# Uniformly Continuous proof verification

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 $$(0,1)$$?

• Indeed, you are right, it is not correct. – Alexey Oct 28 '19 at 22:03
• If you have found a counterexample for $\delta$,then it is also a counterexample for $\delta' > \delta$. So, you can always assume that $\frac{\delta}{1+\delta}+\delta < 1$. – amsmath Oct 28 '19 at 22:12
• @amsmath is the negation correct though? – monoidaltransform Oct 28 '19 at 22:13
• The "or there exists" does not make sense. Think about it. – amsmath Oct 28 '19 at 22:18
• @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? – monoidaltransform Oct 28 '19 at 22:19

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.

• 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. – monoidaltransform Oct 28 '19 at 22:31
• Yes, I will add to my answer. – Matematleta Oct 28 '19 at 22:32

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.

• please re read what I wrote, please and comment. – monoidaltransform Oct 28 '19 at 22:10
• I meant the negation part. – monoidaltransform Oct 28 '19 at 22:17
• @topologicalmagician yes the negation is correct. – Marios Gretsas Oct 28 '19 at 22:17
• thank you so much – monoidaltransform Oct 28 '19 at 22:18
• @topologicalmagician in the negation you must say that there exist x AND y...not x OR y – Marios Gretsas Oct 28 '19 at 22:20