Mixing
Example of algebra which is not σ-algebra.
$begingroup$
I have troubles with constructing of an example of algebra of sets which is not σ-algebra. Could you please help me with this?
measure-theory
asked Nov 9 '12 at 17:18
Mihran HovsepyanMihran Hovsepyan
224127
$endgroup$
add a comment |
$begingroup$
I have troubles with constructing of an example of algebra of sets which is not σ-algebra. Could you please help me with this?
measure-theory
asked Nov 9 '12 at 17:18
Mihran HovsepyanMihran Hovsepyan
224127
$endgroup$
add a comment |
$begingroup$
I have troubles with constructing of an example of algebra of sets which is not σ-algebra. Could you please help me with this?
measure-theory
asked Nov 9 '12 at 17:18
Mihran HovsepyanMihran Hovsepyan
224127
$endgroup$
I have troubles with constructing of an example of algebra of sets which is not σ-algebra. Could you please help me with this?
measure-theory
measure-theory
asked Nov 9 '12 at 17:18
Mihran HovsepyanMihran Hovsepyan
224127
asked Nov 9 '12 at 17:18
Mihran HovsepyanMihran Hovsepyan
224127
asked Nov 9 '12 at 17:18
Mihran HovsepyanMihran Hovsepyan
224127
asked Nov 9 '12 at 17:18
Mihran HovsepyanMihran Hovsepyan
224127
asked Nov 9 '12 at 17:18
Mihran HovsepyanMihran Hovsepyan
224127
224127
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Let $X$ be an infinite set, and $mathcal A$ be the collection of all subsets of $X$ which are finite or have finite complement. Then $mathcal A$ is an algebra of sets which is not a $sigma$-algebra.
answered Nov 9 '12 at 17:33
AndrewAndrew
9,28311931
$endgroup$
$begingroup$
But proving this is not sigma algebra would require that every infinite set has a countable subset(as much I know proof) but which require Axiom of Countable choice(CC). Or is there proof without using CC
$endgroup$
– Sushil
Oct 5 '14 at 15:26
$begingroup$
Let $X$ be the set of natural numbers. Let $A_n$ be the set of even numbers up to and including $2n$. Then the countable union of the $A_n$ is the set of all even numbers, which is infinite and has an infinite complement: the set of all odd numbers. Cool :)
$endgroup$
– Michael
Oct 15 '15 at 15:39
add a comment |
$begingroup$
Let $L$ be the collection of all finite disjoint unions of all intervals of the form:
$(−infty, a], (a, b], (b, infty), emptyset, mathbf{R}$.
Then $L$ is an algebra over $mathbf{R}$, but not a σ-algebra because
union of sets $left{(0,frac{i-1}{i}]right}$ for all $i ge 1 = (0, 1) notin L $.
edited Jan 20 at 14:48
pointguard0
1,50211021
answered Oct 5 '14 at 15:23
SushilSushil
1,035927
$endgroup$
add a comment |
$begingroup$
For example, let $mathbb{R}^{[0,+infty)}$ denote the set of all the real value functions on $[0,+infty)$. Consider the so called n-dimensional cylinder set, defined as
begin{equation*}
C_{t_1,t_2,ldots, t_n,B}={fin mathbb{R}^{[0,+infty)}:(f(t_1),f(t_2),ldots, f(t_n))in B},
end{equation*}
where $Binmathcal{B}^d$ is a Borel set and $Csubseteq mathbb{R}^{[0,+infty)}$. Then the set $mathcal{C}'$ of all the cylinders $C$, that is,
begin{equation*}
mathcal{C}'={C_{t_1,t_2,ldots t_n,B}, forall ngeq 1, forall (t,t_1,t_2,ldots,t_n), forall Bin mathcal{B}^n }
end{equation*}
is an algebra but not $sigma$ - algebra.
