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?
Best Answer
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.