The $sigma$-algebra generated by $F = {A subseteq mathbb{R} : 0 in A^{circ} text{or } 0 in (A^c)^circ}$...
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I am trying to show that the $sigma$-algebra generated by $F = {A subseteq mathbb{R} : 0 in A^{circ} text{or } 0 in (A^c)^circ}$ contains all singleton sets ${x} in mathbb{R}$.
I know that $F$ is an algebra. I believe $F$ itself is not a $sigma$-algebra, since, if we take an infinite sequence of $A_n in F$ such that $0 in ((A_n)^c)^circ$ for each $n in mathbb{N}$ then we do not necessarily have that $0 in cup_{n=1}^{infty}A_n$ (since $0 in cap_{n=1}^{infty}(A_n^c)^circ supseteq (cap_{n=1}^{infty}((A_n)^c)^circ = (cup_{n=1}^{infty}A_n)^circ$).
So, I am not exactly sure how to explicitly construct and/or describe the $sigma$-algebra generated by $F$. (From there, I am pretty sure I should be able to show that the singleton sets are in $sigma(F)$. I just don't know how to get to that point in the first place.)
This link gave an explanation about constructing a $sigma$-algebra from a collection of sets, but it ended up being more confusing than helpful.
real-analysis elementary-set-theory
asked Nov 25 at 19:13
Jane Doe
14712
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up vote
2
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I am trying to show that the $sigma$-algebra generated by $F = {A subseteq mathbb{R} : 0 in A^{circ} text{or } 0 in (A^c)^circ}$ contains all singleton sets ${x} in mathbb{R}$.
I know that $F$ is an algebra. I believe $F$ itself is not a $sigma$-algebra, since, if we take an infinite sequence of $A_n in F$ such that $0 in ((A_n)^c)^circ$ for each $n in mathbb{N}$ then we do not necessarily have that $0 in cup_{n=1}^{infty}A_n$ (since $0 in cap_{n=1}^{infty}(A_n^c)^circ supseteq (cap_{n=1}^{infty}((A_n)^c)^circ = (cup_{n=1}^{infty}A_n)^circ$).
So, I am not exactly sure how to explicitly construct and/or describe the $sigma$-algebra generated by $F$. (From there, I am pretty sure I should be able to show that the singleton sets are in $sigma(F)$. I just don't know how to get to that point in the first place.)
This link gave an explanation about constructing a $sigma$-algebra from a collection of sets, but it ended up being more confusing than helpful.
real-analysis elementary-set-theory
asked Nov 25 at 19:13
Jane Doe
14712
$F$ is not a $sigma$-algebra because $A=(-1,1)in F$ and $B=(1,2)in F$ but $Acup Bnot in F.$
– DanielWainfleet
Nov 25 at 20:08
@DanielWainfleet yes, I mentioned that $F$ is not a $sigma$-algebra in my post. I am trying to figure out how to describe the $sigma$-algebra generated by $F$ and show that each singleton is contained in that $sigma$-algebra.
– Jane Doe
Nov 25 at 20:11
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I am trying to show that the $sigma$-algebra generated by $F = {A subseteq mathbb{R} : 0 in A^{circ} text{or } 0 in (A^c)^circ}$ contains all singleton sets ${x} in mathbb{R}$.
I know that $F$ is an algebra. I believe $F$ itself is not a $sigma$-algebra, since, if we take an infinite sequence of $A_n in F$ such that $0 in ((A_n)^c)^circ$ for each $n in mathbb{N}$ then we do not necessarily have that $0 in cup_{n=1}^{infty}A_n$ (since $0 in cap_{n=1}^{infty}(A_n^c)^circ supseteq (cap_{n=1}^{infty}((A_n)^c)^circ = (cup_{n=1}^{infty}A_n)^circ$).
So, I am not exactly sure how to explicitly construct and/or describe the $sigma$-algebra generated by $F$. (From there, I am pretty sure I should be able to show that the singleton sets are in $sigma(F)$. I just don't know how to get to that point in the first place.)
