I don't think that there is a useful way to do what you ask (namely to work with
limits/convergence); as other answers have explained, the non-Hausdorff nature of the Zariski topology (among other things) makes this difficult.
On the other hand, most basic lemmas in topology/analysis which can be proved via
a consideration of convergent sequences can normally also be proved via arguments with open sets instead, and so your intuition for the topology of metric spaces can to some extent be carried over, if you are willing to make these sort of translations. At some point (if you practice), and with a bit of luck, the translation will become automatic (or at least close to automatic).
(Although you may
think of non-Hausdorffness as a serious pathology that invalidates what I've just said, in the end it's less serious psychologically than it seems at first
--- at least in my experience.)
Speaking for myself, I certainly regard the Zariski topology as a topology, just like any other (meaning that I don't think of it as some other thing which happens to satisfy the axioms of a topology; I think of it as a topology in a genuine sense). It is just that it doesn't allow many closed sets: only those which can be cut out as the zero locus of a polynomial.
So a good way to practice thinking about the Zariski topology is to more generally practice thinking about topologies in terms of what kinds of closed sets are allowed, or, more precisely, what kinds of functions are allowed to cut out closed sets as their zero loci.
Thinking in terms of functions is a way of bridging the analytic intuition that you seem to like and the general topological formalism that underlies the Zariski topology. What I mean is: in standard real analysis, if you have a continuous function on $\mathbb R^n$ (or a subset thereof), its zero locus is closed. One way to think about this is via sequences (this is a way that you
seem to like): if $f(x_n) = 0$ for each member of a convergent sequence,
then $f(\lim_n x_n) = 0$ too, as long as $f$ is continuous.
Now, when working with the Zariski topology, you have to throw away the argument with sequences, but you can still keep the consequence: the zero locus of a "continuous" function is closed. The key point is that now the only functions that you are allowed to think of as being continuous are polynomials. This may take some getting used to, but is not so bad (after all, polynomials are continuous in the usual topology on $\mathbb C^n$ as well!).
Summary: It doesn't seem possible to work rigorously with a sequence/convergence point of view, but (a) it is not so misleading for very basic topological facts; and (b) another analytic view-point that is very helpful is to think about the topology in terms of its closed sets being zero-loci of continuous functions --- you just have to restrict the functions that you call "continuous" to be polynomials.
what are (dis)advantages of the net vs filter languages.
nets:
- Some statements are easier with nets. e.g.
If $X$ is a topological space and $A\subset X$ then $a\in \overline A$ iff some net on $A$ converges to $a$.
with filters one has to define what convergence of a filter on $A$ in $X$ means.
- Because nets are more intuitive and we can think of a net as a collection of points (somehow ordered), some natural questions may be inspired; for example:
Suppose $X$ is a topological space and every net on $A\subseteq X$ has
a convergent
subnet. Is
$\overline A$ compact?
(the answer is yes in Tychonoff spaces)
Almost all statements about sequences in analysis, can be translated to nets on topological or uniform spaces. e.g. this or this or even this with nets.
Net theorems will stick in mind, especially if you have studied analysis, because they can be imagined. But filters are more abstract. This is why I prefer nets.
filters:
With filters some proofs about compactness are easier. Even Tychonoff Theorem can be proved with filters.
Any (diagonal) uniformity is a filter. Before studying uniform spaces one should study filters.
Filter has something to do with Bornology.
Convergence of a filter controls the convergence of all nets which correspond to that filter. This says filters only have the necessary features for convergence while nets have features that are hardly pertinent to convergence.
Unlike superfilters, there are several definitions for subnets. So before using the word subnet you should clarify what you mean by that.
More about the so-called equivalence of filters and nets can be found in last pages of this pdf.
Best Answer
If you look at Munkres' Topology textbook (2000 edition p.98), a definition of a convergent sequence in an arbitrary topological space is given as follows. A sequence $x_1, x_2, \ldots$ of points in a space $X$ converges to a point $x \in X$ if for each neighborhood $U$ of $x$, $\exists N$ such that $\forall n \geqslant N$, $x_n \in U$.
The topology here is arbitrary and there is no mention of nets.