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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?

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Notice that the Zariski topology is not metrizable--it is not Haudorff--so the concept of boundedness does not fit here. –  AQP Mar 5 '12 at 22:53
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If I recall correctly the Zariski topology is the co-finite topology. This means that a set is closed if and only if it is finite. $(0,1)$ is far from finite, its complement is not finite either. We have that this set is neither open nor closed. While at it, since this is a non-Hausdorff topology compact sets need not be closed. –  Asaf Karagila Mar 5 '12 at 22:57
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Why don't you try the cover by sets { $R-{x} \cap(0,1)$: x in (0,1)}? Can you find a finite subcover for it? –  AQP Mar 5 '12 at 23:04

1 Answer 1

up vote 7 down vote accepted

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.

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Thank you! Your answer really helped! –  user26069 Mar 6 '12 at 0:11

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