While solving:
Prove $\mathbb{R}$ with the Zariski topology is compact
it seems like the Zariski topology should make a compact space with whatever set it is definied on:
To solve this case, I took a cover $\mathcal{U}$ of $\mathbb{R}$, and a $U_1 \in \mathcal{U}$ which is inherently open, so $\mathbb{R}\smallsetminus U_1 $ is closed, and by the Zariski topology this is finite. We then pcik points out of $\mathbb{R} \smallsetminus U_1$ and find a set in $\mathcal{U}$ which covers that point. Since there are finite points, there are finite sets to cover them, so a subcover made of $U_1$ and this finite number of subsets is also finite, i.e. compact.
But this method seems independent of being in $\mathbb{R}$. In any finite set with the Zariski topology, any cover will have an obvious finite subcover? In the infinte case, the above argument should still hold?
So, is any set with the Zariski topology compact?
In the Zariski topology on $F^n$, for $F$ a field, the closed sets are the zeros of polynomials in $n$ indeterminates.
In the case $n=1$, these sets are finite, except for the whole set. So the Zariski topology on $\mathbb{R}$ is the cofinite topology, which is compact.
The Zariski topology is likewise compact on $F^n$, because the space is Noetherian anyway.
It's unclear what you mean with “any set with the Zariski topology”.