0
$\begingroup$

A question in MIT archives goes like this:

a) Let $f : X\to Y$ be a uniformly continuous function between metric spaces. Show that if $(x_n)_{n=1}^{\infty}$ is a Cauchy sequence in $X$, then $(f(x_n))_{n=1}^{\infty}$ is a Cauchy sequence in $Y$ . Show, therefore, that the function $f(x)=\frac{1}{x^2}$ defined on $(0,\infty)$ is not uniformly continuous.

I just want to make sure what I have done is correct since I am using this material to teach myself.

My Attempt: Since, $\{x_n\}$ is a Cauchy sequence in $X$, it follows $\forall\ \ n,m\ge N; N\in\mathbb{N}$, $d_X(x_n-x_m)<\delta$. Given that $f(x)$ is uniformly continuous, it follows that $$\exists\epsilon>0,\forall\ \ \delta>0;d_Y(f(x_m),f(x_n))<\epsilon \text{ given that } d_X(x_m,x_n)<\delta$$ It follows therefore, that $f(x_n)$ is a Cauchy sequence in $Y$.

I am not sure if I have the right idea about proving the second part though, i.e., proving $f(x)=\frac{1}{x^2}$ is not uniformly continuous on $(0,\infty)$ using Cauchy sequences. My reasoning is this: Say a sequence $\{x_n\}=\frac{1}{n}, n\in\mathbb{N}$ spanning $(0,1]$ is Cauchy. Hence, for $n,m\ge N,\ \ |x_n-x_m|<\delta$. However, $|f(x_n)-f(x_m)|=|m^2-n^2|=|(m-n)(m+n)|\ge (m+n)$. We can choose $m,n$ to make $|x_n-x_m|$ arbitrarily small while $fx_n-fx_m$ is arbitrarily larger. Therefore, $\{f(x_n)\}$ is not Cauchy for Cauchy sequence $\{x_n\}$ and hence, not uniformly continuous.

Is that OK? Or is there a way to disprove uniform continuity more directly using Cauchy sequences?

$\endgroup$

1 Answer 1

1
$\begingroup$

You're sort of doing the whole $\varepsilon$-$\delta$ thing backwards. Start from the definitions. You want to show that $(f(x_n))_{n\in \mathbb{N}}$ is Cauchy. That means, you should let $\varepsilon>0$ be given and attempt to parry it.

Now, you can use that $f$ is uniformly continuous to say that there exists $\delta>0$ such that $d_X(x,y)<\delta$ implies $d_Y(f(x),f(y))<\varepsilon$, and then, you can use that $(x_n)_{n\in\mathbb{N}}$ is Cauchy to get some $N$ such that $n,m\geq N$ implies $d_X(x_n,x_m)<\delta$. Adding it all together, for $n,m\geq N$, we have $d_Y(f(x_n),f(x_m))<\varepsilon$. Since $\varepsilon$ was arbitrary, we get the result.

You've got the right idea for showing that $f(x)=\frac{1}{x^2}$ isn't Cauchy. Here again, your argument is sort of messy. You want to show that $(x_n)_{n\in\mathbb{N}}$ is Cauchy. This follows, since, given $\varepsilon>0,$ we have $|x_n-x_m|\leq \frac{1}{n}+\frac{1}{m}\leq \frac{2}{\min\{n,m\}}<\varepsilon$ for $n,m\geq \frac{2}{\varepsilon}$. Now, you want to argue that $(f(x_n))_{n\in\mathbb{N}}$ is not Cauchy. It's probably easiest to just argue that $|f(x_n)-f(x_{n+1})|=2n+1\geq 1,$ which is never smaller than $\varepsilon=\frac{1}{2}$, so the sequence can't be Cauchy.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.