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A function $f:A \subseteq \mathbb{R} \rightarrow \mathbb{R}$ is uniformly continuous if $\forall \epsilon>0$ $\exists \delta>0$ such that $\forall x \in A and \forall y \in A$ with $|x-y|\leq \delta$ we have $|f(x)-f(y)|\leq \epsilon$

Is the negation:

There exists and $\epsilon>0$ such that $\forall \delta>0$ there exists $x\in A$ or there exists $y\in A$ such that $|x-y|\leq \delta$ and $|f(x)-f(y)|\geq \epsilon$?

Show that $f:(0,1)\rightarrow \mathbb{R}$ given by $f(x)=\frac{1}{x}$ is not uniformly continuous.

proof: Let $\epsilon=1$ let $\delta>0$ arbitrary. Set $x\in (0,1)$ to be such that $x<\frac{\delta}{1+\delta}$ and $y= x+\delta$ . Then $|\frac{1}{x}-\frac{1}{x+\delta}|$ $=$ $|\frac{\delta}{x(x+\delta)}|$ $\geq$ $\frac{\delta}{x(1+\delta)}>1$

Is the proof correct? The only question I have is why is $y=x+\delta$ guaranteed to be in $(0,1)$?

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  • $\begingroup$ Indeed, you are right, it is not correct. $\endgroup$
    – Alexey
    Oct 28, 2019 at 22:03
  • $\begingroup$ If you have found a counterexample for $\delta$,then it is also a counterexample for $\delta' > \delta$. So, you can always assume that $\frac{\delta}{1+\delta}+\delta < 1$. $\endgroup$
    – amsmath
    Oct 28, 2019 at 22:12
  • $\begingroup$ @amsmath is the negation correct though? $\endgroup$
    – user643073
    Oct 28, 2019 at 22:13
  • $\begingroup$ The "or there exists" does not make sense. Think about it. $\endgroup$
    – amsmath
    Oct 28, 2019 at 22:18
  • $\begingroup$ @amsmath ($\forall x \in A and \forall y \in A$ ) What would be the negation of that statement then? $\exists x\in A or y\in A$, no? $\endgroup$
    – user643073
    Oct 28, 2019 at 22:19

2 Answers 2

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You basically have it. You just need to be a little more careful. You are claiming that for $\epsilon=1$, and for every $\delta>0$, there are $x,y\in (0,1)$ such that$|x-y|<\delta$ and $|f(x)-f(y)|>1.$ You may assume without loss of generality that $\delta<1/2$ because it the claim is true for all such $\delta,$ it will be true for any value of $\delta$ larger than $1/2.$

(Remember, all you need to do is find two numbers in $(0,1)$ whose difference is less than $\delta$ in absolute value. The ones that work for $\delta<1/2$ will also work for $\textit{any}\ \delta\ge 1/2.$ Example: suppose we have $\delta=15$ and you can find $x,y$ such that $|x-y|<1/2$ and $|f(x)-f(y)|>1.$ Then, the $x,y$ work for $\textit{both}$ values of $\delta$ simultaneously because if $|x-y|<1/2$ it is also $<15$).

Now, $|f(x)-f(y)|=\left|\frac{x-y}{xy}\right|$ and we want to choose $x$ and $y$ so that $|x-y|<\delta$ but $\left|\frac{x-y}{xy}\right|>1$, so take $x=\delta$ and $y=2\delta.$ Then, $x$ and $y$ are actually in $(0,1)$ and $\left|\frac{x-y}{xy}\right|=\frac{2}{\delta}>1$, and you are done.

It may be easier to do it with sequences: with $\epsilon=1/2,$ take $\delta_n=1/n$ and find sequences $(x_n)$ and $(y_n)$ such that $|x_n-y_n|\to 0$ but $|f(x_n)-f(y_n)|>1/2.$ Choose $x_n=1/n$ and $y_n=1/n+1$ and check that this assignment works.

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  • $\begingroup$ May you elaborate on why I might assume $\delta<\frac{1}{2}$, please? The negation of U.C is to show that it holds for all delta, I cant just show it for some delta. $\endgroup$
    – user643073
    Oct 28, 2019 at 22:31
  • $\begingroup$ Yes, I will add to my answer. $\endgroup$ Oct 28, 2019 at 22:32
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It is almost correct.Nice work.

To overcome the situation whether $x+\delta \in (0,1)$ you can do this.

Assume that it is uniformly continuous.

Let $x \in (0,1)$

Then for $\epsilon=1$ exists $\delta>0$ such that..e.t.c.

So the ''e.t.c'' part of the proof will be true also for every $\delta_0<\min\{\frac{x-1}{2},\frac{x}{2},\delta\}$

So you can work for $\delta_0$'s the same way you workded since. $x+\delta_0 \in (0,1), \forall \delta_0<\min\{\frac{x-1}{2},\frac{x}{2},\delta\}$

You can also use sequences to prove the statement.

Take $x_n=\frac{1}{n+1}$ and $y_n=\frac{1}{n+2}$

Then $x_n-y_n \to 0$ but $|f(y_n)-f(x_n)|=1 \to 1 \neq 0$

So $f$ is not uniformly continuous.

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  • $\begingroup$ please re read what I wrote, please and comment. $\endgroup$
    – user643073
    Oct 28, 2019 at 22:10
  • $\begingroup$ I meant the negation part. $\endgroup$
    – user643073
    Oct 28, 2019 at 22:17
  • $\begingroup$ @topologicalmagician yes the negation is correct. $\endgroup$ Oct 28, 2019 at 22:17
  • $\begingroup$ thank you so much $\endgroup$
    – user643073
    Oct 28, 2019 at 22:18
  • $\begingroup$ @topologicalmagician in the negation you must say that there exist x AND y...not x OR y $\endgroup$ Oct 28, 2019 at 22:20

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