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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!

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  • $\begingroup$ 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 $\phi(0)$. $\endgroup$ Feb 26, 2021 at 14:14
  • $\begingroup$ 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 $\endgroup$ Feb 26, 2021 at 14:17
  • $\begingroup$ @SamoGrecco That should be in the question! $\endgroup$ Feb 26, 2021 at 22:00

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$f$ 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 $f(0) = l$.

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