How do I think about finding limits of sequences in Zariski Topology? We have the set $\mathbb{C}$ with Zariski Topology. I was thinking about limit points of sequence, say $\{b_n\}$, where $b_n$ is itself the sequence of solution of some polynomial equations $f_n(z)=z^3 - \dfrac{1}{n^3}$.  
This is How I was thinking about it:
By definition, $a$ is a limit point of the sequence if every open set containing $a$ must have all but finitely many points of the sequence. 
In Zariski topology, open sets are of the form $X-\{x_1, x_2,\ldots, x_n\}$, where x$_{i}$s are the root of polynomials $f_n(z)$. So any open set in Zariski Topology.
Now if my given point a is not a solution of given $f_n(z)$ then it is a limit point because all the open sets contain it only miss finitely many points from the sequence. Is that true? After a long hard thinking, I could not think anything else concrete.
 A: For $\mathbb{C}$ and the closed sets being the solution sets of polynomials $f(z) = \sum_{i=1}^m a_i z^i $, we have a very simple description of the topology in terms of closed sets: 
A set is $F \subseteq \mathbb{C}$ closed in the Zariski topology iff $F = \mathbb{C}$ or $F$ is finite.
Proof: left to right: either $f(z)$ is the zero polyomial (all coefficients are $0$) and then $\mathbb{C}$ is its solution set, or it has a non-zero coefficient, so is of some finite degree $n \ge 0$, and in any field the solution set will have at most $n$ members, so solution sets are finite if they're not $\mathbb{C}$.
Right to left: $\mathbb{C}$ is the solution set for the $0$-polynomial (and $\emptyset$ of the constantly $1$-polynomial, e.g.), and if $F = \{a_1,\ldots,a_m\}$ is finite, it's closed in the Zariski topology because it's the solution set (or zero-set) of $f(z) = (z - a_1)\cdot\ldots\cdot(z-a_m)$.
This topology has a name in topology: it's the cofinite topology (or the finite-closed) topology on a set.
Convergence is quite simple in this topology. The open sets are indeed the empty set and all sets $\mathbb{C}\setminus F$, where $F$ is finite.
If $x_n \rightarrow x$, then this means that for any finite set $F$, such that $x \notin F$ (so $x \in \mathbb{C}\setminus F$, which is a general open neighbourhood of $x$), for all but finitely many points of the sequence, also $x_n \notin F$ as well (because then $x_n \in \mathbb{C} \setminus F$ for all but finitely many $n$).
So in your case, the points in the sequence $(x_n)$ are all distinct, and this means that the above holds for any $x$ because for any finite set, only finitely many of them can be in the sequence at all. So your sequence converges to all points of $\mathbb{C}$ at the same time (which is very strange if you're used to metric spaces, where limits of sequences are unique).
