# Striking examples to show the non-hausdorffness of the Zariski toplogy?

We are told that the Zariski topology is not Hausdorff, but I have rarely seen "concrete examples" of the dramatic failures this can induce.

### A concrete example I have in mind:

For example, one way I know to show that the Zariski topolgy is weird is as follows:

We claim that on $$\mathbb R$$ equipped with the zariski topology, two points $$x, y \in \mathbb R, x \not= y$$ cannot be separated by open sets $$X, Y \subseteq \mathbb R$$ such that $$x \in X, y \in Y, X \cap Y = \emptyset$$.

The idea is that since the closed sets induced by the Zariski topology are of the form $$\{ x \in \mathbb R : \forall i \in I, f_i(x) = 0, f_i(x) \in \mathbb R[X] \}$$

Hence the open sets will be of the form $$\{x \in \mathbb R: \exists i \in I, f_i(x) \neq 0, f_i(x) \in \mathbb R[X]$$ }.

However, in general, polynomials are zero at finitely many locations. Hence, the sets where polynomials are not zero are extremely large. Therefore, it is hard to separate two points with open sets, since the non-zero sets of polynomials will likely intersect in a large region of space.

I don't understand this handway example very well either: I would like to see a formal proof of this "fact" that I have learnt from the folklore.

### Examples of dramatic failures I would like:

• concrete functions becoming continuous that we do not think are continuous.
• concrete functions becoming discontinuous that we do not think are discontinuous (I think this cannot happen, since the Zariski topology is very coarse. However, I don't have a good intuition for the topology, so if somoene can correct me on this, I would be glad)
• A concrete sequence having multiple limits
• Well, the "formal" proof of the non-Hausdorffness require just a bit of quantification instead of imprecise words like "extremely large" : the closed sets are the zeros of polynomials, hence are finite ! (or the whole space). It follows that the union of two closed set is not the whole space (unless one of them is the whole space). Taking complements : the intersection of two open sets cannot be empty (unless one is empty). So you cannot find $X\ni x$ and $Y\ni y$ such that $X\cap Y=\emptyset$. – Roland Jan 20 at 17:47
• The MathWorld entry can be interesting for you. In particular, it states that there are very few continuous functions of $(\mathbb C^n, \mathrm{Zariski})$ into $\mathbb C$ (equipped with its natural topology, of course). – Giuseppe Negro Jan 23 at 14:33

On $$\mathbb R$$ (or any other field, actually), the Zariski topology is exactly the cofinite topology, in which the closed sets are the finite ones. Example 18-19 of Steen and Seebach's "Counterexamples in topology" is dedicated to this.
One fact that you may find interesting is that every infinite subset is dense. An explicit example of this is that the sequence of positive integers converges to any point of $$\mathbb R$$. Indeed, if $$x\in\mathbb R$$ and $$U$$ is an open neighborhood of it, then the complement of $$U$$ is finite, hence it has a maximum $$M$$. If $$n\in\mathbb N$$ is sufficiently big, then $$n>M$$ and so $$n\in U$$. Thus $$n\to x$$.
In comments, Noah rightfully points out that in higher dimension the Zariski topology is more complicated than the cofinite one. On $$\mathbb R^2$$ with Zariski topology, for example, it is no longer the case that all infinite sets are dense. For example, the zero set of any polynomial is closed and a proper subset of $$\mathbb R^2$$, hence not a dense set.
However, any infinite set that is not contained in the zero set of a polynomial is dense. For example, $$\{(x, y)\in\mathbb R^2 \ :\ y=\log\, \lvert x \rvert \}$$ is dense in $$\mathbb R^2$$.
• Note that the first paragraph does not generalize: the Zariski topology on $\mathbb{R}^2$ is much more complicated than the cofinite topology. – Noah Schweber Jan 20 at 21:30
• @NoahSchweber Can you please link to a resource that provides a deep dive into the Zariski topology of $\mathbb R^2$, if any such exist? – Siddharth Bhat Jan 23 at 9:57