Anyway it is well known that there exists the smallest $sigma$ - algebra containing $mathcal{C}'$, but this is another story.
answered Nov 15 '12 at 21:33
the_elderthe_elder
1048
$endgroup$
add a comment |
$begingroup$
As another example which supports the first answer above, let $Omega $ $=$ $(0,1]$, and let $mathcal{F}$
contain $varnothing $, all $(a,b]$ with $a,b$ $in $ $mathbb{Q}$, $a,b in [0,1]$ $a$ $<$ $b$, and all finite unions of $(a,b]$. Let $[z]$ round $z$ to the nearest integer. Then $mathcal{F}$ is a
field by straightforward verification. It is not a $sigma $-field: let $A_{n}=(a_{n},1]$ with $a_{n}$ $=$ $%
10^{n}/[10^{n}pi ]$, hence $A_{n}$ $in $ $mathcal{F}$ but $cup
_{n=1}^{infty }A_{n}$ $=$ $(pi ,1]$ $notin $ $mathcal{F}$.
answered Aug 22 '13 at 13:18
JBHJBH
212
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $X$ be an infinite set, and $mathcal A$ be the collection of all subsets of $X$ which are finite or have finite complement. Then $mathcal A$ is an algebra of sets which is not a $sigma$-algebra.
answered Nov 9 '12 at 17:33
AndrewAndrew
9,28311931
$endgroup$
$begingroup$
But proving this is not sigma algebra would require that every infinite set has a countable subset(as much I know proof) but which require Axiom of Countable choice(CC). Or is there proof without using CC
$endgroup$
– Sushil
Oct 5 '14 at 15:26
$begingroup$
Let $X$ be the set of natural numbers. Let $A_n$ be the set of even numbers up to and including $2n$. Then the countable union of the $A_n$ is the set of all even numbers, which is infinite and has an infinite complement: the set of all odd numbers. Cool :)
$endgroup$
– Michael
Oct 15 '15 at 15:39
add a comment |
$begingroup$
Let $X$ be an infinite set, and $mathcal A$ be the collection of all subsets of $X$ which are finite or have finite complement. Then $mathcal A$ is an algebra of sets which is not a $sigma$-algebra.
answered Nov 9 '12 at 17:33
AndrewAndrew
9,28311931
$endgroup$
$begingroup$
But proving this is not sigma algebra would require that every infinite set has a countable subset(as much I know proof) but which require Axiom of Countable choice(CC). Or is there proof without using CC
$endgroup$
– Sushil
Oct 5 '14 at 15:26
$begingroup$
Let $X$ be the set of natural numbers. Let $A_n$ be the set of even numbers up to and including $2n$. Then the countable union of the $A_n$ is the set of all even numbers, which is infinite and has an infinite complement: the set of all odd numbers. Cool :)
$endgroup$
– Michael
Oct 15 '15 at 15:39
add a comment |
$begingroup$
Let $X$ be an infinite set, and $mathcal A$ be the collection of all subsets of $X$ which are finite or have finite complement. Then $mathcal A$ is an algebra of sets which is not a $sigma$-algebra.
answered Nov 9 '12 at 17:33
AndrewAndrew
9,28311931
$endgroup$
Let $X$ be an infinite set, and $mathcal A$ be the collection of all subsets of $X$ which are finite or have finite complement. Then $mathcal A$ is an algebra of sets which is not a $sigma$-algebra.
