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If we consider $\mathbb{R}$ together with the Zariski topology, what is the closure, interior and boundary of $(0,1)$?

A set is closed iff it is either finite or $\mathbb{R}$ under this topology, so since the closure of $(0,1)$ must be closed am I right in thinking that this must be either $\{0,1\}$, $\emptyset$ or $\mathbb{R}$? I think if I can correctly understand the closure of this set I would be able to get the latter parts of my question.

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A set is contained in its closure. Therefore the closure cannot be $\{0,1\}$ or $\emptyset$. Hence it must be $\mathbb R$, since $\mathbb{R}$ is the only closed set containing $(0,1)$.

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The closure of $A\subset X$ where $X$ is a topological space is the smallest closed subset that contains $A$. In our case it is $\mathbb{R}$. The interior is the largest open set contained in $A$ so in our case it is $\varnothing$. The boundary is the complement of the interior relative to the closure so in our case it is again $\mathbb{R}$

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The closure is the smallest closed superset, so in the case is $\Bbb R$. The interior is the greatest oper subset, so in this case is... And the border is...

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