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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}?


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Can you think of a polynomial that vanishes on just $[0, 1]$? –  Dylan Moreland Mar 4 '12 at 16:20
Remember: $\mathbb{R}[X]$ is a PID, so every ideal is generated by a single polynomial. How many roots can a polynomial have? –  M Turgeon Mar 4 '12 at 16:31
To follow up on Dylan's comment: To prove that a continuous function $f:\mathbb R\to\mathbb R$ is zero, it suffices to prove that it vanishes on a dense subset (that is, dense for the usual topology). But if $f$ is polynomial, it suffices to prove that it vanishes on an infinite subset. So, try to construct a topology on $\mathbb R$ whose dense subsets are precisely the infinite ones. –  Pierre-Yves Gaillard Mar 4 '12 at 16:35
@DylanMoreland I think it's likely that user26069 has encountered this definition in a first course on topology, and not in the context of algebraic geometry. I may be wrong, but your comment might make no sense to him/her. The same of course applies to M Turgeon's comment –  Daniel Freedman Mar 4 '12 at 18:01

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up vote 9 down vote accepted

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 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.

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I don't think it's such a good idea to call these spaces "separated", considering this adjective is generally used for Hausdorff (aka $T_2$) spaces. The "$T_1$" terminology is the one to use I think, considering there's no possible ambiguity in this case (unlike $T_3$ for example: some authors use it for "$T_2$ regular space", the most widely accepted use, other just for "regular space"). –  Najib Idrissi Mar 4 '12 at 20:41
@zulon: In French the translation "separe" is used for Hausdorff spaces, but in English so far as I know everyone calls Hausdorff spaces...Hausdorff. Sometimes you hear phases like "complete and separated for the I-adic topology", but always in a situation in which the space is uniformizable in which case separated and Hausdorff coincide. –  Pete L. Clark Mar 4 '12 at 21:04
Note also that it is standard terminology to say that two subsets $A,B$ of a topological space $X$ are separated if each is disjoint from the other's closure. Thus a space is separated iff its singleton sets are separated. It seems pretty logical to me...(What one cannot do is simply slip in "separated" without explaining what you mean. Note that I didn't do that!) –  Pete L. Clark Mar 4 '12 at 21:07

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