# Real Analysis: Cauchy, continuous sequence examples

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

a continuous function $f: [0,1] \to\mathbb{R}$ and a Cauchy sequence $(x_n)$ such that $f(x_n)$ is not a Cauchy sequence

a continuous function $f: [0,\infty ) \to\mathbb{R}$ and a Cauchy sequence $(x_n)$ such that $f(x_n)$ is not a Cauchy sequence

• What have you tried? What do you know about uniformly continuous functions and how they treat cauchy sequences? – Prahlad Vaidyanathan Dec 3 '13 at 4:40
• Just think of f(x)=1/xforx∈(0,1)and x_n =1/n for n∈N. f(x_n) =n. Does it work? For question 2 and 3 my example will fail because it is undefined at 0. Are those two possibles have a function? – afsdf dfsaf Dec 3 '13 at 4:58

1. You are correct for the first example.

2. It is not possible, because if $f:[0,1]\to \mathbb{R}$ is continuous, then it is uniformly continuous. Suppose $(x_n)$ is Cauchy, and $\epsilon > 0$, there is $\delta > 0$ such that $$|x-y| < \delta \Rightarrow |f(x) - f(y)| <\epsilon \quad\forall x,y\in [0,1]$$ For this $\delta > 0, \exists N\in \mathbb{N}$ such that $$|x_n - x_m| < \delta \quad\forall n,m \geq N$$ and so conclude that $f(x_n)$ is a Cauchy sequence.

3. Again, this is not possible, because if $(x_n)$ is Cauchy in $[0,\infty)$, then it converges in $\mathbb{R}$, since $\mathbb{R}$ is complete. Since $[0,\infty)$ is closed, there is a point $x \in [0,\infty)$ such that $x_n \to x$. Since $f$ is continuous, $f(x_n) \to f(x)$ and so $f(x_n)$ must be a Cauchy sequence.

• @MhenniBenghorbal : Could you give an example/explain why my explanation is wrong? Maybe I am missing something, but a Cauchy sequence is bounded, so the function is effectively uniformly continuous on the (large enough) closed interval containing the sequence, right? – Prahlad Vaidyanathan Dec 3 '13 at 6:45
• for (3) you need the function to be uniformly continuous. – Mhenni Benghorbal Dec 3 '13 at 8:48
• if $(x_n)$ is cauchy in $[0,\infty)$ then shouldn't it be case that it converges in $[0,\infty)$ too(coverges in $\mathbb{R}$ is too weak statement) By order limit theorem? – viru Jul 6 at 17:00