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 $n \in \mathbb{N}$.

  • 1
    $\begingroup$ 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 $(x_\delta, u_\delta)$. $\endgroup$
    – davidlowryduda
    Apr 22, 2020 at 15:14
  • $\begingroup$ 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? $\endgroup$
    – Evan Kim
    Apr 22, 2020 at 15:17
  • $\begingroup$ 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 $\delta < \lvert x_0 - y_0 \rvert$. $\endgroup$
    – davidlowryduda
    Apr 22, 2020 at 15:19

1 Answer 1


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.

Hope this helps.

PS: the result is not true if you stay within a compact set.


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