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
Best Answer
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.