# Affine Algebraic Sets: Open and Closed Sets?

I was hoping you'd might be able to clear something that should be minor out for me since I've been getting a bit confused over seemingly contradictory statements as to whether varieties are open or closed sets, a situation not made any easier by the fact that it is perfectly possible for them to be both.

Such as I was taught basic algebraic geometry, affine algebraic sets are the closed sets of the Zariski topology. Nevertheless, on Wikipedia, I read that "every open subset of a variety is a variety".

I decided to try to look at a simple example, but matters just got more confusing there. For example, consider the affine algebraic set $$V_1 = V((x^2+2x-1)(y-1)-z) \subset \mathbb{C}^3$$ with the algebraic subset $$V_2 = V(y, x^2+2x+z-1)$$. In as far as I can tell, $$V_2$$ would have to be open, since it is a neighbourhood for every point in it (unless I am making some serious mistake). But it would have to be simultaneously closed since it is an algebraic subset?

Am I right, wrong? In either case, how?

If neither $$V_1$$ or $$V_2$$ are open, can someone please provide me with some reasonably good examples of open subsets of $$\mathbb{C}^3$$ in the Zariski topology.

Neither $$V_1$$ nor $$V_2$$ are open in $$\Bbb C^3$$. They're closed (since they're $$V$$ of some ideal), so if either of them were open, they would be a connected component of $$\Bbb C^3$$ with nontrivial complement (the origin isn't in either of them). This would imply that $$\Bbb C^3$$ is disconnected in the Zariski topology, which is false ($$\Bbb C^3$$ has coordinate algebra $$\Bbb C[x,y,z]$$, which is an integral domain, so it's irreducible and in particular connected).

As for an example of an open subset of $$\Bbb C^3$$ in the Zariski topology, here are a few:

• $$\{(x,y,z)\in\Bbb C^3 \mid x\neq 0\}$$
• $$\{(x,y,z)\in\Bbb C^3 \mid x^2+y^2+z^2\neq 0\}$$
• $$\{(x,y,z)\in\Bbb C^3 \mid xyz\neq 0\}$$
• $$\{(x,y,z)\in\Bbb C^3 \mid (x,y,z)\neq (0,0,0)\}$$.

As for what's going on with wikipedia, I'd say you're meeting up with a couple different issues at the same time. First, the concept of a variety as you're describing it in your post (closed algebraic set in $$k^n$$ for an integer $$n$$ and a field $$k$$) gets upgraded pretty quickly to something more sophisticated, so some things one reads about "varieties" make more or less sense (or, indeed aren't totally correct) depending on where the reader and the writer are on the continuum of definitions for varieties. Secondly, sometimes people have actually different definitions - this shows up a lot once people start phrasing variety as "scheme + adjectives". For instance, in much of Hartshorne, a variety is (or is equivalent to) an integral separated scheme of finite type over a field, while other books or papers drop any number of those assumptions, allowing for example reducible varieties, non-reduced varieties, etc. Thirdly, people often aren't super precise about what their definition of "variety" actually is, which can lead to more confusion amongst readers, especially those who are newer to the subject. I'll just say that once you adopt the idea that a variety is a scheme with some adjectives, the statement on Wikipedia is correct and very quick from the definitions (assuming you've chosen the right adjectives).

• This is excellent! A final question, have I understood it correctly in that $\{ 1 \}$ is both an open and a closed subset in the Zariski topology of the complex numbers? It is closed in that it may be written as $Z(x-1)$, and open in that it is an open subset of $Z((x-1)(x-3))$ as $Z((x-1)(x-3)) \backslash Z(x-3)$? – StormyTeacup Oct 14 '19 at 14:55
• Not so fast! A set is not open or closed on it's own, it's open or closed depending on the topological space it's embedded it. Here, $\{1\}$ is a clopen subset of $V((x-1)(x-3))$, but as a subset of $\Bbb C$ it's closed and not open (the same proof from the answer applies). – KReiser Oct 14 '19 at 15:27
• Awesome! I kind of figured it might be something like that, but I felt a bit hesitant, since I had never actually come across any textbook stating it explicitly, or any direct example of such a case. Glad you sorted it out! :) FINAL final question then, and I'll leave you alone, when we are discussing "open subsets of an algebraic set $X \subset \mathbb{A}^n$", like, for example, we are discussing sets that are open subsets of $X$, but not necessarily $\mathbb{A}^n$, correct? – StormyTeacup Oct 14 '19 at 15:44
• Yes, that's correct. (Recall that $X$ gets the induced topology from $\Bbb A^n$, so while the open subsets of $X$ are usually not open in $\Bbb A^n$, they do come from open sets in $\Bbb A^n$.) – KReiser Oct 14 '19 at 19:48
• Splendid! I thank you and wish you a pleasant evening. :) – StormyTeacup Oct 14 '19 at 20:08