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?


1 Answer 1


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.


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