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 $X=(-\infty, \sqrt{2}) \cup (\sqrt{2}, \infty)$.


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.

  • $\begingroup$ Is $X$ a Hausdorff ($T_2$) space? Are the connected components of $X$ countable? $\endgroup$ – Jack D'Aurizio Sep 12 '16 at 16:23
  • $\begingroup$ 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 $\endgroup$ – De Yang Sep 12 '16 at 16:25
  • $\begingroup$ 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: math.stackexchange.com/questions/1872170/… $\endgroup$ – Maik Pickl Sep 12 '16 at 16:38
  • $\begingroup$ In addition to closed and disjoint, you want $X_1$ and $X_2$ to be non-empty $\endgroup$ – marlu Sep 12 '16 at 16:43
  • $\begingroup$ @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. $\endgroup$ – De Yang Sep 12 '16 at 16:50

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).

  • $\begingroup$ thanks! guess I will think about my question more before asking next time. $\endgroup$ – De Yang Sep 12 '16 at 16:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.