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Let $X$ be a noetherian connected topological space $X$. If $U$ is an open subset of $X$, must $U$ be connected? This is not true if $X$ is not noetherian e.g take $X=\mathbb{R}$ in the usual topology.

If $X$ is irreducible instead of merely connected, then I have the following proof showing $U$ is connected: if $U=C_1 \sqcup C_2 $ with $C_1, C_2 $ nonempty, closed and open in $U$, then since $U$ is open, $C_1, C_2$ are open in $X$. Then since $X$ is irreducible, two non-empty sets must intersect, contradicting $C_1, C_2$ being disjoint.

Any connected topological space is the union of its irreducible components...perhaps my proof can be adapted.

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  • $\begingroup$ A nonempty open subset of an irreducible space is even irreducible. $\endgroup$ – Hoot May 24 '16 at 22:05
  • $\begingroup$ @Hoot Oh, yeah, that's much easier than my proof. $\endgroup$ – usr0192 May 24 '16 at 22:07
  • $\begingroup$ I wouldn't say easier — I just wanted you to have the right statement. $\endgroup$ – Hoot May 24 '16 at 22:07
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Let $X=\{0,1,2\}$ with open sets $\varnothing,\{0\},\{1\},\{0,1\}$, and $X$. $X$ is Noetherian and connected, but the open set $\{0,1\}$ is not connected.

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In the affine plane $\mathbb A^2_k=\operatorname { Spec}k[x,y]$ over a field $k$ the closed affine subset ("the cross") $$X:=V(x\cdot y)=\{x\cdot y=0\}\subset \mathbb A^2_k$$ is a connected noetherian topological space (in the Zariski topology), but its open subset $$X_0:=X\setminus V(x,y)=X\setminus \{(0,0)\}$$ is no longer connected.
Indeed it is the disjoint union of two non empty open subsets: $$X_0=(X\cap D(x))\coprod (X\cap D(y))$$

Edit
As an amusing remark, notice that if you localize the ring $k[\bar x,\bar y]=k[x,y]/(x\cdot y)$ at its maximal ideal $(\bar x,\bar y)$ you obtain a three point subspace B$ =\operatorname { Spec}k[\bar x,\bar y]_{(\bar x, \bar y)}\subset X=\operatorname { Spec}k[\bar x,\bar y]$ such that B is homeomorphic to Brian's example.

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