For a finite complement topology , to which point or points does the sequence converge? For a finite complement topology on $R$  to which point or points does the sequence ${\frac{1
}{n}}$ converge?
For a finite complement topology on real numbers the only set including the limit point of sequence is $R$ so the sequence converges to every point of $R$. Am I right?
 A: Perhaps the following is what you tried to convey in your question:
Take any point $\;r\in\Bbb R\;$ and  let $\;U_r\;$ be any open neighborhood of it, which means that $\;X:=\Bbb R\setminus U_r\;$ is finite and thus there exists $\;M\in\Bbb N\;$ such that $\;m>M\implies \frac1m\notin X\;$, which means that
$$\forall\,m>M\,,\,\,\frac1m\in U_r\implies r=\lim_{n\to\infty}\frac1n$$
A: Let $X$ be a topological space, $\langle x_n:n\in\Bbb N\rangle$ a sequence of points of $X$, and $x\in X$. 

Definition: $\langle x_n:n\in\Bbb N\rangle$ converges to $x$ if and only if for each open nbhd $U$ of $x$ there is an $m_U\in\Bbb N$ such that $x_n\in U$ whenever $n\ge m_U$. 

Now let $x\in\mathbb R$ be an arbitrary element and let $U$ be an open nbhd of $x$. 
Then $U^c$ is finite so some $m_U\in\mathbb N$ exists such that $\frac1n\notin U^c$ whenever $n\geq m_U$.
That can be refrased as: some $m_U\in\mathbb N$ exists such that $\frac1n\in U$ whenever $n\geq m_U$.
So according to the definition sequence $\langle \frac1n:n\in\Bbb N\rangle$ converges to $x$. 
Here $x$ was taken arbitrary so $\langle \frac1n:n\in\Bbb N\rangle$ converges to every $x\in\mathbb R$.
A: Yes, any sequence of distinct
elements converges to every point.  A sequence with a
finite number of distinct elements, converges iff it's
eventually constant.
