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The problem goes like this:

Determine whether the following sequences are convergent and find their limits (if they exist):

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

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

  3. $a_n=n$ 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

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    The definition of convergence is a statement about the relationship between the sequence and the open sets of the topology. So in each case think about the pen sets and try to fit that into the definition of convergence. If you get stuck ask a particular question here. – 2017-01-11
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    In the lower limit topology the open sets are $[a,b>, a,b \in \mathbb{R}$ and a sequence $(a_n)$ converges to $a \in \mathbb{R}$ if for each open ngbh $A$ of $a$ there egzists $N\in \mathbb{N}$ such that fore each $n \in \mathbb{N} , n>N , a_n \in A$. So we asume there egzists $a \in \mathbb{R}$ such that $(a_n)$ converges to $a$. Then there egzists an open nbgh of $a$, $A$ such that for each $n>=N a_n \in A$ but no such ngbh exists since for any open set $[a,b>$ there egzists $a_n$ such that $b – 2017-01-11
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    Please edit your comment into your answer and ask if it's right. It may not be . The definition you quote says "for each open neighborhood" so to prove nonconvergence you must find a particular open neighborhood that fails. You may have the right idea - I can't quite tell. – 2017-01-11
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    @ostochast You have two options. If you think the sequence converges, then you need to find a particular point $x \in \mathbb{R}$ which you think is the limit, and show that $(a_n)$ is eventually in every neighborhood of $x$. If you think it doesn't converge, then you need to show that for *every* $x \in \mathbb{R}$, there's a neighborhood of $x$ which $(a_n)$ doesn't eventually stay in. In the case of 1, you did the latter. But note that for some topologies, limits aren't unique, that is a single sequence can have multiple limits (sometimes every point in the space is a limit...). – 2017-01-11
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    HINT for (2) and (3): Show that *every* real number is a limit of the sequence in the finite complement topology, and *no* real number is a limit in the countable complement topology. – 2017-01-11

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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.