$sigma(mathcalA) = $ the set of countable unions of countable intersections of elements or complements of elements of $mathcalA$$inf_kgeq n f_k$ is measurable.Identify the smallest sigma-algebra of subsets of $mathbbR$ that contains the set [0, 1]Show that the class of countable unions and intersections of the elements of an algebra is an algebra.(Countable) partition generated $sigma$-algebraShow that $mathcalO$ forms a $sigma$-algebraproduct $sigma$-algebra equals the set of all countable unions of disjoint measurable rectangles?If $Omega = [0,1]$ and $mathcalF$ is the set of subsets $A$ such that either $A$ or $A^c$ are countable, is $mathcalF$ an ($sigma$) algebra?Proof of all finite subsets on $mathbbR$ together with complements is not a $sigma$ algebraWhy isn't minimal sigma-algebra simply the set containing all countable unions and complementsIs a monotone class which is closed under complements and finite disjoint unions a sigma-algebra?
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$sigma(mathcalA) = $ the set of countable unions of countable intersections of elements or complements of elements of $mathcalA$
$inf_kgeq n f_k$ is measurable.Identify the smallest sigma-algebra of subsets of $mathbbR$ that contains the set [0, 1]Show that the class of countable unions and intersections of the elements of an algebra is an algebra.(Countable) partition generated $sigma$-algebraShow that $mathcalO$ forms a $sigma$-algebraproduct $sigma$-algebra equals the set of all countable unions of disjoint measurable rectangles?If $Omega = [0,1]$ and $mathcalF$ is the set of subsets $A$ such that either $A$ or $A^c$ are countable, is $mathcalF$ an ($sigma$) algebra?Proof of all finite subsets on $mathbbR$ together with complements is not a $sigma$ algebraWhy isn't minimal sigma-algebra simply the set containing all countable unions and complementsIs a monotone class which is closed under complements and finite disjoint unions a sigma-algebra?
$begingroup$
Let $mathcalA subseteq mathcalP(Omega)$, $Omega$ a set. Then isn't the set of call countable unions of countable intersections of elements or complements of elements of $mathcalA$ equal to $sigma(mathcalA)$? More precisely, $sigma(mathcalA) = cup_i=1^infty cap_j=1^infty A_j^i : A = A_j^i$ or $A = A_j^i$ for some $A in mathcalA$, for all $i,j$. If that's not true, then I'm pretty sure that if you also contain the countable $cap$ of countable $cup$ 's, then it is true.
I found it difficult trying to prove that $cap_i=1^infty cup_j=1^infty A_j^i = cap_j=1^infty cup_i=1^infty A_j^k(i,j)$, where $k :mathbbNtimesmathbbNrightarrowmathbbN$ is some onto function. So that's why I've also said it might have to include the $capcup$'s.
What do you think, is it true?
measure-theory elementary-set-theory
asked Jul 19 '13 at 2:45
BananaCatsBananaCats
9,35052659
$endgroup$
add a comment |
$begingroup$
Let $mathcalA subseteq mathcalP(Omega)$, $Omega$ a set. Then isn't the set of call countable unions of countable intersections of elements or complements of elements of $mathcalA$ equal to $sigma(mathcalA)$? More precisely, $sigma(mathcalA) = cup_i=1^infty cap_j=1^infty A_j^i : A = A_j^i$ or $A = A_j^i$ for some $A in mathcalA$, for all $i,j$. If that's not true, then I'm pretty sure that if you also contain the countable $cap$ of countable $cup$ 's, then it is true.
I found it difficult trying to prove that $cap_i=1^infty cup_j=1^infty A_j^i = cap_j=1^infty cup_i=1^infty A_j^k(i,j)$, where $k :mathbbNtimesmathbbNrightarrowmathbbN$ is some onto function. So that's why I've also said it might have to include the $capcup$'s.
What do you think, is it true?
measure-theory elementary-set-theory
asked Jul 19 '13 at 2:45
BananaCatsBananaCats
9,35052659
$endgroup$
2
$begingroup$
Not in general.
$endgroup$
– Andrés E. Caicedo
Jul 19 '13 at 2:49
add a comment |
$begingroup$
Let $mathcalA subseteq mathcalP(Omega)$, $Omega$ a set. Then isn't the set of call countable unions of countable intersections of elements or complements of elements of $mathcalA$ equal to $sigma(mathcalA)$? More precisely, $sigma(mathcalA) = cup_i=1^infty cap_j=1^infty A_j^i : A = A_j^i$ or $A = A_j^i$ for some $A in mathcalA$, for all $i,j$. If that's not true, then I'm pretty sure that if you also contain the countable $cap$ of countable $cup$ 's, then it is true.
I found it difficult trying to prove that $cap_i=1^infty cup_j=1^infty A_j^i = cap_j=1^infty cup_i=1^infty A_j^k(i,j)$, where $k :mathbbNtimesmathbbNrightarrowmathbbN$ is some onto function. So that's why I've also said it might have to include the $capcup$'s.
What do you think, is it true?
measure-theory elementary-set-theory
asked Jul 19 '13 at 2:45
BananaCatsBananaCats
9,35052659
$endgroup$
Let $mathcalA subseteq mathcalP(Omega)$, $Omega$ a set. Then isn't the set of call countable unions of countable intersections of elements or complements of elements of $mathcalA$ equal to $sigma(mathcalA)$? More precisely, $sigma(mathcalA) = cup_i=1^infty cap_j=1^infty A_j^i : A = A_j^i$ or $A = A_j^i$ for some $A in mathcalA$, for all $i,j$. If that's not true, then I'm pretty sure that if you also contain the countable $cap$ of countable $cup$ 's, then it is true.
I found it difficult trying to prove that $cap_i=1^infty cup_j=1^infty A_j^i = cap_j=1^infty cup_i=1^infty A_j^k(i,j)$, where $k :mathbbNtimesmathbbNrightarrowmathbbN$ is some onto function. So that's why I've also said it might have to include the $capcup$'s.
What do you think, is it true?
measure-theory elementary-set-theory
measure-theory elementary-set-theory
asked Jul 19 '13 at 2:45
BananaCatsBananaCats
9,35052659
asked Jul 19 '13 at 2:45
BananaCatsBananaCats
9,35052659
asked Jul 19 '13 at 2:45
BananaCatsBananaCats
9,35052659
asked Jul 19 '13 at 2:45
BananaCatsBananaCats
9,35052659
asked Jul 19 '13 at 2:45
BananaCatsBananaCats
9,35052659
9,35052659
2
$begingroup$
Not in general.
$endgroup$
– Andrés E. Caicedo
Jul 19 '13 at 2:49
add a comment |
2
$begingroup$
Not in general.
$endgroup$
– Andrés E. Caicedo
Jul 19 '13 at 2:49
2
2
$begingroup$
Not in general.
$endgroup$
– Andrés E. Caicedo
Jul 19 '13 at 2:49
$begingroup$
Not in general.
$endgroup$
– Andrés E. Caicedo
Jul 19 '13 at 2:49
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $Sigma$ be the described set. $Omega$ is in $Sigma$ since you can construct a trivial $cupcap$ consisting of any $AinmathcalA$ and its complement. If $AinSigma$, then $A = cupcap A_j^i$, then Demorgan's law gives $A = capcup (A_j^i)^c$, which equals a $cupcap$ possibly, or if it doesn't then say we're working with the second described set. Let $(A_n)_ngeq 1$ be a sequence in $Sigma$. Then each $A_n = cupcap A_n,j^i$. A countable union of a countable union is a countable union, so $cup_n=1^inftyA_n = $ Oops!
I see now where my mistake is, we can't let the $capcup$'s in or my last statement gets ruined. But I think $cupcap =$ some $capcup$. If that's true, then this proof will work.
answered Jul 19 '13 at 3:02
BananaCatsBananaCats
9,35052659
$endgroup$
add a comment |
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$begingroup$
Let $Sigma$ be the described set. $Omega$ is in $Sigma$ since you can construct a trivial $cupcap$ consisting of any $AinmathcalA$ and its complement. If $AinSigma$, then $A = cupcap A_j^i$, then Demorgan's law gives $A = capcup (A_j^i)^c$, which equals a $cupcap$ possibly, or if it doesn't then say we're working with the second described set. Let $(A_n)_ngeq 1$ be a sequence in $Sigma$. Then each $A_n = cupcap A_n,j^i$. A countable union of a countable union is a countable union, so $cup_n=1^inftyA_n = $ Oops!
I see now where my mistake is, we can't let the $capcup$'s in or my last statement gets ruined. But I think $cupcap =$ some $capcup$. If that's true, then this proof will work.
answered Jul 19 '13 at 3:02
BananaCatsBananaCats
9,35052659
$endgroup$
add a comment |
$begingroup$
Let $Sigma$ be the described set. $Omega$ is in $Sigma$ since you can construct a trivial $cupcap$ consisting of any $AinmathcalA$ and its complement. If $AinSigma$, then $A = cupcap A_j^i$, then Demorgan's law gives $A = capcup (A_j^i)^c$, which equals a $cupcap$ possibly, or if it doesn't then say we're working with the second described set. Let $(A_n)_ngeq 1$ be a sequence in $Sigma$. Then each $A_n = cupcap A_n,j^i$. A countable union of a countable union is a countable union, so $cup_n=1^inftyA_n = $ Oops!
I see now where my mistake is, we can't let the $capcup$'s in or my last statement gets ruined. But I think $cupcap =$ some $capcup$. If that's true, then this proof will work.
answered Jul 19 '13 at 3:02
BananaCatsBananaCats
9,35052659
$endgroup$
add a comment |
$begingroup$
Let $Sigma$ be the described set. $Omega$ is in $Sigma$ since you can construct a trivial $cupcap$ consisting of any $AinmathcalA$ and its complement. If $AinSigma$, then $A = cupcap A_j^i$, then Demorgan's law gives $A = capcup (A_j^i)^c$, which equals a $cupcap$ possibly, or if it doesn't then say we're working with the second described set. Let $(A_n)_ngeq 1$ be a sequence in $Sigma$. Then each $A_n = cupcap A_n,j^i$. A countable union of a countable union is a countable union, so $cup_n=1^inftyA_n = $ Oops!
I see now where my mistake is, we can't let the $capcup$'s in or my last statement gets ruined. But I think $cupcap =$ some $capcup$. If that's true, then this proof will work.
answered Jul 19 '13 at 3:02
BananaCatsBananaCats
9,35052659
$endgroup$
Let $Sigma$ be the described set. $Omega$ is in $Sigma$ since you can construct a trivial $cupcap$ consisting of any $AinmathcalA$ and its complement. If $AinSigma$, then $A = cupcap A_j^i$, then Demorgan's law gives $A = capcup (A_j^i)^c$, which equals a $cupcap$ possibly, or if it doesn't then say we're working with the second described set. Let $(A_n)_ngeq 1$ be a sequence in $Sigma$. Then each $A_n = cupcap A_n,j^i$. A countable union of a countable union is a countable union, so $cup_n=1^inftyA_n = $ Oops!
I see now where my mistake is, we can't let the $capcup$'s in or my last statement gets ruined. But I think $cupcap =$ some $capcup$. If that's true, then this proof will work.
answered Jul 19 '13 at 3:02
BananaCatsBananaCats
9,35052659
answered Jul 19 '13 at 3:02
BananaCatsBananaCats
9,35052659
answered Jul 19 '13 at 3:02
BananaCatsBananaCats
9,35052659
answered Jul 19 '13 at 3:02
BananaCatsBananaCats
9,35052659
9,35052659
add a comment |
add a comment |
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$begingroup$
Not in general.
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– Andrés E. Caicedo
Jul 19 '13 at 2:49