answered Nov 9 '12 at 17:33
AndrewAndrew
9,28311931
answered Nov 9 '12 at 17:33
AndrewAndrew
9,28311931
answered Nov 9 '12 at 17:33
AndrewAndrew
9,28311931
answered Nov 9 '12 at 17:33
AndrewAndrew
9,28311931
9,28311931
$begingroup$
But proving this is not sigma algebra would require that every infinite set has a countable subset(as much I know proof) but which require Axiom of Countable choice(CC). Or is there proof without using CC
$endgroup$
– Sushil
Oct 5 '14 at 15:26
$begingroup$
Let $X$ be the set of natural numbers. Let $A_n$ be the set of even numbers up to and including $2n$. Then the countable union of the $A_n$ is the set of all even numbers, which is infinite and has an infinite complement: the set of all odd numbers. Cool :)
$endgroup$
– Michael
Oct 15 '15 at 15:39
add a comment |
$begingroup$
But proving this is not sigma algebra would require that every infinite set has a countable subset(as much I know proof) but which require Axiom of Countable choice(CC). Or is there proof without using CC
$endgroup$
– Sushil
Oct 5 '14 at 15:26
$begingroup$
Let $X$ be the set of natural numbers. Let $A_n$ be the set of even numbers up to and including $2n$. Then the countable union of the $A_n$ is the set of all even numbers, which is infinite and has an infinite complement: the set of all odd numbers. Cool :)
$endgroup$
– Michael
Oct 15 '15 at 15:39
$begingroup$
But proving this is not sigma algebra would require that every infinite set has a countable subset(as much I know proof) but which require Axiom of Countable choice(CC). Or is there proof without using CC
$endgroup$
– Sushil
Oct 5 '14 at 15:26
$begingroup$
But proving this is not sigma algebra would require that every infinite set has a countable subset(as much I know proof) but which require Axiom of Countable choice(CC). Or is there proof without using CC
$endgroup$
– Sushil
Oct 5 '14 at 15:26
$begingroup$
Let $X$ be the set of natural numbers. Let $A_n$ be the set of even numbers up to and including $2n$. Then the countable union of the $A_n$ is the set of all even numbers, which is infinite and has an infinite complement: the set of all odd numbers. Cool :)
$endgroup$
– Michael
Oct 15 '15 at 15:39
$begingroup$
Let $X$ be the set of natural numbers. Let $A_n$ be the set of even numbers up to and including $2n$. Then the countable union of the $A_n$ is the set of all even numbers, which is infinite and has an infinite complement: the set of all odd numbers. Cool :)
$endgroup$
– Michael
Oct 15 '15 at 15:39
add a comment |
$begingroup$
Let $L$ be the collection of all finite disjoint unions of all intervals of the form:
$(−infty, a], (a, b], (b, infty), emptyset, mathbf{R}$.
Then $L$ is an algebra over $mathbf{R}$, but not a σ-algebra because
union of sets $left{(0,frac{i-1}{i}]right}$ for all $i ge 1 = (0, 1) notin L $.
edited Jan 20 at 14:48
pointguard0
1,50211021
answered Oct 5 '14 at 15:23
SushilSushil
1,035927
$endgroup$
add a comment |
$begingroup$
Let $L$ be the collection of all finite disjoint unions of all intervals of the form:
$(−infty, a], (a, b], (b, infty), emptyset, mathbf{R}$.
Then $L$ is an algebra over $mathbf{R}$, but not a σ-algebra because
union of sets $left{(0,frac{i-1}{i}]right}$ for all $i ge 1 = (0, 1) notin L $.
edited Jan 20 at 14:48
pointguard0
1,50211021
answered Oct 5 '14 at 15:23
SushilSushil
1,035927
$endgroup$
add a comment |
$begingroup$
Let $L$ be the collection of all finite disjoint unions of all intervals of the form:
$(−infty, a], (a, b], (b, infty), emptyset, mathbf{R}$.
Then $L$ is an algebra over $mathbf{R}$, but not a σ-algebra because
union of sets $left{(0,frac{i-1}{i}]right}$ for all $i ge 1 = (0, 1) notin L $.
edited Jan 20 at 14:48
pointguard0
1,50211021
answered Oct 5 '14 at 15:23
SushilSushil
1,035927
$endgroup$
Let $L$ be the collection of all finite disjoint unions of all intervals of the form:
$(−infty, a], (a, b], (b, infty), emptyset, mathbf{R}$.
Then $L$ is an algebra over $mathbf{R}$, but not a σ-algebra because
union of sets $left{(0,frac{i-1}{i}]right}$ for all $i ge 1 = (0, 1) notin L $.
