Consider the Zariski topology on the set $\mathbb{R}$.
1) Is the set $(0,1)$ compact in this topology?
I said that it was because under the Zariski topology it was closed as there are infinitely many points and as it's clearly bounded it's compact.
2) Is the set $X = (0,1) \cup (2,3)$ connected?
I said no as if you let $A = \mathbb{R}\backslash (0,1)$ and $B = \mathbb{R} \backslash (2,3)$ then clearly $X$ is a subset of the union and the intersection $X \cap A \cap B \neq \emptyset$.
Have I got the reasoning right?
Best Answer
Your answer to (1) is correct, but the reasoning is not. The closed sets in the Zariski topology on $\Bbb{R}$ are $\Bbb{R}$ itself and the finite sets; that is, the Zariski topology is the cofinite topology. In this topology every set is compact, but boundedness has nothing to do with it. Let $A$ be a subset of $\Bbb{R}$, and let $\mathbb{U}$ be an open cover of $A$. Let $a$ be any point of $A$; there must be some $U_a\in\mathbb{U}$ that contains $a$. Now $\Bbb{R}\setminus U_a$ is finite, so $A\setminus U_a$ is finite. For each point $x\in A\setminus U_a$ let $U_x\in\mathscr{U}$ contain $x$. Then $\{U_a\}\cup\{U_x:x\in A\setminus U_a\}$ is a finite subset of $\mathscr{U}$ that covers $A$.
The Heine-Borel theorem that a subset of $\Bbb{R}^n$ is compact iff it is closed and bounded applies specifically to the spaces $\Bbb{R}^n$ with their usual Euclidean topologies. Indeed, it doesn’t even make sense for the Zariski topology, because the notion of boundedness employed in the theorem is metric boundedness: a set is bounded if there is a finite upper bound on the distances between points of the set. The Zariski topology, however, is not metrizable: it isn’t even Hausdorff.
Your answer to (2) is incorrect: every infinite subset of $\Bbb{R}$ is connected in the Zariski topology. Let $A$ be an infinite subset of $\Bbb{R}$, and suppose that $A=U\cup V$, where $U\cap V=\varnothing$, and $U$ and $V$ are both relatively open in $A$. This means that there are open sets $G$ and $H$ in $\Bbb{R}$ such that $U=G\cap A$ and $V=H\cap A$. But $A$ is infinite, so at least one of $U$ and $V$ must be infinite, say $U$. Clearly $U\cap H=\varnothing$, so the complement of $H$ is infinite. $H$ is open, so $\Bbb{R}\setminus H$ is an infinite closed set. But the only infinite closed set in the Zariski topology on $\Bbb{R}$ is $\Bbb{R}$ itself, so $\Bbb{R}\setminus H$ must be $\Bbb{R}$, and that means that $H=\varnothing$. In other words, there is no decomposition of $A$ into two disjoint non-empty open sets, and $A$ must be connected.