Mixing
Proof that a singleton set must be closed or open when its ambient space is connected and given subspace is...
I have been asked to prove the following:
Let $X$ be a connected topological space, let $pin X$, and let $X-{p}$ be disconnected. Prove that ${p}$ must be either open or closed in $X$, but cannot be both open and closed in $X$.
I understand that ${p}$ cannot be both open and closed in $X$ as this would imply that ${p}$ and $X-{p}$ are both open in $X$, which implies that $(X-{p}) cup{p}=X$ is disconnected, as it is the union of two open disjoint sets.
So far I have been unable to establish why the case of ${p}$ being neither closed nor open is untenable. If you have any input or suggestions, it would, as always, be greatly appreciated.
general-topology connectedness
asked Nov 21 '18 at 22:48
Heinrich WagnerHeinrich Wagner
24919
add a comment |
I have been asked to prove the following:
Let $X$ be a connected topological space, let $pin X$, and let $X-{p}$ be disconnected. Prove that ${p}$ must be either open or closed in $X$, but cannot be both open and closed in $X$.
I understand that ${p}$ cannot be both open and closed in $X$ as this would imply that ${p}$ and $X-{p}$ are both open in $X$, which implies that $(X-{p}) cup{p}=X$ is disconnected, as it is the union of two open disjoint sets.
So far I have been unable to establish why the case of ${p}$ being neither closed nor open is untenable. If you have any input or suggestions, it would, as always, be greatly appreciated.
general-topology connectedness
asked Nov 21 '18 at 22:48
Heinrich WagnerHeinrich Wagner
24919
add a comment |
I have been asked to prove the following:
Let $X$ be a connected topological space, let $pin X$, and let $X-{p}$ be disconnected. Prove that ${p}$ must be either open or closed in $X$, but cannot be both open and closed in $X$.
I understand that ${p}$ cannot be both open and closed in $X$ as this would imply that ${p}$ and $X-{p}$ are both open in $X$, which implies that $(X-{p}) cup{p}=X$ is disconnected, as it is the union of two open disjoint sets.
So far I have been unable to establish why the case of ${p}$ being neither closed nor open is untenable. If you have any input or suggestions, it would, as always, be greatly appreciated.
general-topology connectedness
asked Nov 21 '18 at 22:48
Heinrich WagnerHeinrich Wagner
24919
I have been asked to prove the following:
Let $X$ be a connected topological space, let $pin X$, and let $X-{p}$ be disconnected. Prove that ${p}$ must be either open or closed in $X$, but cannot be both open and closed in $X$.
I understand that ${p}$ cannot be both open and closed in $X$ as this would imply that ${p}$ and $X-{p}$ are both open in $X$, which implies that $(X-{p}) cup{p}=X$ is disconnected, as it is the union of two open disjoint sets.
So far I have been unable to establish why the case of ${p}$ being neither closed nor open is untenable. If you have any input or suggestions, it would, as always, be greatly appreciated.
general-topology connectedness
general-topology connectedness
asked Nov 21 '18 at 22:48
Heinrich WagnerHeinrich Wagner
24919
asked Nov 21 '18 at 22:48
Heinrich WagnerHeinrich Wagner
24919
asked Nov 21 '18 at 22:48
Heinrich WagnerHeinrich Wagner
24919
asked Nov 21 '18 at 22:48
Heinrich WagnerHeinrich Wagner
24919
asked Nov 21 '18 at 22:48
Heinrich WagnerHeinrich Wagner
24919
24919
add a comment |
add a comment |
1 Answer
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Since $Xsetminus{p}$ is disconnected, there are open non-empty subsets $A$ and $B$ of $Xsetminus{p}$ whose intersection is empty and whose union is $Xsetminus{p}$. Since $A$ and $B$ are open subsets of $Xsetminus{p}$, there are open subsets $A^star$ and $B^star$ of $X$ such that $A=A^starcap(Xsetminus{p})$ and that $B=B^starcap(Xsetminus{p})$. Now, there are $4$ possibilities:
$pnotin A^star$ and $pnotin B^star$: then $A=A^star$ and $B=B^star$. So, ${p}$ is a closed set, since $Xsetminus{p}=A^starcup B^star$, which is an open set.
$pin A^star$ and $pnotin B^star$: this is impossible, because then $X$ would be disconnected, since $A^starcap B^star=Acap B=emptyset$, $A^starcup B^star=X$, and $A^star,B^starneqemptyset$.
$pnotin A^star$ and $pin B^star$: its like the previous case.
$pin A^star$ and $pin B^star$: then ${p}=A^starcap B^star$ and therefore ${p}$ is an open subset of $X$.
answered Nov 21 '18 at 23:13
José Carlos SantosJosé Carlos Santos
152k22123226
Your proof is elegant and clear, and I understand your reasoning fully. Thank you for taking the time to help.
– Heinrich Wagner
Nov 22 '18 at 1:24
add a comment |
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1 Answer
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Since $Xsetminus{p}$ is disconnected, there are open non-empty subsets $A$ and $B$ of $Xsetminus{p}$ whose intersection is empty and whose union is $Xsetminus{p}$. Since $A$ and $B$ are open subsets of $Xsetminus{p}$, there are open subsets $A^star$ and $B^star$ of $X$ such that $A=A^starcap(Xsetminus{p})$ and that $B=B^starcap(Xsetminus{p})$. Now, there are $4$ possibilities:
$pnotin A^star$ and $pnotin B^star$: then $A=A^star$ and $B=B^star$. So, ${p}$ is a closed set, since $Xsetminus{p}=A^starcup B^star$, which is an open set.
