# Confusion about convergent sequences getting mapped to convergent sequences by a continuous function.

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?

• $(0,1)$ is not a complete metric space. Jul 8, 2020 at 16:03
• So is the correct statement as follows? In a complete metric space, a continuous function maps a convergent sequence to a convergent sequence. Jul 8, 2020 at 16:50
• In any metric space, a continuous function always maps convergent sequences to convergent sequences. Jul 8, 2020 at 17:02
• 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. Jul 8, 2020 at 17:04
• That's right: $(0,1)$ is not complete. Jul 8, 2020 at 17:05

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.