1
$\begingroup$

In the exercises of Abbott's Understanding Analysis, we are asked to provide an example or a justification for the following claims:

  1. A continuous function $f : (0 ,1) \to R$ and a Cauchy sequence $(x_n)$ such that $f(x_n)$ is not a Cauchy sequence.

  2. A continuous function $f : [0 ,∞) → R$ and a Cauchy sequence $(x_n)$ such that $f(x_n)$ is not a Cauchy sequence.

Considering it true that a continuous function maps convergent sequences to convergent sequences and that a sequence in $\mathbb{R}$ is convergent iff it is Cauchy, I thought that both the claims are false.

For the second claim, I argued that since $f$ is continuous, for a sequence $(x_n)$ converging to $x$, $f(x_n)$ converges to $f(x)$. Because $f(x_n)$ converges, $f(x_n)$ is Cauchy.

But it turns out that the first claim has a valid example according to the solutions. $f(x) = 1/x$ on $(0, 1)$ and consider $(x_n) = 1/n$. Clearly, $f(x_n)$ is not convergent and hence not Cauchy.

Why is it that a convergent sequence is not being mapped to a convergent sequence? Does this have something to do with the domain being open or closed?

$\endgroup$
9
  • $\begingroup$ $(0,1)$ is not a complete metric space. $\endgroup$ Jul 8, 2020 at 16:03
  • $\begingroup$ So is the correct statement as follows? In a complete metric space, a continuous function maps a convergent sequence to a convergent sequence. $\endgroup$ Jul 8, 2020 at 16:50
  • $\begingroup$ In any metric space, a continuous function always maps convergent sequences to convergent sequences. $\endgroup$ Jul 8, 2020 at 17:02
  • $\begingroup$ Sorry, but I don't understand. Then why is it not the case with the sequence in (0, 1)? You said that it's not complete. $\endgroup$ Jul 8, 2020 at 17:04
  • $\begingroup$ That's right: $(0,1)$ is not complete. $\endgroup$ Jul 8, 2020 at 17:05

1 Answer 1

1
$\begingroup$

"A continuous map of metric spaces takes convergent sequences to convergent sequences". Let's write the precise statement of this:

Theorem $1$: Let $f: X \to Y$ be a continuous function of metric spaces. If $(s_n)$ is a sequence in $X$, converging to $x_0 \in X$, then $(f(s_n))$ is a sequence in $Y$ converging to $f(x_0)$.

In the setting of your first claim, we have a continuous function $f: X \to Y$ with $X = (0, 1)$ and $Y = \mathbf{R}$.

Sure $(x_n) = 1/n$ is a sequence in $X$ for any positive integer $n$. However, $(x_n)$ converges to $0$ which is not in $X$. So Theorem 1 above does not hold.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.