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If T is the set $\mathbb{R}$ with the zariski topology then is the set $X=\{0,1\}$ connected?

I think it is connected because the only nonempty subsets are {0,1}, {0}, {1} which are all closed under this topology so therefore there doesn't exist a decomposition into open sets so it is connected. Is this the right reasoning?

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No, $\{0,1\}$ is disconnected: the set $U=\mathbb{R}-\{0\}$ is open (it is the complement of the zero set of $(x)$), and $V=\mathbb{R}-\{1\}$ is open (the complement of the zero set of $(x-1)$). Note that $X\subseteq U\cup V$, and $X\cap U\cap V = \emptyset$, so $U$ and $V$ disconnect $X$. – Arturo Magidin Mar 5 '12 at 17:39
No, you're supposed to consider the subspace topology on $\{ 0, 1 \}$, and there are certainly decompositions of $\{ 0, 1 \}$ into non-empty disjoint open subsets in the subspace topology. – Zhen Lin Mar 5 '12 at 17:41
P.S.: As soon as you find two disjoint, nonempty, complementary closed sets, you've proven your space is not connected; you correctly identified $\{0\}$ and $\{1\}$ as closed sets, but did not notice they were disjoint, nonempty, and complementary. They provide you with a disconnection. – Arturo Magidin Mar 5 '12 at 17:50
up vote 3 down vote accepted

No, it is not connected. Since $\{0\},\{1\}$ are closed, their complements are open and intersected with $X$ these are just $\{1\},\{0\}$, which are disjoint and cover $X$.

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