As someone who is very fond of analysis, I feel most comfortable working in topological spaces via the notion of convergence of sequences (or nets, in infinite-dimensional Banach spaces, etc.). In every metric space (or first-countable topological space) anything you can say about the topology can be said in terms of convergent sequences (though perhaps less elegantly). In arbitrary topological spaces, one can use nets instead. However, if the space is not Hausdorff then nets don't have unique limits, which makes it quite difficult to work with them the way one is used to working with sequences (of real numbers, say). This is the case with the Zariski topology on $\mathbb{C}^n$ (I have no urgent desire to consider more general algebraic varieties). Is there some convenient way to express the Zariski topology (that is: its closed sets, the continuous functions it allows,…) in terms of some "convergence structure" which satisfies properties similar to those we have in any sequential/metric/first-countable space?
General Topology – Working with Zariski Topology and Limits
algebraic-geometryconvergence-divergencegeneral-topology
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The spaces characterized by the property that a subset is closed if and only if it is closed w.r.t. limits of transfinite sequences are called pseudoradial spaces.
I would like to give an example of a space that is not pseudoradial together with sketch of a proof. (I hope I am not doing something very easy in a too complicated way.) I would also like to mention a few references and some properties of these spaces.
Before presenting the example, one short remark. Pseudoradial spaces are represented by the convergence of net on well-ordered nets. Obviously, it is sufficient to take the nets on ordinals. We can go a little further - regular cardinals are sufficient. Indeed, if we have a cofinal subset of an ordinal, we can use this cofinal subset to get another convergent net.
Let us consider the following example. Each arrow in the picture bellow represents a convergent sequence. I.e., this is a topological space homeomorphic to $\{0\}\cup\{\frac1n;n\in\mathbb N\}$ taken as a subspace of real line. Equivalently, this is precisely the ordinal $\omega+1$ taken with the order topology.
We take all these sequences and identify some of the points as in the picture. (I.e., we make a quotient space of some of these spaces.) Let us call the resulting space $S_2$. Then we take the subspace of this space as shown in the picture. This subspace will be called $S_2^-$. (I've taken the notation $S_2$ and $S_2^-$ from this paper: Franklin S.P., Rajagopalan M., On subsequential spaces, Topology Appl. 35 (1990), 1-19. But you can notice that this space is very very similar to Arens-Fort space mentioned in Brian's answer.)
Now we want to show that $S_2^-$ is not pseudoradial.
Note that the space $S_2^-$ has only one non-isolated point. Let us call it $\omega$ . So we ask whether there is a transfinite sequence, consisting only of points different from $\omega$, which converges to $\omega$.
First, let us show that this is not possible for a regular cardinal $\alpha>\omega$. Suppose that $(x_\eta)_{\eta<\alpha}$ is an $\alpha$-sequence of points of $S_2^-\setminus\{\infty\}$, which converges to $\infty$. Let us denote $n_\eta$ the "column" to which $x_\eta$ belongs. In we use the notation the notation from the picture bellow $n_\eta$ is the first coordinate of ordered pair $x_\eta$.
We can see that $n_\eta$ converges to $\omega$. (E.g. by noticing that $(x,y)\mapsto x$ and $\omega\to\omega$ is a quotient map from $S_2^-$ to $\omega$ with order topology.)
Now this is not possible, since the we would be able to construct an increasing $\alpha$-sequence converging to $\omega$ and using this sequence we would be able to show that cofinality of $\alpha$ is $\omega$.
So the only possibility is to take a sequence in the usual sense, i.e., a sequence of length $\omega$. Perhaps with a little handwaving, but it is more-or-less clear that general situation is similar to the situation when the $n$-term of the sequence is in the $n$-th column. So we have a sequence $x_n=(n,y_n)$. Obviously $\{\omega\}\cup\bigcup\limits_{n\in\omega} \{n\}\times(y_n,\infty)$ is a neighborhood of the point $\omega$ containing no terms of this sequence.
Pseudoradial spaces were introduced by H. Herrlich. Quotienten geordneten Räume und Folgenkonvergenz. Fund. Math., 61:79–81, 1967; pdf. They were later studied by A.V. Arhangelskii and many others.
The class of pseudoradial spaces is closed under the formation of closed subspaces, quotients and topological sums. They are a coreflective subcategory of the category Top of all topological spaces. This means that for each topological space we have pseudoradial coreflection; a pseudoradial space which is, in some sense, close to this space. The pseudoradial coreflection is obtained simply by taking sets closed under limits of transfinite sequences as closed sets in a new topology on the same set. (E.g. the pseudoradial coreflection of $S_2^-$ is discrete.)
The same thing can be done with any class $\mathbb P$ of directed sets instead of ordinals. This is called $\mathbb P$-net spaces in P. J. Nyikos. Convergence in topology. (In M. Hušek and J. van Mill, editors, Recent Progress in General Topology, pages 537–570, Amsterdam 1992. North-Holland.) The properties of pseudoradial spaces which I mentioned in the preceding paragraph are true for $\mathbb P$-net spaces, too.
Interestingly, if we take the linearly ordered sets, we obtain the same class of spaces as from well-ordered sets, see James R. Boone: A note on linearly ordered net spaces. Pacific J. Math. Volume 98, Number 1 (1982), 25-35; link.
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.
Best Answer
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.