Cauchy preserving implies continuity of function

Question

Let $f:(0,1)\rightarrow (0,1)$. Prove or disprove :

if $\{f(a_n)\}_{n=1}^{\infty}$ is a Cauchy sequence whenever $\{a_n\}_{n=1}^{\infty}$ is then $f$ is continuous.

Suggested solution from the book is the following:

If $f$ is not continuous there is in $(0,1)$ an $x_0$ and a sequence $\{x_n\}_{n=1}^{\infty}$ such that $x_n\rightarrow x_0$ and $f(x_n)\not\rightarrow f(x_0)$. If $y_{2n}=x_n$, $y_{2n-1}=x_0$ for $n=1,2,3...$ then $\{y_n\}_{n=1}^{\infty}$ is a Cauchy sequence and $\{f(y_n)\}_{n=1}^{\infty}$ is not!

I am really confused why $(y_n)$ is Cauchy and $\{f(y_n)\}$ is not. We do not even know how $f$ looks like, so how can we justify that $\{f(y_n)\}$? Thank you for all explanation in advance!

Answer 1

Because since $f(x_n)$ does not converge to $f(x_0)$ exists $s>0$ such that forall $m \in \Bbb{N}$ exists $n \geq m$ such that $|f(x_n)-f(x_0)| \geq s$

Using this and the sequence $y_n$ you defined ,you can reach a contradiction.

Answer 2

The sequence $(y_n)_{n\in\mathbb N}$ converges to $x_0$ and every convergent sequence is a Cauchy sequence.

On the other hand, $\bigl(f(y_n)\bigr)_{n\in\mathbb N}$ is not convergent, and therefore it is not a Cauchy sequence.