# Proof verification of uniform continuity

I wanted to know if my proof is correct or not, and if not, what I need to change. My question given to me is to show that $f:\bigcup\limits_{n=1}^{\infty}\left[n,n+\frac{1}{n^{2}}\right]\rightarrow\mathbb{R}$ is uniformly continuous on $\bigcup\limits_{n=1}^{\infty}\left[n,n+\frac{1}{n^{2}}\right]$, where $f\left(x\right)=x^{2}$. Here is what I have.

If $x,y\in\left[n,n+\frac{1}{n^{2}}\right]$, then $\left\lvert x+y\right\rvert\le\frac{2\left(n^{3}+1\right)}{n^{2}}$. Let $\epsilon>0$ and $\delta=\epsilon\cdot\frac{n^{2}}{2\left(n^{3}+1\right)}$. Then, $$\left\lvert f\left(x\right)-f\left(y\right)\right\rvert=\left\lvert x^{2}-y^{2}\right\rvert=\left\lvert x+y\right\rvert\left\lvert x-y\right\rvert$$ $$\le\frac{2\left(n^{3}+1\right)}{n^{2}}\left\lvert x-y\right\rvert<\frac{2\left(n^{3}+1\right)}{n^{2}}\delta$$ $$=\frac{2\left(n^{3}+1\right)}{n^{2}}\epsilon\cdot\frac{n^{2}}{2\left(n^{3}+1\right)}=\epsilon.$$ Thus, for all $\epsilon>0$, there exists $\delta=\epsilon\cdot\frac{n^{2}}{2\left(n^{3}+1\right)}>0$ such that if $\left\lvert x-y\right\rvert<\delta$, then $\left\lvert f\left(x\right)-f\left(y\right)\right\rvert<\epsilon$, so $f$ is uniformly continuous.

• It looks like you picking $\delta$ based $x,y$ (the dependence is implicit through $n$). Uniform continuity requires your $\delta$ to be independent of $x,y$. – parsiad Nov 27 '17 at 18:55
• Note that for $N > 2/\epsilon$ any $x,y$ picked from the $N^\mathrm{th}$ closed interval will be such that $|f(x) - f(y)| < \epsilon$. So you don't need to keep shrinking $\delta$ indefinitely, as you are doing here. – gj255 Nov 27 '17 at 19:16
• Ah okay I see. Thanks to you both! – Jake Nov 27 '17 at 19:39
• (I should have perhaps said something like $N > 4/\epsilon$, however). – gj255 Nov 27 '17 at 20:40
• @gj255, right, that is what I came to. At the moment, I have $\left\lvert f\left(x\right)-f\left(y\right)\right\rvert\le\frac{4}{n}\le\frac{4}{N}\le\frac{4}{\frac{\epsilon}{4}}=\epsilon$, but does this mean that I do not need to specify a delta at all when I am in the intermediate step $\left\lvert x+y\right\rvert\left \lvert x-y\right\rvert\le\frac{2\left(n^{3}+1\right)}{n^{2}}\frac{1}{n^{2}}=\frac{2\left(n^{3}+1\right)}{n^{4}}\le2\frac{2n^{3}}{n^{4}}=\frac{4}{n}$? Because I thought I needed to find a $\delta$, but it seems like I don't need to in this case. – Jake Nov 27 '17 at 21:47

First show that for any $\epsilon > 0$ there exists $N$ depending only on $\epsilon$ and $\delta(\epsilon) > 0$ such that for all $x,y \in [N, \infty)$, if $|x - y| < \delta(\epsilon)$ then $|x^2 - y^2| < \epsilon$.

The appropriate choice to make this work is $N = \max(\lceil \sqrt{2} \rceil, 4/\epsilon)$ and any $\delta(\epsilon) > 0$ such that $\delta(\epsilon) < 1 /N^2.$

Now we need to verify that two conditions are satisfied when $x,y \in [N, \infty)$ and $|x - y| < \delta(\epsilon):$

$$x,y \in [M,M + 1/M^2] \text{ where } M \geqslant N, \\ |x^2 - y^2 | < \epsilon.$$

Suppose that $x \in [M_1,M_1 + 1/M_1^2]$ and $y \in [M_2,M_2 + 1/M_2^2]$ where WLOG $M_2 > M_1 \geqslant N$. In this case,

$$|x-y| \geqslant M_2 - M_1 - 1/M_1^2 > 1 - 1/M_1^2 \geqslant 1 - 1/N^2.$$

However, since $N > \sqrt{2}$, it follows that $1 - 1/N^2 > 1/2 > 1/N^2 > \delta(\epsilon).$ This implies $|x- y| > \delta(\epsilon)$, a contradiction.

Hence, we must have $x, y \in [M,M + 1/M^2]$ with $M \geqslant N$ and

$$|x^2 - y^2| = |x-y||x + y| < \frac{1}{M^2}\left(2M + \frac{2}{M^2} \right) = \frac{2}{M}\left(1 + \frac{1}{M^3}\right)< \frac{4}{N} < \epsilon.$$

You should be able to finish the proof by using the fact that $\displaystyle \bigcup_{n=1}^{N-1}[n,n + 1/n^2]$ is compact, producing a single value of $\delta(\epsilon)$ where $|x^2 - y^2| < \epsilon$ for all $x,y$ in the domain with $|x-y|< \delta(\epsilon)$.

• I think it is a little simpler to observe that if $|x-y|<1$ the x and y are in the same interval and MVT can be used inside $[n,n+ \frac 1 n^{2}$ to get the bound $2(n+ \frac 1 n^{2} \{ \frac 1 n^{2}\}$ for $|f(x)-f(y)$. This has to be combined with uniform continuity on $[0,N]$ for any N. This seems to reduce the amount of calculations one has to make. – Kavi Rama Murthy Nov 28 '17 at 6:16
• @KaviRamaMurthy: " ... if $|x-y|<1$ the $x$ and $y$ are in the same interval ..." So $|11 - 10.01| < 1$ but $10.01 \in [10,10+1/10^2]$ and $11 \in [11,11 + 1/11^2]$ which are not the same intervals. – RRL Nov 28 '17 at 6:45
• The point of my proof is to handle that annoying detail of ensuring that $x$ and $y$ are in the same interval. Everything else is relatively easy and it matters not if we use MVT or a direct estimate. – RRL Nov 28 '17 at 6:47