Prove or disprove: If $(x_n)$ is a Cauchy sequence in $X$ and $f$ is a continuous function on $X$, then $(f(x_n))$ is a Cauchy sequence.

Question

Note that $X$ is a metric space and $f:(X,\rho)\rightarrow (X,\sigma)$.

Disproving: Suppose $(x_n)$ is a Cauchy sequence, $f$ is continuous and $(f(x_n))$ is not a Cauchy sequence, then $\exists\epsilon'>0$ such that $\forall M\in\Bbb{N}$, there exists $i,j\geq M$ where $\sigma(f(x_i),f(x_j))>\epsilon'$. Also, knowing that $f$ is continuous, then we know that $\exists\delta'>0$ such that if $\rho(x_i,y)<\delta'$, such that $y\in X$, then $\sigma(f(x_i),f(y))<\epsilon'$. However, $y$ is not necessarily a term of the Cauchy sequence $(x_n)$. Hence, it does not necessarily mean that $(f(x_n))$ is also a Cauchy sequence.

Now, I just noticed after typing that this may get tagged as duplicate since the same question has been asked here: If $X = \{x_n:n \in \mathbb N\}$ is a cauchy sequence in a metric space $S$ and $f : S \rightarrow T$ is continuous , is $f(x_n)$ a cauchy sequence?, but would my argument be valid in general? Thank you!

Answer 1

Another counterexample that goes into a slightly different direction.

Choose $X=\mathbb Q$,

$$f(x)= \left\{ \begin{matrix} 0 & \text{, if } x < \sqrt{2}\\ 1 & \text{, if } x > \sqrt{2}. \\ \end{matrix} \right. $$

This is a continuous function $\mathbb Q \to \mathbb Q$.

If you choose a sequence $\{x_n\}$ of rationals that tends to $\sqrt{2}$ from both sides (infinitely many terms both above and below $\sqrt{2}$), then $\{x_n\}$ is Cauchy, but $\{f(x_n)\}$ is not, as it will contain infinitely many 0's and 1's.

The main idea is to realize that your proposition becomes true when $X$ is a complete metric space, so both my example and that of Salahamam Fatima use a non-complete space.

Answer 2

No, it is not correct. You start talking about $y$ without telling what $y$ is. But, above all, your approach only tells us why you think that the sequence $\bigl(f(x_n)\bigr)_{n\in\mathbb N}$ might not be a Cauchy sequence. The best approach, in this case, consists in providing a counterexample.

Answer 3

Take $$X=(0,+\infty)$$

$$x_n=\frac 1 n$$

$$f:x\mapsto \frac{1}{x}$$

$$(x_n) \text{ is Cauchy}$$

$$f \text{ is continuous}$$

$$f(x_n)=n \text{ is not Cauchy}$$

Your statement will be true if $f$ is UNIFORMLY CONTINUOUS.