# Show that $x \sin x$ is not uniformly continuous

I have seen a lot of posts that have solved this already by taking two sequences $$(a_n), (b_n)$$, then showing $$\lim(a_n - b_n) = 0$$ but $$|f(a_n) - f(b_n)| > \epsilon$$, but I would like to show nonuniform continuity with a different approach.

My question is how would you pick two points (not sequences), $$x_{\delta}$$ and $$u_{\delta}$$ such that $$|x_{\delta} - u_{\delta}| < \delta$$ FOR ALL $$\delta > 0$$?

In particular using this criteria for nonuniform continuity to show that xsinx is not uniform continuous:

ii) There exists an $$\epsilon_0 > 0$$ such that for every $$\delta > 0$$ there are points $$x_{\delta}, u_{\delta}$$ in $$A$$ such that $$|x_{\delta} - u_{\delta}| < \delta$$ and $$|f(x_{\delta}) - f(u_{\delta})| \geq \epsilon_0$$ for all $$n \in \mathbb{N}$$.

• This is fundamentally the same as the sequence approach. Namely, no pair of points work for all delta. By taking a sequence of $\delta \to 0$, the pairs $(x_\delta, u_\delta)$ form a sequence of the form you are trying to give. Thus taking an expression for your desired $(x_\delta, u_\delta)$ pairs gives an expression for a sequence (say, by taking $\delta = 1/n \to 0$), and conversely an expression for such a sequence gives you pairs $(x_\delta, u_\delta)$. Apr 22, 2020 at 15:14
• wait I thought I was trying to show that ALL pairs of points work for ALL delta, but you are saying it's the opposite, that I need to show that NO pair of points works for delta? Apr 22, 2020 at 15:17
• No, you need to show for each $\delta$, there exists a pair $(x_\delta, u_\delta)$. I have emphasized the word each, since I believe this is where the confusion lies. As an immediate followup, if should be clear that no single pair $(x_0, y_0)$ can work for all $\delta$, since it necessarily doesn't work for $\delta < \lvert x_0 - y_0 \rvert$. Apr 22, 2020 at 15:19

You don't need $$|x_\delta-u_\delta|<\delta$$ for all $$\delta>0$$. According to the $$\varepsilon-\delta$$ definition, what you need is to find $$\varepsilon_0$$ such that for any $$\delta$$ such a pair of points $$x_\delta,u_\delta$$ exists, but the values can change with $$\delta$$.
You may choose for instance $$\varepsilon_0=1$$. Then for any $$\delta>0$$ take $$x_\delta=2k\pi$$ for some integer $$k=k(\delta)$$ sufficiently large thanks to the oscillation of the cosine there will exist some
$$u_\delta\in(2k\pi-\delta,2k\pi+\delta)$$ such that $$|u_\delta|\geq1$$ as desired.