A question about converging sequences in topological spaces The problem goes like this:
Determine whether the following sequences are convergent and find their limits (if they exist):


*

*$a_n=n$ in $\mathbb{R}$ with the lower limit toplology.

*$a_n=n$ in $\mathbb{R}$ with finite complement topology.

*$a_n=n$ in $\mathbb{R}$ with countable complement topology.
I know the definition of convergence in a topological space but for some reason I can't seem to apply it. Can somebody give me a hint or solve one of the upper problems so I get the general idea?
Here is my attempt at 1.
In the lower limit topology the open sets are $[a,b)$ where $a,b \in \mathbb{R}$ 
and a sequence $(a_n)$ converges to $a \in \mathbb{R}$ if for each open nbhd $A$ of $a$ there exists $N \in \mathbb{N}$ such that fore each $n \in \mathbb{N}, n>N, a_n \in A$. So we assume there exists $a \in \mathbb{R}$ such that $(a_n)$ converges to $a$. Then there exists an open nbhd of $a$, $A$ such that for each $n \geq N, a_n \in A$ but no such nbhd exists, since for any open set $[a,b)$ there exists $a_n$ such that $b<a_n$ so the sequence does not converge.
Is my reasoning correct?
 A: You mixed up your quantifiers in your attempt at number 1. I'll try to point out all the mistakes, and see if you can take it from there.
The half-open intervals $[a, b)$ are a basis for the lower-limit topology. The open sets are in fact arbitrary unions of such intervals.
In order to show that no limit exists, you use a proof by contradiction, which is the right approach. So suppose there is an $a \in \mathbb{R}$ to which your sequence converges.
Now you say that this means there exists a neighborhood of $a$ that contains a tail of the sequence (a tail is all elements past a certain point). This is true, but isn't strong enough to show $a$ isn't the limit, because as mentioned above the open sets are in fact unions of half-open intervals, so $[a, \infty)$ is an open neighborhood of $a$, and it does contain a tail of the sequence. So no contradiction is to be found here.
But in fact, if the sequence converges to $a$, then all open neighborhoods of $a$ contain a tail of the sequence. So in order to find a contradiction, you just have to find one neighborhood of $a$ that fails to contain a tail. Indeed, any basis element containing $a$ will do.
So the whole proof would go as follows:
We proceed by contradiction. Suppose the sequence converges to $a \in \mathbb{R}$. Then every neighborhood of $a$ must contain a tail of the sequence. But if we let $b > a$, the half-open interval $[a, b)$ is a neighborhood of $a$, and $a_n$ is eventually greater than $b$, hence $[a, b)$ does not contain a tail of the sequence. This is a contradiction, so the sequence does not converge to $a$. Since $a$ was arbitrary, it does not converge at all.
