0
$\begingroup$

I need to show that the expression 'is a cauchy sequence' is not a topological expression.

I could show this by finding a homeomorphic function $\phi$ such that a Cauchy sequence is mapped onto a sequence which is not Cauchy.

I have no idea how to find this example. Could someone give me some pointers?

$\endgroup$
2
  • 2
    $\begingroup$ $(-\pi/2,\pi/2)$ is homeomorphic to $\Bbb R$. $\tan$ witnesses this. Look there. $\endgroup$ Commented Jun 5, 2015 at 11:18
  • $\begingroup$ @DavidMitra Would something like $\{ \frac{\pi}{2} - \frac{1}{n}\}_{n\in \mathbb{N}}$ with $\phi: ]\frac{-\pi}{2}, \frac{\pi}{2}[ \to \mathbb{R}; x\mapsto \tan(x)$ do? $\endgroup$
    – dietervdf
    Commented Jun 5, 2015 at 16:43

1 Answer 1

1
$\begingroup$

Let $X = (0,\infty)$, and consider: $\Phi: X \longrightarrow X$, $\Phi(t) = 1/t$.

Then, $\Phi$ is a homeomorphism. However, $\{1/n\}_n$ is cauchy in X and $\{f(x_n) = n \}_n$ is not Cauchy.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .