Is the family of sets is algebra, sigma-algebra? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Tail $sigma$-algebra for a sequence of sets$sigma$-algebra of well-approximated Borel setsGenerate the smallest $sigma$-algebra containing a given family of setsAlgebra vs. Sigma-Algebra ConditionWhen is the image of a $sigma$-algebra a $sigma$-algebra?Union of ascending chain of $sigma$-algebras is not necessarily a $sigma$-algebra, need help concluding.Pi-system and the union of $sigma$-algebrasFind the sigma algebra generated by the following setsset of points in $E_n$ infinitely often and almost always are in $sigma$-algebraAn algebra containing a countable family of subsets is countable as well.
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Is the family of sets is algebra, sigma-algebra?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Tail $sigma$-algebra for a sequence of sets$sigma$-algebra of well-approximated Borel setsGenerate the smallest $sigma$-algebra containing a given family of setsAlgebra vs. Sigma-Algebra ConditionWhen is the image of a $sigma$-algebra a $sigma$-algebra?Union of ascending chain of $sigma$-algebras is not necessarily a $sigma$-algebra, need help concluding.Pi-system and the union of $sigma$-algebrasFind the sigma algebra generated by the following setsset of points in $E_n$ infinitely often and almost always are in $sigma$-algebraAn algebra containing a countable family of subsets is countable as well.
$begingroup$
$mathcal A = A:= bigcup_i=1^m I_i : I_1,...,I_m in mathbb I, I_i cap I_j = emptyset, i not = j.$
$mathbb I= (a, b], (- infty,b], (a, + infty), (- infty, + infty).$
Is $mathcal A$ algebra, sigma-algebra?
I started from:
1) $mathbb R in mathcal A$
2) Let $A = bigcup_i=1^m I_i in mathcal A.$ Then $A^c = (bigcup_i=1^m I_i)^c = bigcap_i=1^m (I_i)^c = I_i^ccap...cap I_m^c in mathcal A.$
3) Let $A, B in mathcal A.$ Then $A cup B = (cup_i=1^m I_i') bigcup (cup_i=1^m I_i'')=bigcup_i=1^m (I_i' cup I_i'') in mathcal A.$
general-topology measure-theory
asked Apr 1 at 7:59
PhilipPhilip
917
$endgroup$
add a comment |
$begingroup$
$mathcal A = A:= bigcup_i=1^m I_i : I_1,...,I_m in mathbb I, I_i cap I_j = emptyset, i not = j.$
$mathbb I= (a, b], (- infty,b], (a, + infty), (- infty, + infty).$
Is $mathcal A$ algebra, sigma-algebra?
I started from:
1) $mathbb R in mathcal A$
2) Let $A = bigcup_i=1^m I_i in mathcal A.$ Then $A^c = (bigcup_i=1^m I_i)^c = bigcap_i=1^m (I_i)^c = I_i^ccap...cap I_m^c in mathcal A.$
3) Let $A, B in mathcal A.$ Then $A cup B = (cup_i=1^m I_i') bigcup (cup_i=1^m I_i'')=bigcup_i=1^m (I_i' cup I_i'') in mathcal A.$
general-topology measure-theory
asked Apr 1 at 7:59
PhilipPhilip
917
$endgroup$
add a comment |
$begingroup$
$mathcal A = A:= bigcup_i=1^m I_i : I_1,...,I_m in mathbb I, I_i cap I_j = emptyset, i not = j.$
$mathbb I= (a, b], (- infty,b], (a, + infty), (- infty, + infty).$
Is $mathcal A$ algebra, sigma-algebra?
I started from:
1) $mathbb R in mathcal A$
2) Let $A = bigcup_i=1^m I_i in mathcal A.$ Then $A^c = (bigcup_i=1^m I_i)^c = bigcap_i=1^m (I_i)^c = I_i^ccap...cap I_m^c in mathcal A.$
3) Let $A, B in mathcal A.$ Then $A cup B = (cup_i=1^m I_i') bigcup (cup_i=1^m I_i'')=bigcup_i=1^m (I_i' cup I_i'') in mathcal A.$
general-topology measure-theory
asked Apr 1 at 7:59
PhilipPhilip
917
$endgroup$
$mathcal A = A:= bigcup_i=1^m I_i : I_1,...,I_m in mathbb I, I_i cap I_j = emptyset, i not = j.$
$mathbb I= (a, b], (- infty,b], (a, + infty), (- infty, + infty).$
Is $mathcal A$ algebra, sigma-algebra?
I started from:
1) $mathbb R in mathcal A$
2) Let $A = bigcup_i=1^m I_i in mathcal A.$ Then $A^c = (bigcup_i=1^m I_i)^c = bigcap_i=1^m (I_i)^c = I_i^ccap...cap I_m^c in mathcal A.$
3) Let $A, B in mathcal A.$ Then $A cup B = (cup_i=1^m I_i') bigcup (cup_i=1^m I_i'')=bigcup_i=1^m (I_i' cup I_i'') in mathcal A.$
general-topology measure-theory
general-topology measure-theory
asked Apr 1 at 7:59
PhilipPhilip
917
asked Apr 1 at 7:59
PhilipPhilip
917
asked Apr 1 at 7:59
PhilipPhilip
917
asked Apr 1 at 7:59
PhilipPhilip
917
asked Apr 1 at 7:59
PhilipPhilip
917
917
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$bigcup_ninmathbb Z(2n,2n+1]$ is not in $mathcal A$ so $mathcal A$ is not a sigma algebra.
To show it is an algebra you have to first show that any union of intervals of the type mentioned can be expressed as a disjoint union. The complement of a set in $mathcal A$ is easy to compute.
answered Apr 1 at 8:01
Kavi Rama MurthyKavi Rama Murthy
75.3k53270
$endgroup$
$begingroup$
Can you give some hints how to show that any union of intervals of the type mentioned can be expressed as a disjoint union?
$endgroup$
– Philip
Apr 1 at 8:50
$begingroup$
$I_1cup I_2cup...cup I_n=I_1cup (I_2setminus I_1) cup...cup (I_nsetminus (I_1 cup I_2cup...I_n-1)$. Next see what happens to $I_2setminus I_1$. Can you see that this is a disjoint union of two intervals or a single interval of the type we have?. Next look at $I_3setminus I_1 cup I_2$, etc.
$endgroup$
– Kavi Rama Murthy
Apr 1 at 8:54
$begingroup$
Thank you! And the complement I can get in the similar way?
$endgroup$
– Philip
Apr 1 at 9:02
$begingroup$
@Philip Yes,you can use similar arguments for the complement.
$endgroup$
– Kavi Rama Murthy
Apr 1 at 9:04
add a comment |
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$begingroup$
$bigcup_ninmathbb Z(2n,2n+1]$ is not in $mathcal A$ so $mathcal A$ is not a sigma algebra.
To show it is an algebra you have to first show that any union of intervals of the type mentioned can be expressed as a disjoint union. The complement of a set in $mathcal A$ is easy to compute.
answered Apr 1 at 8:01
Kavi Rama MurthyKavi Rama Murthy
75.3k53270
$endgroup$
$begingroup$
Can you give some hints how to show that any union of intervals of the type mentioned can be expressed as a disjoint union?
$endgroup$
– Philip
Apr 1 at 8:50
$begingroup$
$I_1cup I_2cup...cup I_n=I_1cup (I_2setminus I_1) cup...cup (I_nsetminus (I_1 cup I_2cup...I_n-1)$. Next see what happens to $I_2setminus I_1$. Can you see that this is a disjoint union of two intervals or a single interval of the type we have?. Next look at $I_3setminus I_1 cup I_2$, etc.
$endgroup$
– Kavi Rama Murthy
Apr 1 at 8:54
$begingroup$
Thank you! And the complement I can get in the similar way?
$endgroup$
– Philip
Apr 1 at 9:02
$begingroup$
@Philip Yes,you can use similar arguments for the complement.
$endgroup$
– Kavi Rama Murthy
Apr 1 at 9:04
add a comment |
$begingroup$
$bigcup_ninmathbb Z(2n,2n+1]$ is not in $mathcal A$ so $mathcal A$ is not a sigma algebra.
To show it is an algebra you have to first show that any union of intervals of the type mentioned can be expressed as a disjoint union. The complement of a set in $mathcal A$ is easy to compute.
answered Apr 1 at 8:01
Kavi Rama MurthyKavi Rama Murthy
75.3k53270
$endgroup$
$begingroup$
Can you give some hints how to show that any union of intervals of the type mentioned can be expressed as a disjoint union?
$endgroup$
– Philip
Apr 1 at 8:50
$begingroup$
$I_1cup I_2cup...cup I_n=I_1cup (I_2setminus I_1) cup...cup (I_nsetminus (I_1 cup I_2cup...I_n-1)$. Next see what happens to $I_2setminus I_1$. Can you see that this is a disjoint union of two intervals or a single interval of the type we have?. Next look at $I_3setminus I_1 cup I_2$, etc.
$endgroup$
– Kavi Rama Murthy
Apr 1 at 8:54
$begingroup$
Thank you! And the complement I can get in the similar way?
$endgroup$
– Philip
Apr 1 at 9:02
$begingroup$
@Philip Yes,you can use similar arguments for the complement.
$endgroup$
– Kavi Rama Murthy
Apr 1 at 9:04
add a comment |
$begingroup$
$bigcup_ninmathbb Z(2n,2n+1]$ is not in $mathcal A$ so $mathcal A$ is not a sigma algebra.
To show it is an algebra you have to first show that any union of intervals of the type mentioned can be expressed as a disjoint union. The complement of a set in $mathcal A$ is easy to compute.
answered Apr 1 at 8:01
Kavi Rama MurthyKavi Rama Murthy
75.3k53270
$endgroup$
$bigcup_ninmathbb Z(2n,2n+1]$ is not in $mathcal A$ so $mathcal A$ is not a sigma algebra.
To show it is an algebra you have to first show that any union of intervals of the type mentioned can be expressed as a disjoint union. The complement of a set in $mathcal A$ is easy to compute.
answered Apr 1 at 8:01
Kavi Rama MurthyKavi Rama Murthy
75.3k53270
edited Apr 1 at 8:10
answered Apr 1 at 8:01
Kavi Rama MurthyKavi Rama Murthy
75.3k53270
answered Apr 1 at 8:01
Kavi Rama MurthyKavi Rama Murthy
75.3k53270
answered Apr 1 at 8:01
Kavi Rama MurthyKavi Rama Murthy
75.3k53270
75.3k53270
$begingroup$
Can you give some hints how to show that any union of intervals of the type mentioned can be expressed as a disjoint union?
$endgroup$
– Philip
Apr 1 at 8:50
$begingroup$
$I_1cup I_2cup...cup I_n=I_1cup (I_2setminus I_1) cup...cup (I_nsetminus (I_1 cup I_2cup...I_n-1)$. Next see what happens to $I_2setminus I_1$. Can you see that this is a disjoint union of two intervals or a single interval of the type we have?. Next look at $I_3setminus I_1 cup I_2$, etc.
$endgroup$
– Kavi Rama Murthy
Apr 1 at 8:54
$begingroup$
Thank you! And the complement I can get in the similar way?
$endgroup$
– Philip
Apr 1 at 9:02
$begingroup$
@Philip Yes,you can use similar arguments for the complement.
$endgroup$
– Kavi Rama Murthy
Apr 1 at 9:04
add a comment |
$begingroup$
Can you give some hints how to show that any union of intervals of the type mentioned can be expressed as a disjoint union?
$endgroup$
– Philip
Apr 1 at 8:50
$begingroup$
$I_1cup I_2cup...cup I_n=I_1cup (I_2setminus I_1) cup...cup (I_nsetminus (I_1 cup I_2cup...I_n-1)$. Next see what happens to $I_2setminus I_1$. Can you see that this is a disjoint union of two intervals or a single interval of the type we have?. Next look at $I_3setminus I_1 cup I_2$, etc.
$endgroup$
– Kavi Rama Murthy
Apr 1 at 8:54
$begingroup$
Thank you! And the complement I can get in the similar way?
$endgroup$
– Philip
Apr 1 at 9:02
$begingroup$
@Philip Yes,you can use similar arguments for the complement.
$endgroup$
– Kavi Rama Murthy
Apr 1 at 9:04
$begingroup$
Can you give some hints how to show that any union of intervals of the type mentioned can be expressed as a disjoint union?
$endgroup$
– Philip
Apr 1 at 8:50
$begingroup$
Can you give some hints how to show that any union of intervals of the type mentioned can be expressed as a disjoint union?
$endgroup$
– Philip
Apr 1 at 8:50
$begingroup$
$I_1cup I_2cup...cup I_n=I_1cup (I_2setminus I_1) cup...cup (I_nsetminus (I_1 cup I_2cup...I_n-1)$. Next see what happens to $I_2setminus I_1$. Can you see that this is a disjoint union of two intervals or a single interval of the type we have?. Next look at $I_3setminus I_1 cup I_2$, etc.
$endgroup$
– Kavi Rama Murthy
Apr 1 at 8:54
$begingroup$
$I_1cup I_2cup...cup I_n=I_1cup (I_2setminus I_1) cup...cup (I_nsetminus (I_1 cup I_2cup...I_n-1)$. Next see what happens to $I_2setminus I_1$. Can you see that this is a disjoint union of two intervals or a single interval of the type we have?. Next look at $I_3setminus I_1 cup I_2$, etc.
$endgroup$
– Kavi Rama Murthy
Apr 1 at 8:54
$begingroup$
Thank you! And the complement I can get in the similar way?
$endgroup$
– Philip
Apr 1 at 9:02
$begingroup$
Thank you! And the complement I can get in the similar way?
$endgroup$
– Philip
Apr 1 at 9:02
$begingroup$
@Philip Yes,you can use similar arguments for the complement.
$endgroup$
– Kavi Rama Murthy
Apr 1 at 9:04
$begingroup$
@Philip Yes,you can use similar arguments for the complement.
$endgroup$
– Kavi Rama Murthy
Apr 1 at 9:04
add a comment |
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