This link gave an explanation about constructing a $sigma$-algebra from a collection of sets, but it ended up being more confusing than helpful.
real-analysis elementary-set-theory
asked Nov 25 at 19:13
Jane Doe
14712
I am trying to show that the $sigma$-algebra generated by $F = {A subseteq mathbb{R} : 0 in A^{circ} text{or } 0 in (A^c)^circ}$ contains all singleton sets ${x} in mathbb{R}$.
I know that $F$ is an algebra. I believe $F$ itself is not a $sigma$-algebra, since, if we take an infinite sequence of $A_n in F$ such that $0 in ((A_n)^c)^circ$ for each $n in mathbb{N}$ then we do not necessarily have that $0 in cup_{n=1}^{infty}A_n$ (since $0 in cap_{n=1}^{infty}(A_n^c)^circ supseteq (cap_{n=1}^{infty}((A_n)^c)^circ = (cup_{n=1}^{infty}A_n)^circ$).
So, I am not exactly sure how to explicitly construct and/or describe the $sigma$-algebra generated by $F$. (From there, I am pretty sure I should be able to show that the singleton sets are in $sigma(F)$. I just don't know how to get to that point in the first place.)
This link gave an explanation about constructing a $sigma$-algebra from a collection of sets, but it ended up being more confusing than helpful.
real-analysis elementary-set-theory
real-analysis elementary-set-theory
asked Nov 25 at 19:13
Jane Doe
14712
asked Nov 25 at 19:13
Jane Doe
14712
edited Nov 25 at 20:09
asked Nov 25 at 19:13
Jane Doe
14712
asked Nov 25 at 19:13
Jane Doe
14712
asked Nov 25 at 19:13
Jane Doe
14712
14712
$F$ is not a $sigma$-algebra because $A=(-1,1)in F$ and $B=(1,2)in F$ but $Acup Bnot in F.$
– DanielWainfleet
Nov 25 at 20:08
@DanielWainfleet yes, I mentioned that $F$ is not a $sigma$-algebra in my post. I am trying to figure out how to describe the $sigma$-algebra generated by $F$ and show that each singleton is contained in that $sigma$-algebra.
– Jane Doe
Nov 25 at 20:11
add a comment |
$F$ is not a $sigma$-algebra because $A=(-1,1)in F$ and $B=(1,2)in F$ but $Acup Bnot in F.$
– DanielWainfleet
Nov 25 at 20:08
@DanielWainfleet yes, I mentioned that $F$ is not a $sigma$-algebra in my post. I am trying to figure out how to describe the $sigma$-algebra generated by $F$ and show that each singleton is contained in that $sigma$-algebra.
– Jane Doe
Nov 25 at 20:11
$F$ is not a $sigma$-algebra because $A=(-1,1)in F$ and $B=(1,2)in F$ but $Acup Bnot in F.$
– DanielWainfleet
Nov 25 at 20:08
$F$ is not a $sigma$-algebra because $A=(-1,1)in F$ and $B=(1,2)in F$ but $Acup Bnot in F.$
– DanielWainfleet
Nov 25 at 20:08
@DanielWainfleet yes, I mentioned that $F$ is not a $sigma$-algebra in my post. I am trying to figure out how to describe the $sigma$-algebra generated by $F$ and show that each singleton is contained in that $sigma$-algebra.
– Jane Doe
Nov 25 at 20:11
@DanielWainfleet yes, I mentioned that $F$ is not a $sigma$-algebra in my post. I am trying to figure out how to describe the $sigma$-algebra generated by $F$ and show that each singleton is contained in that $sigma$-algebra.
– Jane Doe
Nov 25 at 20:11
add a comment |
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
It contains ${0}$ because that is the intersection of the sets $A_n = (-frac{1}{n},frac{1}{n})$, which all verify $0in A_n^circ$, hence $A_n in F$. On the other hand, if $xinmathbb{R}setminus {0}$, then we have that ${x}$ is closed and ${x}^c$ is an open set containing $0$, hence ${x} in F$.
answered Nov 25 at 19:51
mlerma54
83438
2
This is correct but unnecessarily complicated. All singletons except ${0}$ are in $F$, so all that's needed is to prove that the $sigma$-algebra contains ${0}$, which is the first sentence of your answer.
– Andreas Blass
Nov 26 at 1:35
You are right, I have edited my solution to make it simpler according to your remark.
