# How do I disprove uniform continuity in this particular case?

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?

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.