# Showing the Zariski topology is in fact a topology

I have a definition of the closed sets of the Zariski topology that is : A subset $$V$$ of $$\Bbb R^{n}$$ is Zariski closed if there exists a set, $$I$$, consisting of polynomials in $$n$$ real variables such that

$$V = \{ r \in \Bbb R^{n}| f(r)=0$$ for all $$f ∈ I \}$$ .

My first question is to wonder isn't any subset of $$\Bbb R^n$$ Zariski closed using the zero polynomial?

Secondly, if I consider the open sets to be the complements of the sets of type $$V$$, I want to show that an arbitrary union of open sets is open. This amounts to showing , by DeMorgan's Law, that an arbitrary intersection of sets of type $$V$$ are closed, meaning for all the elements in the intersection, there has to be polynomials that evaluate to zero for these elements, which I am not sure how to show other than stating that the zero polynomial works, which seems too simple/wrong. Any hints appreciated.

• For the first question: using the $0$ polynomial is how you show $\mathbb{R}^n$ is closed. – Ethan Bolker Nov 16 '18 at 1:31
• @EthanBolker why would this not apply also to a subset? – IntegrateThis Nov 16 '18 at 1:32
• Because (if you denote the right hand side as $Z(I)$) the requirement is that $V$ is exactly equal to $Z(I)$, not that $V \subseteq Z(I)$. – Daniel Schepler Nov 16 '18 at 1:37
• For the second question: suppose you have a family $\{ I_\lambda \mid \lambda \in \Lambda \}$ of sets of polynomials - then what you want to show is that $\bigcap_{\lambda \in \Lambda} Z(I_\lambda) = Z(\bigcup_{\lambda \in \Lambda} I_\lambda)$. – Daniel Schepler Nov 16 '18 at 1:41
• @DanielSchepler what is $Z(I)$ – IntegrateThis Nov 16 '18 at 1:43

You show directly that the set of all $$Z(I) = \{x \in \mathbb{R}^n: \forall f \in I: f(x) = 0\}$$, where $$I$$ is any set of polynomials in $$n$$ variables, obeys the axioms for closed sets:

$$\emptyset$$ is closed, because $$\emptyset = Z(\{1\})$$, where $$1$$ is the constant polynomial with value $$1$$ so there is no zero for it.

$$\mathbb{R}^n$$ is closed, because $$\mathbb{R}^n = Z(\{0\})$$, with $$0$$ the constant polynomial with value $$0$$, so all $$x$$ are zeroes of it, trivially. We could also have used $$\mathbb{R}^n = Z(\emptyset)$$ if you like void truth.

If $$Z(I), Z(J)$$ are two closed sets (for finitely many it's enough to check the case of $$2$$ sets), then form $$IJ = \{fg: f \in I, g \in J\}$$, which is a well-defined set of $$n$$-dimensional polynomials on $$\mathbb{R}^n$$. If $$x \in Z(I)$$, $$x$$ vanishes for all $$f \in I$$, so also for all $$fg \in IJ$$. The same holds for $$x \in Z(J)$$, so $$Z(I) \cup Z(J) \subseteq Z(IJ)$$. If $$x \notin z(I) \cup Z(J)$$ this means there is some $$f \in I$$ such that $$f(x) \neq 0$$ and some $$g \in J$$ such that $$g(x) \neq 0$$. It follows that $$(fg)(x) \neq 0$$ and so $$x \notin Z(IJ)$$. This shows

$$Z(IJ) = Z(I) \cup Z(J)$$

so that the set of $$Z(I)$$ is closed under finite unions.

If $$Z(I_\alpha), \alpha \in A$$ is any collection of such sets, then by the definitions it's clear that

$$\bigcap_{\alpha \in A}Z(I_\alpha) = Z(\bigcup_{\alpha \in A} I_\alpha)$$

and so this collection is closed under arbitrary intersections.

Now de Morgan or a standard theorem in elementary topology tells us that the complements of the sets of the form $$Z(I)$$ indeed form a topology on $$\mathbb{R}^n$$. Note that the argument works for any commutative ring without zero-divisors (I used that for the finite unions).