– mlerma54
Nov 26 at 3:30
add a comment |
up vote
1
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Every $Ssubset Bbb R$ belongs to $sigma(F).$
(I)... $Ssetminus (-1/n,1/n)in F$ for each $nin Bbb N$ so $Ssetminus {0}=cup_{nin Bbb N},(,Ssetminus (-1/n,1/n),)in sigma (F).$
(II)... $(-1/n,1/n)in F$ for each $nin Bbb N$ so ${0}=cap_{nin Bbb N}(-1/n,1/n)in sigma (F).$
(III)... So either $S=Ssetminus {0}in sigma(F)$ or $S=(Ssetminus {0})cup {0}in sigma(F).$
answered Nov 25 at 21:05
DanielWainfleet
33.7k31647
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
It contains ${0}$ because that is the intersection of the sets $A_n = (-frac{1}{n},frac{1}{n})$, which all verify $0in A_n^circ$, hence $A_n in F$. On the other hand, if $xinmathbb{R}setminus {0}$, then we have that ${x}$ is closed and ${x}^c$ is an open set containing $0$, hence ${x} in F$.
answered Nov 25 at 19:51
mlerma54
83438
2
This is correct but unnecessarily complicated. All singletons except ${0}$ are in $F$, so all that's needed is to prove that the $sigma$-algebra contains ${0}$, which is the first sentence of your answer.
– Andreas Blass
Nov 26 at 1:35
You are right, I have edited my solution to make it simpler according to your remark.
– mlerma54
Nov 26 at 3:30
add a comment |
up vote
2
down vote
accepted
It contains ${0}$ because that is the intersection of the sets $A_n = (-frac{1}{n},frac{1}{n})$, which all verify $0in A_n^circ$, hence $A_n in F$. On the other hand, if $xinmathbb{R}setminus {0}$, then we have that ${x}$ is closed and ${x}^c$ is an open set containing $0$, hence ${x} in F$.
answered Nov 25 at 19:51
mlerma54
83438
2
This is correct but unnecessarily complicated. All singletons except ${0}$ are in $F$, so all that's needed is to prove that the $sigma$-algebra contains ${0}$, which is the first sentence of your answer.
– Andreas Blass
Nov 26 at 1:35
You are right, I have edited my solution to make it simpler according to your remark.
– mlerma54
Nov 26 at 3:30
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
It contains ${0}$ because that is the intersection of the sets $A_n = (-frac{1}{n},frac{1}{n})$, which all verify $0in A_n^circ$, hence $A_n in F$. On the other hand, if $xinmathbb{R}setminus {0}$, then we have that ${x}$ is closed and ${x}^c$ is an open set containing $0$, hence ${x} in F$.
answered Nov 25 at 19:51
mlerma54
83438
It contains ${0}$ because that is the intersection of the sets $A_n = (-frac{1}{n},frac{1}{n})$, which all verify $0in A_n^circ$, hence $A_n in F$. On the other hand, if $xinmathbb{R}setminus {0}$, then we have that ${x}$ is closed and ${x}^c$ is an open set containing $0$, hence ${x} in F$.
answered Nov 25 at 19:51
mlerma54
83438
edited Nov 26 at 4:06
answered Nov 25 at 19:51
mlerma54
83438
answered Nov 25 at 19:51
mlerma54
83438
answered Nov 25 at 19:51
mlerma54
83438
83438
2
This is correct but unnecessarily complicated. All singletons except ${0}$ are in $F$, so all that's needed is to prove that the $sigma$-algebra contains ${0}$, which is the first sentence of your answer.
– Andreas Blass
Nov 26 at 1:35
You are right, I have edited my solution to make it simpler according to your remark.
– mlerma54
Nov 26 at 3:30
add a comment |
2
This is correct but unnecessarily complicated. All singletons except ${0}$ are in $F$, so all that's needed is to prove that the $sigma$-algebra contains ${0}$, which is the first sentence of your answer.
– Andreas Blass
Nov 26 at 1:35
You are right, I have edited my solution to make it simpler according to your remark.
– mlerma54
Nov 26 at 3:30
2
2
This is correct but unnecessarily complicated. All singletons except ${0}$ are in $F$, so all that's needed is to prove that the $sigma$-algebra contains ${0}$, which is the first sentence of your answer.
