Vrftsjtryk

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.

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).