$x^{2}$ uniformly continuous on infinite union of closed and bounded intervals

Question

I've been thinking about this problem for some time, but I haven't gotten anywhere with it. I know that $f\left(x\right)=x^{2}$ is uniformly continuous on any bounded interval of $\mathbb{R}$, but how can I show that $f:\bigcup\limits_{n=1}^{\infty}\left[n,n+\frac{1}{n^{2}}\right]\rightarrow\mathbb{R}$ is uniformly continuous?

Answer 1

Observe that $\bigcup\limits_{n=1}^{\infty}\left[n,n+\frac{1}{n^{2}}\right] = [1,2\frac14]\cup\bigcup\limits_{n=3}^{\infty}\left[n,n+\frac{1}{n^{2}}\right]$. We know that $f$ is uniformly continuous on $[1, 2\frac14]$ so I'm going to show it for $\bigcup\limits_{n=3}^{\infty}\left[n,n+\frac{1}{n^{2}}\right]$.

Let $\epsilon > 0$.

The distance from the endpoint of one interval to the start of the next ist then greater than $\frac{3}{4}$.

Choose $\delta = min\{\frac{3}{\epsilon}, \frac{3}{4} \}$. Then we have for $|x-y|< \delta$ (note that per definition of $\delta$ we have that $x$ and $y$ must be in the same interval) :

$$|f(x)-f(y)| \leq (n+\frac{1}{n^2})^2 - n^2 = \frac{2}{n} + \frac{1}{n^2} \leq \frac{3}{n} \leq\epsilon.$$