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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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    $(0,1)$ isn't open in the Zariski topology! (A set $A \subseteq \mathbb R$ is Zariski-open iff it is empty or cofinite).2012-05-15
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    I thought closed set were the finite sets and the whole or $\mathbb{R}$, hence an infinite set is open?2012-05-15
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    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.2012-05-15
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    "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".2012-05-15
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    @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.2012-05-21

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