# Introductory question to uniform continuity [closed]

This was an introductory question in my Analysis course.

Suppose $$f:]0,1] \rightarrow \mathbb{R}$$ is a continuous function.

Prove that if $$f$$ maps Cauchy sequences in $$]0,1]$$ to Cauchy sequences in $$\mathbb{R}$$, then $$f$$ has a continuous expansion to $$[0,1]$$.

We got some hints, but they were not that helpful...

One of them was that we had to prove/explain why the limit of $$\phi(x_n)$$ is independent of the chosen sequence $$(x_n)_n$$ if $$(x_n)_n$$ is any sequence in $$]0,1]$$ that converges to $$0$$. I intuitively get why this should be so, but I had a hard time proving this.

Could someone give this a shot? Thanks!

• Hint: 1) $\overline{]0,1]} = [0,1]$.2) If $x \in \overline{X}$, there is a sequence in $X$ that converges to $x$ 3) Define $\phi: [0,1] \rightarrow \mathbb{R}$, $\phi(x) = f(x)$, if $x \in ]0,1]$ and use 2 to define $\phi(0)$. Feb 26, 2021 at 14:14
• We had as a hint that we had to prove/explain why the limit of $\phi(x_n)$ is independent of the chosen sequence $(x_n)_n$ if $(x_n)_n$ is any sequence in $]0,1]$ that converges to $0$. That's the part I'm stuck on Feb 26, 2021 at 14:17
• @SamoGrecco That should be in the question! Feb 26, 2021 at 22:00

$$f$$ is bounded on $$[0,1]$$ as otherwise, one could find a Cauchy sequence $$\{x_n\}$$ in $$(0,1]$$ with $$\{f(x_n)\}$$ unbounded in contradiction with the fact that a Cauchy sequence is bounded.

Now by contradiction, if $$f$$ wasn't continuous, it would exist two sequences $$\{x_n\}, \{y_n\}$$ converging to zero with

$$\lim\limits_{n \to \infty} f(x_n) = l_x \neq l_y = \lim\limits_{n \to \infty} f(x_n).$$

The sequence

$$z_n = \begin{cases} x_n \text{ for } n \text{ even}\\ y_n \text{ for } n \text{ odd} \end{cases}$$

would be Cauchy while $$\{f(z_n)\}$$ would not be. A contradiction.

Therefore for every sequence $$\{x_n\}$$ converging to zero, $$\{f(z_n)\}$$ converges to a unique value $$l$$. Proving that $$f$$ can be extended to a continuous map on $$[0,1]$$ by setting $$f(0) = l$$.