# Cauchy sequence of a uniformly continuous image

If $$\{fx_n\}$$ is a Cauchy sequence with $$f$$, a continuous self map on a complete metric space, we know that $$\{x_n\}$$ need not be Cauchy. Is it true for a uniformly continuous $$f$$?

Here's my take: Suppose $$\{x_n\}$$ is not Cauchy, then for some $$\alpha>0$$, $$d(x_n,x_m)\geq\alpha$$, and since $$f$$ is uniformly continuous, $$\forall\varepsilon>0$$ there exists $$\delta>0$$ with $$d(fx_n,fx_m)<\varepsilon$$ whenever $$d(x_n,x_m)<\delta$$.

Choosing $$\delta=\alpha$$, we get $$d(fx_n,fx_m)\geq\varepsilon^\prime$$ for some $$\varepsilon^\prime>0$$, a contradiction.

I believe there might be some flaws in this argument.

And one more question, if not, then under what minimum additional condition on $$f$$, like injective?

On $$[-1,1]$$ let $$f(x)=x^{2}$$ and $$x_n=(-1)^{n}$$. Then $$f(x_n)$$ is Cauchy and $$f$$ is uniformly continuous but $$x_n$$ is not Cauchy.
If $$f$$ is injective and $$X$$ is compact then the conclusion holds.
Proof: $$f(x_n)$$ converges to some point $$y$$. If $$x$$ is any limit point of $$(x_n)$$ then there is a subsequence converging to $$x$$ and continuity of $$f$$ gives $$f(x)=y$$. Since $$f$$ is injective this shows that you cannot have more then one limit point. Hence $$(x_n)$$ is convergent, therefore Cauchy.
• Thanks, you have been a God sent for my Cauchy problems. Another querry: If $f$ is bijective and continuous in the original question, then $\{x_n\}$ is Cauchy. Am I right? – mark haokip May 2 at 10:38