– Andreas Blass
Nov 26 at 1:35
This is correct but unnecessarily complicated. All singletons except ${0}$ are in $F$, so all that's needed is to prove that the $sigma$-algebra contains ${0}$, which is the first sentence of your answer.
– Andreas Blass
Nov 26 at 1:35
You are right, I have edited my solution to make it simpler according to your remark.
– mlerma54
Nov 26 at 3:30
You are right, I have edited my solution to make it simpler according to your remark.
– mlerma54
Nov 26 at 3:30
add a comment |
up vote
1
down vote
Every $Ssubset Bbb R$ belongs to $sigma(F).$
(I)... $Ssetminus (-1/n,1/n)in F$ for each $nin Bbb N$ so $Ssetminus {0}=cup_{nin Bbb N},(,Ssetminus (-1/n,1/n),)in sigma (F).$
(II)... $(-1/n,1/n)in F$ for each $nin Bbb N$ so ${0}=cap_{nin Bbb N}(-1/n,1/n)in sigma (F).$
(III)... So either $S=Ssetminus {0}in sigma(F)$ or $S=(Ssetminus {0})cup {0}in sigma(F).$
answered Nov 25 at 21:05
DanielWainfleet
33.7k31647
add a comment |
up vote
1
down vote
Every $Ssubset Bbb R$ belongs to $sigma(F).$
(I)... $Ssetminus (-1/n,1/n)in F$ for each $nin Bbb N$ so $Ssetminus {0}=cup_{nin Bbb N},(,Ssetminus (-1/n,1/n),)in sigma (F).$
(II)... $(-1/n,1/n)in F$ for each $nin Bbb N$ so ${0}=cap_{nin Bbb N}(-1/n,1/n)in sigma (F).$
(III)... So either $S=Ssetminus {0}in sigma(F)$ or $S=(Ssetminus {0})cup {0}in sigma(F).$
answered Nov 25 at 21:05
DanielWainfleet
33.7k31647
add a comment |
up vote
1
down vote
up vote
1
down vote
Every $Ssubset Bbb R$ belongs to $sigma(F).$
(I)... $Ssetminus (-1/n,1/n)in F$ for each $nin Bbb N$ so $Ssetminus {0}=cup_{nin Bbb N},(,Ssetminus (-1/n,1/n),)in sigma (F).$
(II)... $(-1/n,1/n)in F$ for each $nin Bbb N$ so ${0}=cap_{nin Bbb N}(-1/n,1/n)in sigma (F).$
(III)... So either $S=Ssetminus {0}in sigma(F)$ or $S=(Ssetminus {0})cup {0}in sigma(F).$
answered Nov 25 at 21:05
DanielWainfleet
33.7k31647
Every $Ssubset Bbb R$ belongs to $sigma(F).$
(I)... $Ssetminus (-1/n,1/n)in F$ for each $nin Bbb N$ so $Ssetminus {0}=cup_{nin Bbb N},(,Ssetminus (-1/n,1/n),)in sigma (F).$
(II)... $(-1/n,1/n)in F$ for each $nin Bbb N$ so ${0}=cap_{nin Bbb N}(-1/n,1/n)in sigma (F).$
(III)... So either $S=Ssetminus {0}in sigma(F)$ or $S=(Ssetminus {0})cup {0}in sigma(F).$
answered Nov 25 at 21:05
DanielWainfleet
33.7k31647
answered Nov 25 at 21:05
DanielWainfleet
33.7k31647
answered Nov 25 at 21:05
DanielWainfleet
33.7k31647
answered Nov 25 at 21:05
DanielWainfleet
33.7k31647
33.7k31647
add a comment |
add a comment |
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$F$ is not a $sigma$-algebra because $A=(-1,1)in F$ and $B=(1,2)in F$ but $Acup Bnot in F.$
– DanielWainfleet
Nov 25 at 20:08
@DanielWainfleet yes, I mentioned that $F$ is not a $sigma$-algebra in my post. I am trying to figure out how to describe the $sigma$-algebra generated by $F$ and show that each singleton is contained in that $sigma$-algebra.
– Jane Doe
Nov 25 at 20:11