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Let $X$ be a noetherian topological space, $Y$ a subspace having irreducible components $Y_1, \cdots, Y_n$. Prove that the $\bar{Y_i}$ are the irreducible components of $\bar{Y}$.

I think as the irreducible components are the maximal irreducible subsets, it is enough to prove

When a set $Z \subseteq X$ is irreducible, so is $\bar{Z}$.

Is this true? If it is, is it still true when the noetherian condition is dropped?

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Double check your typing. It makes wonders. Specially in titles, which is the very first thing people see of your question! – Mariano Suárez-Alvarez Sep 19 '11 at 3:47
@Mariano Suárez-Alvarez♦: Thank you very much for reminding me. I don't know what careless mistake I have made in my title because it might have been kindly corrected by Arturo Magidin. I will be more careful next time. Thanks again. – ShinyaSakai Sep 19 '11 at 12:30
up vote 1 down vote accepted

This is true in general; you do not need $X$ to be a noetherian space.

Continuing user3296's argument: the only thing left to check is that $C \cap Z$ and $D \cap Z$ are proper in $Z$. You do not need them closed in $X$ -- irreducibility is not a relative property: a space $Z$ is reducible if it is the union of two proper closed -in-itself subsets.

To see that for example $C \cap Z \subsetneq Z$, you have to convince yourself that $C \cap Z = Z$ implies $C = \overline{C} = \overline{C \cap Z} = \overline Z$. The first equality is true because $C$ is closed in the closed set $\overline Z$ (convince yourself that this implies that $C$ is closed). And the second because, on one hand, $C = \overline C$ is contained in $\overline Z = \overline{C \cap Z}$, and on the other hand, $C$ contains $C \cap Z$ so that $\overline C$ contains $\overline{C \cap Z}$.

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I thought of the "closedness" part of the definition for irreducibility in a wrong way. Thank you very much! – ShinyaSakai Oct 23 '11 at 18:08

As to your second gray box:

Suppose $\overline{Z}$ is reducible, say $\overline{Z} = C \cup D$ with $C$ and $D$ proper closed subsets of $\overline{Z}$. Then $C \cap Z$ and $D \cap Z$ are closed in $Z$, and $Z = (C \cap Z) \cup (D \cap Z)$. There is one last thing to check -- what is it, and how do you check it?

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Thanks very much for your answer. I think I should check that both of them are closed and proper. But I have no idea... – ShinyaSakai Sep 19 '11 at 12:37

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