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
Examples for all of the above?
Could anyone please help?
Best Answer
You are correct for the first example.
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.
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.