Characterization of uniform continuity via sequence

Question

For $f(x)$, defined on the interval $X$, $f(x)$ is uniformly continuous in X if and only if for every sequences $x_{n}$, $y_{n}\in X$, when we have $\lim_{n\rightarrow\infty}(x_{n}-y_{n})=0$, then $\lim_{n\rightarrow\infty}[f(x_{n})-f(y_{n})]=0$.

For this characterization, I am confused with three point.

1. It is shown on my textbook that, for $f(x)=x^2$, defined in $[0,+\infty)$, $x_{n}=\sqrt {n+1}$, $y_{n}=\sqrt{n}$ were chosen. I wonder why not to choose some apparent sequence such as $x_{n}=1/n$, $y_{n}=1/2n$.

   In my opinion, I think that $1/n$ and $1/2n$ are not defined on zero point.

2. For function $f(x)=1/x$, defined in $(\xi,1)$, $0\lt\xi\lt1$), I wonder why it's unwarranted to use $x_{n}=1/n$, $y_{n}=1/2n$.

   As far as I am concerned, we need to confirm $\lim_{n\rightarrow\infty}[f(x_{n})-f(y_{n})]$ and if $x_{n}$ is defined in $(\xi,1)$ then $n$ should be in $(1,1/\xi)$. Finally, $n$ can't tend to $\infty$.

3. I noticed that this characterization mention that for all $x_{n}$, $y_{n}$. What if we want to prove uniform continuity for certain function? How can we choose all suited sequences?

thanks.

Answer 1

For 1., your textbook is presumably showing that $f$ is not uniformly continuous on $[0,\infty)$. To prove this, they would show that the property in your characterization of uniform continuity does not hold. Towards this end, sequences $(x_n)$ and $(y_n)$ have to be found such that $x_n-y_n$ tends to $0$ but $f(x_n)-f(y_n)$ does not tend to $0$. The sequences you wish to use do satisfy $f(x_n)-f(y_n)\rightarrow0$ and would not serve to demonstrate that $f$ is not uniformly continuous. The sequences in the text, however, do.

For 2., it's unwarranted to use your sequences simply because not all of its terms are in the given interval $(\zeta,1)$ (for $n$ sufficiently large, $x_n=1/n<\zeta$; so, for such $n$, $x_n\notin(\zeta,1)$). Further, if, as you seem to suggest, you just look at the $x_n$ and $y_n$ that are in $(\zeta,1)$, you no longer have $x_n-y_n\rightarrow0$. In fact, you would no longer have infinite sequences, as needed.

In fact, the function $f(x)=1/x$ is uniformly continuous on $(\zeta,1)$, $1>\zeta>0$. To prove this (and to address your third question), you have to show that whenever $(x_n)$ and $(y_n)$ are sequences in $(\zeta,1)$ with $x_n-y_n\rightarrow0$, we necessarily have $f(x_n)-f(y_n)\rightarrow 0$. You "choose all suited sequences" by fixing (arbitray) sequences $(x_n)$ and $(y_n)$ in $(\zeta,1)$ with $x_n-y_n\rightarrow0$ and then demonstrating that you must have $f(x_n)-f(y_n)\rightarrow0$. So, let $(x_n)$ and $(y_n)$ have this property. Now you have to somehow prove that $f(x_n)-f(y_n)\rightarrow0$. If you can do this, you will have shown what $f$ is indeed uniformly continuous on $(\zeta,1)$.

I presume your text does show that the function $f$ of the preceding paragraph is uniformly continuous on the given interval. If not, a hint here is that the quantity $1/(x_ny_n)$ is bounded away from $0$ on $(\zeta,1)$ and ${1\over x_n}-{1\over y_n}={y_n-x_n\over x_n y_n}$. From this it will follow that $f(x_n)-f(y_n)$ tends to $0$, since $x_n-y_n$ does.