edited Jan 20 at 14:48
pointguard0
1,50211021
answered Oct 5 '14 at 15:23
SushilSushil
1,035927
edited Jan 20 at 14:48
pointguard0
1,50211021
edited Jan 20 at 14:48
pointguard0
1,50211021
edited Jan 20 at 14:48
pointguard0
1,50211021
1,50211021
answered Oct 5 '14 at 15:23
SushilSushil
1,035927
answered Oct 5 '14 at 15:23
SushilSushil
1,035927
answered Oct 5 '14 at 15:23
SushilSushil
1,035927
1,035927
add a comment |
add a comment |
$begingroup$
For example, let $mathbb{R}^{[0,+infty)}$ denote the set of all the real value functions on $[0,+infty)$. Consider the so called n-dimensional cylinder set, defined as
begin{equation*}
C_{t_1,t_2,ldots, t_n,B}={fin mathbb{R}^{[0,+infty)}:(f(t_1),f(t_2),ldots, f(t_n))in B},
end{equation*}
where $Binmathcal{B}^d$ is a Borel set and $Csubseteq mathbb{R}^{[0,+infty)}$. Then the set $mathcal{C}'$ of all the cylinders $C$, that is,
begin{equation*}
mathcal{C}'={C_{t_1,t_2,ldots t_n,B}, forall ngeq 1, forall (t,t_1,t_2,ldots,t_n), forall Bin mathcal{B}^n }
end{equation*}
is an algebra but not $sigma$ - algebra.
Anyway it is well known that there exists the smallest $sigma$ - algebra containing $mathcal{C}'$, but this is another story.
answered Nov 15 '12 at 21:33
the_elderthe_elder
1048
$endgroup$
add a comment |
$begingroup$
For example, let $mathbb{R}^{[0,+infty)}$ denote the set of all the real value functions on $[0,+infty)$. Consider the so called n-dimensional cylinder set, defined as
begin{equation*}
C_{t_1,t_2,ldots, t_n,B}={fin mathbb{R}^{[0,+infty)}:(f(t_1),f(t_2),ldots, f(t_n))in B},
end{equation*}
where $Binmathcal{B}^d$ is a Borel set and $Csubseteq mathbb{R}^{[0,+infty)}$. Then the set $mathcal{C}'$ of all the cylinders $C$, that is,
begin{equation*}
mathcal{C}'={C_{t_1,t_2,ldots t_n,B}, forall ngeq 1, forall (t,t_1,t_2,ldots,t_n), forall Bin mathcal{B}^n }
end{equation*}
is an algebra but not $sigma$ - algebra.
Anyway it is well known that there exists the smallest $sigma$ - algebra containing $mathcal{C}'$, but this is another story.
answered Nov 15 '12 at 21:33
the_elderthe_elder
1048
$endgroup$
add a comment |
$begingroup$
For example, let $mathbb{R}^{[0,+infty)}$ denote the set of all the real value functions on $[0,+infty)$. Consider the so called n-dimensional cylinder set, defined as
begin{equation*}
C_{t_1,t_2,ldots, t_n,B}={fin mathbb{R}^{[0,+infty)}:(f(t_1),f(t_2),ldots, f(t_n))in B},
end{equation*}
where $Binmathcal{B}^d$ is a Borel set and $Csubseteq mathbb{R}^{[0,+infty)}$. Then the set $mathcal{C}'$ of all the cylinders $C$, that is,
begin{equation*}
mathcal{C}'={C_{t_1,t_2,ldots t_n,B}, forall ngeq 1, forall (t,t_1,t_2,ldots,t_n), forall Bin mathcal{B}^n }
end{equation*}
is an algebra but not $sigma$ - algebra.
Anyway it is well known that there exists the smallest $sigma$ - algebra containing $mathcal{C}'$, but this is another story.
answered Nov 15 '12 at 21:33
the_elderthe_elder
1048
$endgroup$
For example, let $mathbb{R}^{[0,+infty)}$ denote the set of all the real value functions on $[0,+infty)$. Consider the so called n-dimensional cylinder set, defined as
begin{equation*}
C_{t_1,t_2,ldots, t_n,B}={fin mathbb{R}^{[0,+infty)}:(f(t_1),f(t_2),ldots, f(t_n))in B},
end{equation*}
where $Binmathcal{B}^d$ is a Borel set and $Csubseteq mathbb{R}^{[0,+infty)}$. Then the set $mathcal{C}'$ of all the cylinders $C$, that is,
begin{equation*}
mathcal{C}'={C_{t_1,t_2,ldots t_n,B}, forall ngeq 1, forall (t,t_1,t_2,ldots,t_n), forall Bin mathcal{B}^n }
end{equation*}
is an algebra but not $sigma$ - algebra.
