Continuous $f\colon(0,1) \to \mathbb{R}$ that is continuos but not Cauchy in $\mathbb{R}$.

Question

I'm trying to think of a function $f\colon (0,1)\to R$ which is continuous in $\mathbb R$ but is not Cauchy in $\mathbb R$. I think I'm picturing what it needs to look like, but can't think of a function that fits the area. I was thinking a function that has a vertical asymptote at $0$ and $1$ ? Maybe $\arcsin(x)$?

Answer 1

You'll want a function that wiggles around a lot, increasingly higher and more frequently as you approach some point (0 is likely easiest). Then you can make the $f(x_i)$ arbitrarily far apart, even though the $x_i$ get close to each other (and 0).

Keep in mind if you want to show a function is not Cauchy continuous, you simply have to find an example when it doesn't hold. That means you get to pick the Cauchy sequence.

Answer 2

It seems that you are looking for a function that is continuous on $(0,1)$ but it fails to send all Cauchy sequences to Cauchy sequences (which is also called Uniformly continuous) in $(0,1)$ .

In that case simply consider $f: (0,1) \to \Bbb R$ defined by, $f(x) = \frac{1}{x}$ . It's continuous on $(0,1)$.

I give 2 arguments :

1. Then consider the Cauchy sequence $\{\frac{1}{n} \}_{n=1}^\infty$ . It's image is simply, $\{n\}_{n=1}^\infty$ a diverging sequence and hence not Cauchy.

2. Or you can simply consider the Continuous extension theorem, which tells that $f$ is uniformly continuous on $(0,1)$ iff $f$ can be extended continuously to the boundary points of the interval i.e. at $x =0 $ and $x=1$ . But it is not continuous at $x=0$ !