The topology of countable complements

Question

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

$1, 1, 1, 1, \ldots$ still converges. – Chris Eagle Jul 14 '13 at 13:33 — Chris Eagle, Jul 14 '13 at 13:33

Stefan Hamcke · Answer 1 · 2013-07-14T13:48:26.447

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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.

edited Jul 14 '13 at 13:48

answered Jul 14 '13 at 13:38

Stefan Hamcke

27,733

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Isn't every non-empty set is sequentially closed? – Asaf Karagila Jul 14 '13 at 13:51
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@AsafKaragila: You right, thank you! But I think even the empty set is sequentially closed :-) – Stefan Hamcke Jul 14 '13 at 13:54
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Well it's closed. – Asaf Karagila Jul 14 '13 at 13:56
Argh! And every closed set is sequentially closed, of course :) – Stefan Hamcke Jul 14 '13 at 13:58
Yes, of course! :-) – Asaf Karagila Jul 14 '13 at 14:00
I see, this is exactly why I asked it, because the space is not hausdorff.. I was thinking only of sequences that doesn't contain their limit.. Thanks! – topsi Jul 14 '13 at 17:26

score 2 · Answer 2 · answered Jul 14 '13 at 13:54

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!

The topology of countable complements

2 Answers2