Anyway it is well known that there exists the smallest $sigma$ - algebra containing $mathcal{C}'$, but this is another story.
answered Nov 15 '12 at 21:33
the_elderthe_elder
1048
answered Nov 15 '12 at 21:33
the_elderthe_elder
1048
answered Nov 15 '12 at 21:33
the_elderthe_elder
1048
answered Nov 15 '12 at 21:33
the_elderthe_elder
1048
1048
add a comment |
add a comment |
$begingroup$
As another example which supports the first answer above, let $Omega $ $=$ $(0,1]$, and let $mathcal{F}$
contain $varnothing $, all $(a,b]$ with $a,b$ $in $ $mathbb{Q}$, $a,b in [0,1]$ $a$ $<$ $b$, and all finite unions of $(a,b]$. Let $[z]$ round $z$ to the nearest integer. Then $mathcal{F}$ is a
field by straightforward verification. It is not a $sigma $-field: let $A_{n}=(a_{n},1]$ with $a_{n}$ $=$ $%
10^{n}/[10^{n}pi ]$, hence $A_{n}$ $in $ $mathcal{F}$ but $cup
_{n=1}^{infty }A_{n}$ $=$ $(pi ,1]$ $notin $ $mathcal{F}$.
answered Aug 22 '13 at 13:18
JBHJBH
212
$endgroup$
add a comment |
$begingroup$
As another example which supports the first answer above, let $Omega $ $=$ $(0,1]$, and let $mathcal{F}$
contain $varnothing $, all $(a,b]$ with $a,b$ $in $ $mathbb{Q}$, $a,b in [0,1]$ $a$ $<$ $b$, and all finite unions of $(a,b]$. Let $[z]$ round $z$ to the nearest integer. Then $mathcal{F}$ is a
field by straightforward verification. It is not a $sigma $-field: let $A_{n}=(a_{n},1]$ with $a_{n}$ $=$ $%
10^{n}/[10^{n}pi ]$, hence $A_{n}$ $in $ $mathcal{F}$ but $cup
_{n=1}^{infty }A_{n}$ $=$ $(pi ,1]$ $notin $ $mathcal{F}$.
answered Aug 22 '13 at 13:18
JBHJBH
212
$endgroup$
add a comment |
$begingroup$
As another example which supports the first answer above, let $Omega $ $=$ $(0,1]$, and let $mathcal{F}$
contain $varnothing $, all $(a,b]$ with $a,b$ $in $ $mathbb{Q}$, $a,b in [0,1]$ $a$ $<$ $b$, and all finite unions of $(a,b]$. Let $[z]$ round $z$ to the nearest integer. Then $mathcal{F}$ is a
field by straightforward verification. It is not a $sigma $-field: let $A_{n}=(a_{n},1]$ with $a_{n}$ $=$ $%
10^{n}/[10^{n}pi ]$, hence $A_{n}$ $in $ $mathcal{F}$ but $cup
_{n=1}^{infty }A_{n}$ $=$ $(pi ,1]$ $notin $ $mathcal{F}$.
answered Aug 22 '13 at 13:18
JBHJBH
212
$endgroup$
As another example which supports the first answer above, let $Omega $ $=$ $(0,1]$, and let $mathcal{F}$
contain $varnothing $, all $(a,b]$ with $a,b$ $in $ $mathbb{Q}$, $a,b in [0,1]$ $a$ $<$ $b$, and all finite unions of $(a,b]$. Let $[z]$ round $z$ to the nearest integer. Then $mathcal{F}$ is a
field by straightforward verification. It is not a $sigma $-field: let $A_{n}=(a_{n},1]$ with $a_{n}$ $=$ $%
10^{n}/[10^{n}pi ]$, hence $A_{n}$ $in $ $mathcal{F}$ but $cup
_{n=1}^{infty }A_{n}$ $=$ $(pi ,1]$ $notin $ $mathcal{F}$.
answered Aug 22 '13 at 13:18
JBHJBH
212
answered Aug 22 '13 at 13:18
JBHJBH
212
answered Aug 22 '13 at 13:18
JBHJBH
212
answered Aug 22 '13 at 13:18
JBHJBH
212
212
add a comment |
add a comment |
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