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Definition: An interval $[a, b]$ has the property $P_{\epsilon}$ if there exists sequences $x_{1}, x_{2}, x_{3}, ...$ and $y_{1}, y_{2}, y_{3}, ...$ with $x_{n},y_{n} \in [a, b], |x_{n} - y_{n}| \lt \frac{1}{n}, |f(x_{n}) - f(y_{n})| \gt \epsilon$ for all indices $n$.

To prove: If f is not uniformly continuous on $[a, b]$, then $[a, b]$ has the property $P_{\epsilon}$ for some $\epsilon \gt 0$.

My try: Since $f$ is not uniformly continuous on $[a, b]$, by definition, $\exists \epsilon \gt 0, \forall \delta \gt 0$ s.t. $ x, y \in [a, b], |x - y| \lt \delta,|f(x) - f(y)| \ge \epsilon.$

I choose $\delta_{n} = \frac{1}{n}$. Stuck. I don't know how to choose the two sequences. Even if I got the two sequences, how to make $\ge \epsilon$ to become $\gt \epsilon$ on the last?

Please give some hints, or if any step is wrong.

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    .... $|f(x)-f(y)|\geq \epsilon$ implies $|f(x)-f(y)|> \tilde\epsilon$ with $\tilde \epsilon = \epsilon /2$.... – 2017-01-26
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    Possibly related http://math.stackexchange.com/a/205747/390226. – 2017-02-27

1 Answers 1

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It's not correct. $f$ is not uniformly continuous if $$\exists \varepsilon>0: \forall \delta>0, \exists x,y\in [a,b]: |x-y|<\delta\quad \text{and}\quad |f(x)-f(y)|>\varepsilon.$$

Which implies that $$\exists \varepsilon>0: \forall n\in \mathbb N^*, \exists x_n,y_n\in [a,b]: |x_n-y_n|<\frac{1}{n}\quad \text{and}\quad |f(x_n)-f(y_n)|>\varepsilon.$$

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    @N1ng: Yes, I mean this part. What you wrote is $$\exists \varepsilon>0: \forall \delta>0,\forall x,y\in [a,b], |x-y|<\delta,|f(x)-f(y)|>\varepsilon$$ which is indeed not correct (and has no real sense in fact). – 2017-01-25
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    @N1ng: Of course, it's my first sentence in my answer ;-) – 2017-01-25
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    @N1ng: I didn't wrote $x,y\in [a,b]$, but $\exists x,y\in [a,b]$, what mean : *there exists $x$ and $y$ in $[a,b]$*. For your last question $\neg( \forall x\in A, P(x))$ is $\exists x\in A: \neg P(x)$. For example, the negation of "all cats are black" is "there is a cat that is not black". – 2017-01-25
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    You'd want $>\epsilon$ rather than $\le\epsilon$ at the end – 2017-01-25
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    yes it's the complete version of the proof. @N1ng – 2017-01-26