3
$\begingroup$

Suppose I have a set with two metrics, which induce distinct topologies, (so neither is contained in the other). There should exist a sequence which converges in both topologies, but to different points (otherwise all sequences converge to the same point in both metrics, which then implies the topologies are equal). I'm having trouble coming up with such an example.

Can anyone kindly provide a hint, or is this a bad question?

Thanks!

$\endgroup$
  • 2
    $\begingroup$ When you say "a hint" - where does this question come from? $\endgroup$ – Yemon Choi Sep 30 '12 at 7:16
  • 1
    $\begingroup$ This is a simple exercise. Please see the FAQ for other sites where problems like this might fit. Also, part of the statement is wrong. $\endgroup$ – Douglas Zare Sep 30 '12 at 8:24
5
$\begingroup$

It is a "bad question", in that the assumption that neither metric topology is contained in the other does not imply that there is a sequence converging in both topologies, but to different points. Example: on the set $\mathbb{Z} \cup \{+\infty\}\cup\{-\infty\}$ consider the metric topology $\tau_1$ where all points are open but $+\infty$ , and $+\infty$ is the limit of the sequence $x_n:=n$; also, the metric topology $\tau_2$ where all points are open but $-\infty$ , and $-\infty$ is the limit of the sequence $x_n:=-n$ . Here, any sequence that converges in both topologies is eventually constant, so the limit is the same, and there are sequences converging in either topology and non-converging in the other.

| cite | improve this answer | |
$\endgroup$
5
$\begingroup$

Take $\mathbb{Q}$ and consider two different norms on it: one is simply Euclidean and another is $p-$ adic. Take the sequence $x_n=\frac{p^n}{p^n-1}.$ Clearly $x_n\to 1$ in Euclidean norm and $x_n\to 0$ in $p-$ adic.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ This is much more complicated than necessary. $\endgroup$ – Douglas Zare Sep 30 '12 at 23:23

Not the answer you're looking for? Browse other questions tagged or ask your own question.