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Let a topological space $X$ have infinitely many connected components. Then why is it true that $X=X_1 \cup X_2$ where $X_1$ and $X_2$ are closed, disjoint nonempty?

For example $X=\mathbb{Q}$ has as its components each point, and $X=(-\infty, \sqrt{2}) \cup (\sqrt{2}, \infty)$.

Motivation:

I am trying to prove that in $\operatorname{Spec}A$, the connected components are clopen sets when $A$ is a noetherian ring. It suffices to show that there are finitely many connected components. A positive answer to this question would let me prove that there are finitely many connected components by contradiction - I would separate $\operatorname{Spec}A$ by two closed subsets and separate the one that is not connected(again by the positive answer to my question), and use that $A$ is noetherian to get a contradiction.

De Yang
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  • Is $X$ a Hausdorff ($T_2$) space? Are the connected components of $X$ countable? – Jack D'Aurizio Sep 12 '16 at 16:23
  • In the application that I am interested in, $X$ will be $T_0$. It won't be $T_2$. I am willing to work with countable components first if that is easier – De Yang Sep 12 '16 at 16:25
  • An affine scheme is quasi compact (no noetherianness needed) and therefore can only have finitely many connected component. This is just a comment to your motivation part. See here: http://math.stackexchange.com/questions/1872170/proof-that-an-affine-scheme-is-quasi-compact – Maik Pickl Sep 12 '16 at 16:38
  • In addition to closed and disjoint, you want $X_1$ and $X_2$ to be non-empty – marlu Sep 12 '16 at 16:43
  • @MaikPickl But that won't help me prove what I want. First it is not true for arbitrary rings that the connected components of $Spec A$ are open. Second, I would have the show that the connected components are open first, (after which it would follow that the connected components form an open cover of $Spec A$). This defeats the purpose because I am trying to show that the connected components are open. – De Yang Sep 12 '16 at 16:50
  • Regarding your motivation, and just as an aside: the "standard" way to deduce this (I think) is to use the fact that even the number of irreducible comps. of Spec $A$ is finite. (This corresponds to a familiar ideal-theoretic statement about Noetherian rings, and can also be proved by Noetherian induction, similarly to your argument about connected comps.) But your direct argument looks like it works too. – tracing Sep 13 '16 at 01:38

1 Answers1

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Connected component is defined as a inclusion-maximal connected subset. As soon as you have a component $X_1\subsetneq X$, it follows that $X$ is not connected (as otherwise $X_1$ would not be maximal).

Hagen von Eitzen
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