Let L/K be a field extension, and let Ln be a vector space associated with the variables x1 through xn. A region R in Ln is an algebraic set if it is S′ for some set of functions S. In the last section we showed such a region is closed. That is, R′′ = R. Conversely, assume R is closed. Now R′′ = R, and if S = R′, then R = S′, and R is an algebraic set. In other words, R is closed iff it is an algebraic set.
So far we have not specified the universal set of functions. It is customary to use polynomials over L. At other times we will restrict attention to polynomials with coefficients in K. Regardless of the universal set, R is closed iff it is an algebraic set. Let's see if this definition makes topological sense.
The function 1 shows the empty set is closed, and the function 0 shows all of Ln is closed. (I'll assume the universal set is a ring, hence it contains 0 and 1.)
Consider the intersection of arbitrarily many closed regions. Let S hold all the corresponding functions, for all the regions, and the intersection becomes an algebraic set. In other words, the intersection of closed sets is closed.
To take the union of R1 and R2, multiply all the functions in S1 by all the functions in S2. The resulting functions vanish on R1∪R2, hence the finite union of closed regions is closed. We have a valid topology.
Any point in Kn is closed via the equations x1 = c1, x2 = c2, x3 = c3, etc. If K = L then the space is half hausdorff.
Take the finite union of points, and any finite subset of Kn is closed.
In L1, which is a pid, S′ is the zeros of the gcd of the polynomials in S. If L = K, a region in L1 is closed iff it is finite (or the entire space). This is the cofinite topology.
Let R be a region in Cn that is closed under the zariski topology. Thus the functions of S vanish on R. Our universe of functions could be those functions that are complex analytic, or any subring thereof, such as the polynomials. Let p be a point not in R, and evaluate each function at p. Some function f in S is nonzero at p, and since f is continuous, it is nonzero throughout a neighborhood about p. This open neighborhood lies outside of R, hence R complement is open, and R is closed.
Conversely, let R be an infinite set of isolated points in C. We showed above that C1 has the cofinite topology, hence R is metric closed, but not zariski closed.