Find limit of sequences in topological space Find limit of sequences $(x_n) = \frac{1}{n}, (y_n) = -\frac{1}{n}, (z_n) = (-1)^n$ in topological space $\tau = \{(a, +\infty), a\in \mathbb{R}\} $
For the first one the limit is A = $(-\infty, 0]$.
For the second one the limit is B = $(-\infty, 0]$
. For the third one the limit is C = $(-\infty, -1]$
I think those are correct answers, but what are the answers if there is a diffrent topology $T = \{U \in \mathbb{R};  \mathbb{R}\setminus U  <\aleph_0\}$?
I think answer for all those sequences is $\mathbb{R}$
 A: The answers for $\tau$ are correct. For the co-finite topology $T$, the answer can be found here.
Consider a sequence $(x_n)_n$. First assume that $|\{n:x_n=c\}|<\infty$ for all $c\in\mathbb R$. For $x\in\mathbb R$ and $U\in T$ with $x\in U$, so $|\mathbb R\setminus  U|<\infty$. Thus, there exists $N$ such that $x_n\in U$ for all $n\ge N$. By definition, $x$ is a limit point and hence the limit is $\mathbb R$.
This covers the first two examples.
For the last example, consider the remaining case that there exists a value $x^*$ which is taken infinitely many times. Further, assume there is a second value $x^\circ$ which is also taken infinitely many times. For $x\in\mathbb R\setminus\{x^*\}$ consider $U=\mathbb R\setminus\{x^*\}$, then for all $N$ there exists $n\ge N$ with $x_n=x^*\not\in U$ by definition, so $x$ is not a limit. By symmetry of $x^*$ and $x^\circ$ we obtain that $x^*$ is not a limit, so there are no limits here. This covers the last example.
For completeness, consider the case that there exists exactly one value $x^*$ that is taken infinitely many times. As before, we see that $\mathbb R\setminus\{x^*\}$ are not limit points. For an open set $U\in T$ with $x^*\in U$ we have $x_n\in U$ for all sufficiently large $n$ since the finitely many points $\mathbb R\setminus U$ are only visitied a finite number of times. This shows that $x^*$ is the unique limit in this case. An example sequence that shows this behavior would be $x_n=\mathbf{1}\{n\in 2\mathbb Z\}\frac{1}{n}$, with unique limit $0$.
A: By $T$ you mean co-countable topology over the reals? Note that in this topology a sequence converges only if it is (eventually) constant: Assume that, for contradiction, $(x_n)$ is a non-constant sequence that converges to some $r\in R$. We always have some open set $U=R-{(x_n)}$ and $U\cup{r}$ is an open set containing $r$ , but not any other terms of $(x_n)$. Thus, $r$ cannot be a limit of the sequence , the sequences you define has no limits as they are not eventually constant.
