Collection of sets which is closed under complementation and disjoint union but not finite union
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Question: Give an example of a collection $mathcal{C}$ of subsets of $Omega$ such that $Omegainmathcal{C}$, if $Ainmathcal{C}$ then $A^cinmathcal{C}$ and if $A,Binmathcal{C}$ are disjoint then also $Acup Binmathcal{C}$, while $mathcal{C}$ is not an algebra.
I have proven that if $Omegain mathcal{A}$ and $Acap B^cin mathcal{A}$ whenever $A,Bin mathcal{A},$ then $mathcal{A}$ is an algebra.
My attempt:
Let $Omega = {1,2,3,4}$ and $mathcal{C} = {emptyset, {1,2}, {2,3}, {3,4}, {1,4},Omega}.$
Clearly $Omegain mathcal{C}$ and the collection is closed under complementation and disjoint union.
Since $ {1,2} cap {1,4}^c = {1,2} cap {2,3} = {2} notin mathcal{C},$ so $mathcal{C}$ is not an algebra.
Is my attempt correct?
real-analysis measure-theory proof-verification
asked Dec 6 '18 at 14:15
IdonknowIdonknow
2,368750113
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add a comment |
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Question: Give an example of a collection $mathcal{C}$ of subsets of $Omega$ such that $Omegainmathcal{C}$, if $Ainmathcal{C}$ then $A^cinmathcal{C}$ and if $A,Binmathcal{C}$ are disjoint then also $Acup Binmathcal{C}$, while $mathcal{C}$ is not an algebra.
I have proven that if $Omegain mathcal{A}$ and $Acap B^cin mathcal{A}$ whenever $A,Bin mathcal{A},$ then $mathcal{A}$ is an algebra.
My attempt:
Let $Omega = {1,2,3,4}$ and $mathcal{C} = {emptyset, {1,2}, {2,3}, {3,4}, {1,4},Omega}.$
Clearly $Omegain mathcal{C}$ and the collection is closed under complementation and disjoint union.
Since $ {1,2} cap {1,4}^c = {1,2} cap {2,3} = {2} notin mathcal{C},$ so $mathcal{C}$ is not an algebra.
Is my attempt correct?
real-analysis measure-theory proof-verification
asked Dec 6 '18 at 14:15
IdonknowIdonknow
2,368750113
$endgroup$
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Yes, your solution is correct!
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– p4sch
Dec 6 '18 at 14:28
add a comment |
$begingroup$
Question: Give an example of a collection $mathcal{C}$ of subsets of $Omega$ such that $Omegainmathcal{C}$, if $Ainmathcal{C}$ then $A^cinmathcal{C}$ and if $A,Binmathcal{C}$ are disjoint then also $Acup Binmathcal{C}$, while $mathcal{C}$ is not an algebra.
I have proven that if $Omegain mathcal{A}$ and $Acap B^cin mathcal{A}$ whenever $A,Bin mathcal{A},$ then $mathcal{A}$ is an algebra.
My attempt:
Let $Omega = {1,2,3,4}$ and $mathcal{C} = {emptyset, {1,2}, {2,3}, {3,4}, {1,4},Omega}.$
Clearly $Omegain mathcal{C}$ and the collection is closed under complementation and disjoint union.
Since $ {1,2} cap {1,4}^c = {1,2} cap {2,3} = {2} notin mathcal{C},$ so $mathcal{C}$ is not an algebra.
Is my attempt correct?
real-analysis measure-theory proof-verification
asked Dec 6 '18 at 14:15
IdonknowIdonknow
2,368750113
$endgroup$
Question: Give an example of a collection $mathcal{C}$ of subsets of $Omega$ such that $Omegainmathcal{C}$, if $Ainmathcal{C}$ then $A^cinmathcal{C}$ and if $A,Binmathcal{C}$ are disjoint then also $Acup Binmathcal{C}$, while $mathcal{C}$ is not an algebra.
I have proven that if $Omegain mathcal{A}$ and $Acap B^cin mathcal{A}$ whenever $A,Bin mathcal{A},$ then $mathcal{A}$ is an algebra.
My attempt:
Let $Omega = {1,2,3,4}$ and $mathcal{C} = {emptyset, {1,2}, {2,3}, {3,4}, {1,4},Omega}.$
Clearly $Omegain mathcal{C}$ and the collection is closed under complementation and disjoint union.
Since $ {1,2} cap {1,4}^c = {1,2} cap {2,3} = {2} notin mathcal{C},$ so $mathcal{C}$ is not an algebra.
Is my attempt correct?
real-analysis measure-theory proof-verification
real-analysis measure-theory proof-verification
asked Dec 6 '18 at 14:15
IdonknowIdonknow
2,368750113
asked Dec 6 '18 at 14:15
IdonknowIdonknow
2,368750113
asked Dec 6 '18 at 14:15
IdonknowIdonknow
2,368750113
asked Dec 6 '18 at 14:15
IdonknowIdonknow
2,368750113
asked Dec 6 '18 at 14:15
IdonknowIdonknow
2,368750113
2,368750113
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Yes, your solution is correct!
$endgroup$
– p4sch
Dec 6 '18 at 14:28
add a comment |
$begingroup$
Yes, your solution is correct!
$endgroup$
– p4sch
Dec 6 '18 at 14:28
$begingroup$
Yes, your solution is correct!
$endgroup$
– p4sch
Dec 6 '18 at 14:28
$begingroup$
Yes, your solution is correct!
$endgroup$
– p4sch
Dec 6 '18 at 14:28
add a comment |
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Yes, your solution is correct!
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– p4sch
Dec 6 '18 at 14:28