I'm having trouble understanding the concept of the Zariski topology on $\mathbb{R}$. My notes say that subsets of $\mathbb{R}$ are closed iff they consist of finitely many points or if they are all of $\mathbb{R}$.
So does that mean no intervals such as [0,1] are not closed? As they consist of infinitely many points in $\mathbb{R}$ and the only 'closed' subsets are ones which only contain points and no intervals such as {1,2,3}?
Thanks!
Best Answer
Yes, you have it exactly right.
What you're observing is that compared to the standard topology on $\mathbb{R}$ -- i.e., the one with a base given by open intervals -- the Zariski topology is very much coarser: every Zariski-closed set is standard-closed, but the converse does not hold.
In fact the Zariski topology on $\mathbb{R}$ -- or on any field $k$ -- is simply the cofinite topology: i.e., the nonempty open sets are those with finite complement. This is the coarsest topology which satisfies the $T_1$ separation axiom*: i.e., that singleton sets are closed. It is not a Hausdorff topology unless the field $k$ is finite....in which case it's the finest possible topology -- the discrete topology.
*: For several years now I have taken to calling such spaces "separated" (rather than "$T_1$", "Frechet-Urysohn"...). So far no one has objected...probably because no one has noticed.