5
$\begingroup$

Take the topology on $\Bbb R$, the real line, which is, $\tau=\{A\subseteq\Bbb R\mid\Bbb R\setminus A\text{ is countable}\}\cup \{\varnothing\}$. Can one find a convergence sequence in this topology? Because, Take a sequence $\{a_n\}$, And suppose $a=\lim a_n$. Take now the set $\Bbb R\setminus\{a_n\}$. Then this is an open neighborhood of $a$ that doesn't contain any of $\{a_n\}$ which is a contradiction...

It feels like I am missing something basic in here but I can't put my finger on it.

Thanks, Shir

$\endgroup$
1
  • 4
    $\begingroup$ $1, 1, 1, 1, \ldots$ still converges. $\endgroup$ Jul 14 '13 at 13:33
5
$\begingroup$

Be careful, if $a=a_n$ for some $n\in\Bbb N$, then the set $\Bbb R-\{a_n\mid n\in\Bbb N\}$ won't contain $a$ itself. But you can still take $(\Bbb R-\{a_n\mid n\in\Bbb N\})\cup\{a\}$. This is always a neighborhood of $a$, and if the sequence isn't eventually constant, then it is not eventually in this neighborhood, so it does not converge to $a$. So the only convergent sequences are eventually constant.

Also note that a convergent sequence has a unique limit, even though the space $(\Bbb R,\tau)$ is not Hausdorff. This means that we have an example of a space where the sequences do not completely determine the topology. For example, the set $A:=[0,\infty)$ is sequentially closed (every convergent sequence in $A$ has its limit within $A$) but not closed.

$\endgroup$
6
  • 2
    $\begingroup$ Isn't every non-empty set is sequentially closed? $\endgroup$
    – Asaf Karagila
    Jul 14 '13 at 13:51
  • 1
    $\begingroup$ @AsafKaragila: You right, thank you! But I think even the empty set is sequentially closed :-) $\endgroup$ Jul 14 '13 at 13:54
  • 1
    $\begingroup$ Well it's closed. $\endgroup$
    – Asaf Karagila
    Jul 14 '13 at 13:56
  • $\begingroup$ Argh! And every closed set is sequentially closed, of course :) $\endgroup$ Jul 14 '13 at 13:58
  • $\begingroup$ Yes, of course! :-) $\endgroup$
    – Asaf Karagila
    Jul 14 '13 at 14:00
2
$\begingroup$

Stefan's answer is great. Here's an exercise relevant to your question which I think is worth thinking about:

Exercise 1: Let $\{a_n\}_{n\geq 1}$ be a sequence in the finite complement topology on $\mathbb{R}$. Under what conditions does $\{a_n\}_{n\geq 1}$ converge and how many limits does it have?

The situation is, in some vague sense, "dual" to the case you're considering of the countable complement topology. I hope this exercise will be fun to think about for you!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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