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 x_{1}, x_{2} , ..., where x_{n} = `n` for any
natural number, `n`. This is just the sequence 1, 2, 3,... etc.

A sequence, x_{1}, x_{2}, ..., 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
x_{j}, `j` > `N`, x_{j} 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 x_{n} = `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.