# How do we prove that uniformly continuous functions map Cauchy sequences onto Cauchy sequences?

Let $$X$$ be a subset of $$\textbf{R}$$, and let $$f:X\rightarrow\textbf{R}$$ be a uniformly continuous function. Let $$(x_{n})_{n=0}^{\infty}$$ be a Cauchy sequence consisting entirely of elements in $$X$$. Then $$(f(x_{n}))_{n=0}^{\infty}$$ is also a Cauchy sequence.

MY ATTEMPT

Since $$f$$ is uniformly continuous, for every $$\varepsilon > 0$$ there is a $$\delta > 0$$ such that \begin{align*} |x - y| \leq \delta \Longrightarrow |f(x) - f(y)| \leq \varepsilon \end{align*}

On the other hand, since $$x_{n}$$ is Cauchy, for every $$\delta > 0$$, there is a natural number $$N\geq 0$$ such that \begin{align*} i,j\geq N \Longrightarrow |x_{i} - x_{j}| \leq \delta \end{align*}

Hence if we substitute $$x = x_{i}$$ and $$y = x_{j}$$, it results that for every $$\varepsilon > 0$$, there is a natural number $$N\geq 0$$ such that \begin{align*} i,j\geq N \Longrightarrow |x_{i} - x_{j}| \leq \delta \Longrightarrow |f(x_{i}) - f(x_{j})| \leq \varepsilon \end{align*} and we conclude that $$(f(x_{n}))_{n=0}^{\infty}$$ is also Cauchy, and we are done.

Could someone please verify if my arguments proceed?

• The proof works – DiegoG7 Apr 30 at 17:54