$pin A^star$ and $pnotin B^star$: this is impossible, because then $X$ would be disconnected, since $A^starcap B^star=Acap B=emptyset$, $A^starcup B^star=X$, and $A^star,B^starneqemptyset$.
$pnotin A^star$ and $pin B^star$: its like the previous case.
$pin A^star$ and $pin B^star$: then ${p}=A^starcap B^star$ and therefore ${p}$ is an open subset of $X$.
answered Nov 21 '18 at 23:13
José Carlos SantosJosé Carlos Santos
152k22123226
Your proof is elegant and clear, and I understand your reasoning fully. Thank you for taking the time to help.
– Heinrich Wagner
Nov 22 '18 at 1:24
add a comment |
Since $Xsetminus{p}$ is disconnected, there are open non-empty subsets $A$ and $B$ of $Xsetminus{p}$ whose intersection is empty and whose union is $Xsetminus{p}$. Since $A$ and $B$ are open subsets of $Xsetminus{p}$, there are open subsets $A^star$ and $B^star$ of $X$ such that $A=A^starcap(Xsetminus{p})$ and that $B=B^starcap(Xsetminus{p})$. Now, there are $4$ possibilities:
$pnotin A^star$ and $pnotin B^star$: then $A=A^star$ and $B=B^star$. So, ${p}$ is a closed set, since $Xsetminus{p}=A^starcup B^star$, which is an open set.
$pin A^star$ and $pnotin B^star$: this is impossible, because then $X$ would be disconnected, since $A^starcap B^star=Acap B=emptyset$, $A^starcup B^star=X$, and $A^star,B^starneqemptyset$.
$pnotin A^star$ and $pin B^star$: its like the previous case.
$pin A^star$ and $pin B^star$: then ${p}=A^starcap B^star$ and therefore ${p}$ is an open subset of $X$.
answered Nov 21 '18 at 23:13
José Carlos SantosJosé Carlos Santos
152k22123226
Your proof is elegant and clear, and I understand your reasoning fully. Thank you for taking the time to help.
– Heinrich Wagner
Nov 22 '18 at 1:24
add a comment |
Since $Xsetminus{p}$ is disconnected, there are open non-empty subsets $A$ and $B$ of $Xsetminus{p}$ whose intersection is empty and whose union is $Xsetminus{p}$. Since $A$ and $B$ are open subsets of $Xsetminus{p}$, there are open subsets $A^star$ and $B^star$ of $X$ such that $A=A^starcap(Xsetminus{p})$ and that $B=B^starcap(Xsetminus{p})$. Now, there are $4$ possibilities:
$pnotin A^star$ and $pnotin B^star$: then $A=A^star$ and $B=B^star$. So, ${p}$ is a closed set, since $Xsetminus{p}=A^starcup B^star$, which is an open set.
$pin A^star$ and $pnotin B^star$: this is impossible, because then $X$ would be disconnected, since $A^starcap B^star=Acap B=emptyset$, $A^starcup B^star=X$, and $A^star,B^starneqemptyset$.
$pnotin A^star$ and $pin B^star$: its like the previous case.
$pin A^star$ and $pin B^star$: then ${p}=A^starcap B^star$ and therefore ${p}$ is an open subset of $X$.
answered Nov 21 '18 at 23:13
José Carlos SantosJosé Carlos Santos
152k22123226
Since $Xsetminus{p}$ is disconnected, there are open non-empty subsets $A$ and $B$ of $Xsetminus{p}$ whose intersection is empty and whose union is $Xsetminus{p}$. Since $A$ and $B$ are open subsets of $Xsetminus{p}$, there are open subsets $A^star$ and $B^star$ of $X$ such that $A=A^starcap(Xsetminus{p})$ and that $B=B^starcap(Xsetminus{p})$. Now, there are $4$ possibilities:
$pnotin A^star$ and $pnotin B^star$: then $A=A^star$ and $B=B^star$. So, ${p}$ is a closed set, since $Xsetminus{p}=A^starcup B^star$, which is an open set.
$pin A^star$ and $pnotin B^star$: this is impossible, because then $X$ would be disconnected, since $A^starcap B^star=Acap B=emptyset$, $A^starcup B^star=X$, and $A^star,B^starneqemptyset$.
$pnotin A^star$ and $pin B^star$: its like the previous case.
$pin A^star$ and $pin B^star$: then ${p}=A^starcap B^star$ and therefore ${p}$ is an open subset of $X$.
answered Nov 21 '18 at 23:13
José Carlos SantosJosé Carlos Santos
152k22123226
answered Nov 21 '18 at 23:13
José Carlos SantosJosé Carlos Santos
152k22123226
answered Nov 21 '18 at 23:13
José Carlos SantosJosé Carlos Santos
152k22123226
answered Nov 21 '18 at 23:13
José Carlos SantosJosé Carlos Santos
152k22123226
152k22123226
Your proof is elegant and clear, and I understand your reasoning fully. Thank you for taking the time to help.
– Heinrich Wagner
Nov 22 '18 at 1:24
add a comment |
Your proof is elegant and clear, and I understand your reasoning fully. Thank you for taking the time to help.
– Heinrich Wagner
Nov 22 '18 at 1:24
Your proof is elegant and clear, and I understand your reasoning fully. Thank you for taking the time to help.
– Heinrich Wagner
Nov 22 '18 at 1:24
Your proof is elegant and clear, and I understand your reasoning fully. Thank you for taking the time to help.
– Heinrich Wagner
Nov 22 '18 at 1:24
add a comment |
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