If $X$ is a noetherian topological space, then any union connected components of $X$ is clopen.

This is exercise 3.6P of Vakil. I can see that a union of connected components is closed. This is so because connected components are closed, and, in a Noetherian topological space, an arbitrary union of closed sets is closed.

My guess is that in order to show that a single connected component is open (which is clearly a sufficient and necessary condition) one should use the fact that, in a Noetherian space, a connected component is a finite union of irreducible components. If we could extend this union to a larger union of irreducible components that covers all of $X$, we'd be done, but I'm not sure if we can do this.

  • $\begingroup$ What is, structurally, the complement of a union of connected components? $\endgroup$ – Daniel Fischer Nov 24 '13 at 18:40
  • $\begingroup$ "In a Noetherian topological space, an arbitrary union of closed sets is closed." This statement is not true, think for example to the complex affine line with the Zariski topology: the union of intinitely many closed points is not a closed set. $\endgroup$ – Pgatti May 22 '15 at 6:34
  • $\begingroup$ Haha...I made the same mistake before. Noetherian topological spaces have DCC not ACC.... $\endgroup$ – No One Jan 19 '18 at 22:36

HINT: Show that $X$ has only finitely many connected components. Let $\mathscr{C}$ be the family of closed subsets of $X$ that have infinitely many connected components. If $\mathscr{C}\ne\varnothing$, $\mathscr{C}$ must have a minimal element $C$ (minimal with respect to inclusion). Use $C$ to get a contradiction, thereby showing that $\mathscr{C}=\varnothing$.


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