\epsilon>0$ let $N=\lfloor\sqrt{3\over \epsilon}\rfloor+1$ therefore we obtain $$\text{for }x,y \in [m,m+{1\over m^3}], m\ge N\to |x^2-y^2|<\epsilon$$and for $x,y \in [m,m+{1\over m^3}]$ , $ m< N$ let $\delta_m(\epsilon)>0$ be such that if $|x-y|<\delta_m(\epsilon)$ then $|x^2-y^2|<\epsilon$ therefore we finally write the general $\delta $ as$$\delta= \min_{1\le m\le \lfloor\sqrt{3\over \epsilon}\rfloor}\delta_m(\epsilon)$$ which is a function of $\epsilon$ and the proof is complete.
