# If $f:X\to \mathbb R$ is a continuous mapping, then it maps cauchy sequences into cauchy sequences.

Let $$(X,d)$$ be a compact metric space. Is the following statement true?

If $$f:X\to \mathbb R$$ is a continuous mapping, then it maps cauchy sequences into cauchy sequences.

I think this statement is incorrect. Consider $$X=[0,1]$$ and $$d=|x-y|$$, then $$(X,d)$$ is a compact metric space. Now, consider the function $$f(x)=\frac{1}{x}$$, which is continuous and consider the sequence $$x_n=\frac{1}{n}$$. Then $$x_n$$ is cauchy sequence in $$X$$, but $$f(x_n)=n$$ in $$\mathbb R$$ is a diveregent sequence, hence not cauchy.

Is my argument correct? Also, if it is true can you please prove it.

• Thou shalt not divide by zero. – Angina Seng Oct 11 '20 at 7:17
• $f(x)=1/x$ is not continuous on $[0,1]$. – Martin R Oct 11 '20 at 7:18
• Thanks for the clarification @AnginaSeng and Matrin R, i have realised my mistake. I think the soln below is correct. – s1mple Oct 11 '20 at 7:19

## 3 Answers

Let $$\{x_n\}$$ be a cauchy sequence in $$X$$. Since $$X$$ is complete, let $$x_n \to x$$. Now, since $$f$$ is continuous, $$f(x_n) \to f(x)$$, which implies $$\{f(x_n)\}$$ is cauchy.

Since $$f$$ is continous and $$X$$ a compact metric space $$f$$ is uniformly continuos.

Let $$\epsilon>0$$ be given.

There is a $$\delta >0$$ s.t. $$|x-x'|<\delta$$

implies $$|f(x)-f(x')|<\epsilon.$$

Since $$x_n$$ is Cauchy there is a $$n_0$$ s.t.

$$|x_n-x_m|<\delta$$ for $$m\ge n\ge n_0$$

implies $$|f(x_n)-f(x_m)| < \epsilon,$$ i. e.

$$f(x_n)$$ is Cauchy.

There is the following interesting parallelism. While a continuous function between metric spaces transforms convergent sequences into convergent sequences, a uniformly continuous function between metric spaces transforms Cauchy sequences into Cauchy sequences.

Indeed, let $$(X,d), (X',d')$$ be metric spaces, and $$\;f : X\to X'\;$$ a unifomly continuous function. Then, along the lines of what Peter Szilas already observed,

$$\forall\,\epsilon>0,\;\exists\,\delta>0\; \text{ s.t. }\; d(x,y)\leq \delta \;\implies\; d'(f(x),f(y))\leq \epsilon, \tag{*}\label{formula}$$

and therefore, if $$(a_n)$$ is a Cauchy sequence in $$X$$, as

$$\forall\,\epsilon>0,\;\exists\,\nu\in\Bbb N \;\text{ s.t. }\; k,h\geq\nu \;\implies\; d(a_k,a_h)\leq\delta,$$

it follows, by $$\eqref{formula}$$:

$$\forall\,\epsilon>0, \,\exists\,\nu\in\Bbb N \;\text{ s.t. }\;k,h\geq\nu\;\implies\; d'(f(a_k),f(a_h))\leq\epsilon,\qquad \text{q.e.d.}$$

In your case, just note that a continuous function defined on a compact set is uniformly continuous.