Uniform Continuity of functions

[pasmith]

The OP may wish to note that

• a function which is continuous on a closed bounded interval is uniformly continuous, and
• a function which is (uniformly) continuous on a closed, bounded interval is uniformly continuous on any subinterval.

If these facts are not known to you, then you should prove them.

 

[WWGD]

The map $x\longmapsto 1/x$ is not uniformly continuous on $(0,1]$, as by the way, I already mentioned in my first reply in post #3. But that has nothing to do with Cauchy sequences.

Uniform continuity implies continuity which is equivalent to
$$
\displaystyle{\lim_{n \to \infty}f(x_n)=f(\lim_{n \to \infty}x_n)}
$$
that is equivalent to the use of Cauchy sequences instead of converging sequences over the reals. Your example just says that the first implication isn't reversible, as by the way, I already mentioned in my first reply in post #3. It is the non-existing limit at $0$ that matters.

The use of Cauchy sequences in this thread is simply off-topic. Yes, it makes things more convenient as you automatically have $|x_n-x_m|<\delta$ without bothering the limit $x_0.$ However, the triangle inequality solves this problem, as by the way, I already mentioned in my reply in post #25.

It does have to see with Cauchy sequences as it's the weakest condition needed to take Cauchy sequences to.Cauchy sequences. If continuity alone is enough, why doesn't $f(x)=1/x$ take the Cauchy sequence $\{1/n\}$ to a Cauchy sequence, rather than taking it to $\{ 1,2,...\}$? Issue here is that $f$ isn't even defined at $x=0$

 

[FactChecker]

This statement has nothing to do with what I said. Cauchy sequences and converging series are the same thing for real numbers, and convergence compatibility of a function is equivalent to simple continuity. There is no such thing as a "Cauchyness".

IMO, if the domain is an open set, a Cauchy sequence is simpler to talk about than convergent sequences. The OP example of $\tan(x)$ on $[0, \pi/2)$ is such a case. Furthermore, the text book's theorem, examples, and the proof attempt in post #15 are all in terms of Cauchy sequences.
I believe that it is important for a person to learn to use the proven theorems and lemmas, as stated.

 

[Mark44]

2) I had to edit your Latex to read your post.

I edited the post you referred to, so the LaTeX is fixed now.

 

[Bling Fizikst]

I am sorry for my late response .
I think i have gained some experience over time on this topic.
For the second problem :
$f(x)=\frac{\sin^2 x}{x}$

Consider the extension of $f$: $$g(x)= 0 , x=0$$ $$g(x)=\frac{\sin^2 x}{x} , x\in(0,\pi]$$
Since $g(x)$ is continuous on $[0,\pi]$ this implies $g$ is uniformly continuous on $[0,\pi]$ this implies $f$ is uniformly continuous on $(0,\pi]$

For the third problem :
$f(x)=\frac{1}{x-3}$

in $(0,3)$:

Consider $x_n=3-\frac{1}{n}$ and $y_n=3-\frac{1}{n+1}$ such that $\lim |x_n-y_n|=0$ but $$\lim |f(x_n)-f(y_n)|=|n-(n+1)|=1\neq 0$$
As we know that if there exists two sequences $x_n$ and $y_n$ in a given $S$ such that $\lim |x_n-y_n|= 0$ and $\lim|f(x_n)-f(y_n)|\neq 0$ then $f$ is NOT unifomrly continuous on $S$
Hence , $f$ is NOT uniformly continuous here .

in $(4,\infty)$:
this means $\frac{1}{x-3}\in(0,1)$:
Choose $\delta=\epsilon$
$$|\frac{1}{x-3}-\frac{1}{y-3}|=\frac{|x-y|}{|x-3||y-3|}<|x-y|<\epsilon$$
Hence , $f$ is uniformly continuous in $(4,\infty)$

in $(3,\infty)$:
Consider $x_n=3+\frac{1}{n}$ and $y_n=3+\frac{1}{n+1}$
$$\lim |x_n-y_n|=0$$
$$\lim|f(x_n)-f(y_n)|=|n-(n+1)|=1\neq 0$$
Hence , $f$ is NOT uniformly continuous in $(3,\infty)$

 

[PeroK]

I am sorry for my late response .
I think i have gained some experience over time on this topic.
For the second problem :
$f(x)=\frac{\sin^2 x}{x}$

Consider the extension of $f$: $$g(x)= 0 , x=0$$ $$g(x)=\frac{\sin^2 x}{x} , x\in(0,\pi]$$
Since $g(x)$ is continuous on $[0,\pi]$ this implies $g$ is uniformly continuous on $[0,\pi]$ this implies $f$ is uniformly continuous on $(0,\pi]$

For the third problem :
$f(x)=\frac{1}{x-3}$

in $(0,3)$:

Consider $x_n=3-\frac{1}{n}$ and $y_n=3-\frac{1}{n+1}$ such that $\lim |x_n-y_n|=0$ but $$\lim |f(x_n)-f(y_n)|=|n-(n+1)|=1\neq 0$$
As we know that if there exists two sequences $x_n$ and $y_n$ in a given $S$ such that $\lim |x_n-y_n|= 0$ and $\lim|f(x_n)-f(y_n)|\neq 0$ then $f$ is NOT unifomrly continuous on $S$
Hence , $f$ is NOT uniformly continuous here .

in $(4,\infty)$:
this means $\frac{1}{x-3}\in(0,1)$:
Choose $\delta=\epsilon$
$$|\frac{1}{x-3}-\frac{1}{y-3}|=\frac{|x-y|}{|x-3||y-3|}<|x-y|<\epsilon$$
Hence , $f$ is uniformly continuous in $(4,\infty)$

in $(3,\infty)$:
Consider $x_n=3+\frac{1}{n}$ and $y_n=3+\frac{1}{n+1}$
$$\lim |x_n-y_n|=0$$
$$\lim|f(x_n)-f(y_n)|=|n-(n+1)|=1\neq 0$$
Hence , $f$ is NOT uniformly continuous in $(3,\infty)$

That all looks good to me.