If $Y$ has irreducible components $Y_1, cdots, Y_n$, then the $overline{Y_i}$ are the irreducible components...












0















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?






general-topology







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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-Álvarez
    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
















0















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?






general-topology







share|cite|improve this question
























  • Double check your typing. It makes wonders. Specially in titles, which is the very first thing people see of your question!
    – Mariano Suárez-Álvarez
    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














0












0








0








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?






general-topology







share|cite|improve this question
















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?






general-topology




general-topology






share|cite|improve this question















share|cite|improve this question













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share|cite|improve this question








edited Sep 19 '11 at 3:57









Arturo Magidin

260k32584904




260k32584904










asked Sep 19 '11 at 3:41









ShinyaSakai

2,96512450




2,96512450












  • Double check your typing. It makes wonders. Specially in titles, which is the very first thing people see of your question!
    – Mariano Suárez-Álvarez
    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


















  • Double check your typing. It makes wonders. Specially in titles, which is the very first thing people see of your question!
    – Mariano Suárez-Álvarez
    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
















Double check your typing. It makes wonders. Specially in titles, which is the very first thing people see of your question!
– Mariano Suárez-Álvarez
Sep 19 '11 at 3:47




Double check your typing. It makes wonders. Specially in titles, which is the very first thing people see of your question!
– Mariano Suárez-Álvarez
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




@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










2 Answers
2






active

oldest

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1





+50









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 $overline{C cap Z}$.






share|cite|improve this answer





















  • 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



















1














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?






share|cite|improve this answer





















  • 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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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1





+50









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 $overline{C cap Z}$.






share|cite|improve this answer





















  • 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
















1





+50









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 $overline{C cap Z}$.






share|cite|improve this answer





















  • 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














1





+50







1





+50



1




+50




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 $overline{C cap Z}$.






share|cite|improve this answer












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 $overline{C cap Z}$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Oct 23 '11 at 7:00









sibilant

467212




467212












  • 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


















  • 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
















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




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











1














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?






share|cite|improve this answer





















  • 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
















1














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?






share|cite|improve this answer





















  • 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














1












1








1






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?






share|cite|improve this answer












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?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Sep 19 '11 at 4:51









Daniel McLaury

15.5k32977




15.5k32977












  • 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


















  • 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
















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




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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