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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.

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

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