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On $\mathbb{R}$ under the Zariski topology I don't understand why the closure of $(0,1)$ is $\mathbb{R}$? Clearly this is the only closed set containing $(0,1)$ but I thought we're looking for an open set containing $(0,1)$ under the definition of closure.

Also, I don't understand why the interior of $(0,1)$ is $\emptyset$. Under the definition of the Zariski topology surely $(0,1)$ is open so the interior would be itself?

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    $\begingroup$ $(0,1)$ isn't open in the Zariski topology! (A set $A \subseteq \mathbb R$ is Zariski-open iff it is empty or cofinite). $\endgroup$
    – martini
    May 15, 2012 at 10:30
  • $\begingroup$ I thought closed set were the finite sets and the whole or $\mathbb{R}$, hence an infinite set is open? $\endgroup$
    – user26069
    May 15, 2012 at 10:35
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    $\begingroup$ No, a set with finite complement is open. $(0,1)$ is neither open nor closed in the Zariski topology. The definition of closure of $(0,1)$ is the smallest closed set containing $(0,1)$, so the fact that $\mathbb{R}$ is the only such set is enough. $\endgroup$
    – mdp
    May 15, 2012 at 10:37
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    $\begingroup$ "I thought closed set were the finite sets and the whole of $\mathbb{R}$, hence an infinite set is open?" Always remember that "open" is not the same thing as "not closed". A lot of sets are "ajar". $\endgroup$
    – rschwieb
    May 15, 2012 at 10:56
  • $\begingroup$ @morphism You are right in that all finite sets are closed. The other direction is false though. Consider for example the set of all even numbers, $2 \mathbb Z$. Then this set is infinite so it's not closed and its complement is also infinite hence it's not open. $\endgroup$ May 21, 2012 at 11:53

2 Answers 2

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Zariski closed sets are defined by the vanishing of polynomials, so the closure of (0,1) in the Zariski topology is the smallest set containing (0,1) that can be defined by the vanishing of polynomials. But any polynomial that vanishes on (0,1) must be zero. Hence the closure of (0,1) must be the whole space.

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The interior of a set $S$ is the set of points $s$ in $S$ for which you can find an open set $O$ such that $s \in O \subset S$.

So let $x_0$ be a point in $(0,1)$ and let $O$ be an open set containing $x_0$. Then by the definition of the Zariski topology you can write $O$ as $O = \bigcup_{p \in P} p^{-1} (\mathbb R \setminus \{0\})$ where $P$ is a subset of $\mathbb R [x]$, the set of polynomials with real coefficients. Since this means that $x_0 \in p^{-1} (\mathbb R \setminus \{0\})$ for some $p$ in $P$ so it's enough to find one polynomial $p$ such that $p(x_0) \neq 0$ and $p(x) = 0$ on all of $\mathbb R \setminus (0,1) = (-\infty , 0] \cup [1,\infty)$. The second condition needs to hold in order for $p^{-1}(\mathbb R \setminus \{0\})$ to be contained in $(0,1)$.

But if $p(x) = 0$ at any point $z$ then by continuity we can find a $\delta$ ball $B(z,\delta)$ on which $p(x) = 0$ which implies that $p$ is the zero polynomial. Hence we get a contradiction to $p(x_0) \neq 0$ and conclude that there can be no open sets contained in $(0,1)$ and hence its interior has to be empty.


For the closure: note that the closure of a set $S$ is all points $x_0$ such that for every open set $O$ containing $x_0$, $O \cap S \neq \varnothing$.

Now let $x_0$ be any point in $\mathbb R$ and let $O$ be an open set containing $x_0$. Then $O = \bigcup_{p \in P} p^{-1} (\mathbb R \setminus \{0\})$ for some set $P \subset \mathbb R [x]$. Assume $O$ does not intersect with $(0,1)$. Then all $p \in P$ are zero on $(0,1)$. But then again using continuity we get that all $p$ in $P$ have to be the zero polynomial. Hence $O = \mathbb R$ and hence $O \cap (0,1)$ is non-empty. Since we chose $x$ in $\mathbb R$ arbitrary we get that the closure of $(0,1)$ is all of $\mathbb R$.

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