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The Zariski topology is weird. I don't pretend to understand noether's writeup, above, but I have read that, on the real numbers, R, the open sets that make up Zariski topology are all those sets whose complements in R are finite, together with the empty set.

We can see that it is a topology by considering its closed sets - all finite sets. Since the intersection of any number of finite sets is itself finite and the union of any finite number of finite sets is finite, the conditions for a topology are met.

Consider the sequence x1, x2 , ..., where xn = n for any natural number, n. This is just the sequence 1, 2, 3,... etc.

A sequence, x1, x2, ..., is said to converge on a point p, in a topological space, T, if and only if there's some term in the sequence beyond which all subsequent terms are members of all open sets in T of which p is a member, in other words:

for all open sets, V in T which contain p, there is some positive natural number N such that for all xj, j > N, xj is in V.
Suppose r is a real in the Zariski topology on R, and is a member of an open set, V.

Since the complement of V is finite, it has a largest member, v. Consequently, there is a natural number, N > v, which is greater than any member of V's complement. Since xn = n, any term of our sequence where n > N is not in the complement of V, and must therefore be in V, so without specifying r we've shown the sequence converges on r. Hence, the sequence converges on any real number, r, that we care to choose, with respect to the Zariski topology on the reals.

This is enough to show the Zariski topology on the reals is not a Hausdorff space (and therefore not a metric space), since in a Hausdorff space, any convergent sequence must converge on a single point.