Does $sigma$ commute with pullback.
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Question. Let $X$ be a set and $T:Xto X$ be a surjective map. Let $mathcal F$ be a collection of subsets of $X$. Is it true that
$$sigma(T^{-1}mathcal F) = T^{-1}(sigma(mathcal F))$$
(Here $sigma(mathcal F)$ denote the $sigma$-algebra generated by $mathcal F$, $T^{-1}(mathcal F)={T^{-1}F:Fin mathcal F}$ etc).
It is clear that the LHS above is contained in the RHS. This is because the RHS is a $sigma$-algebra which contains $T^{-1}mathcal F$.
I am able to prove the reverse containment under the additional assumption that $T$ be bijective.
To do this, let us write $Sigma_{mathcal C}$ to denote the collection of all the $sigma$-algebras on $X$ which contain a given collection of subsets $mathcal C$ of $X$. Then we have a natural map $Sigma_{mathcal C}to Sigma_{T^{-1}mathcal C}$ which takes $mathcal Ain Sigma_{mathcal C}$ to $T^{-1}mathcal A$ and another natural map $Sigma_{T^{-1}mathcal C}to Sigma_{mathcal C}$ which is the inverse of the previous map.
The equivalence follows because $sigma(mathcal F)=bigcap Sigma_{mathcal F}$ and thus $T^{-1}(sigma(mathcal F))= bigcap T^{-1}Sigma_{mathcal F}=bigcapSigma_{T^{-1}mathcal F} = sigma(T^{-1}mathcal F)$.
measure-theory elementary-set-theory
$endgroup$
add a comment |
$begingroup$
Question. Let $X$ be a set and $T:Xto X$ be a surjective map. Let $mathcal F$ be a collection of subsets of $X$. Is it true that
$$sigma(T^{-1}mathcal F) = T^{-1}(sigma(mathcal F))$$
(Here $sigma(mathcal F)$ denote the $sigma$-algebra generated by $mathcal F$, $T^{-1}(mathcal F)={T^{-1}F:Fin mathcal F}$ etc).
It is clear that the LHS above is contained in the RHS. This is because the RHS is a $sigma$-algebra which contains $T^{-1}mathcal F$.
I am able to prove the reverse containment under the additional assumption that $T$ be bijective.
To do this, let us write $Sigma_{mathcal C}$ to denote the collection of all the $sigma$-algebras on $X$ which contain a given collection of subsets $mathcal C$ of $X$. Then we have a natural map $Sigma_{mathcal C}to Sigma_{T^{-1}mathcal C}$ which takes $mathcal Ain Sigma_{mathcal C}$ to $T^{-1}mathcal A$ and another natural map $Sigma_{T^{-1}mathcal C}to Sigma_{mathcal C}$ which is the inverse of the previous map.
The equivalence follows because $sigma(mathcal F)=bigcap Sigma_{mathcal F}$ and thus $T^{-1}(sigma(mathcal F))= bigcap T^{-1}Sigma_{mathcal F}=bigcapSigma_{T^{-1}mathcal F} = sigma(T^{-1}mathcal F)$.
measure-theory elementary-set-theory
$endgroup$
add a comment |
$begingroup$
Question. Let $X$ be a set and $T:Xto X$ be a surjective map. Let $mathcal F$ be a collection of subsets of $X$. Is it true that
$$sigma(T^{-1}mathcal F) = T^{-1}(sigma(mathcal F))$$
(Here $sigma(mathcal F)$ denote the $sigma$-algebra generated by $mathcal F$, $T^{-1}(mathcal F)={T^{-1}F:Fin mathcal F}$ etc).
It is clear that the LHS above is contained in the RHS. This is because the RHS is a $sigma$-algebra which contains $T^{-1}mathcal F$.
I am able to prove the reverse containment under the additional assumption that $T$ be bijective.
To do this, let us write $Sigma_{mathcal C}$ to denote the collection of all the $sigma$-algebras on $X$ which contain a given collection of subsets $mathcal C$ of $X$. Then we have a natural map $Sigma_{mathcal C}to Sigma_{T^{-1}mathcal C}$ which takes $mathcal Ain Sigma_{mathcal C}$ to $T^{-1}mathcal A$ and another natural map $Sigma_{T^{-1}mathcal C}to Sigma_{mathcal C}$ which is the inverse of the previous map.
The equivalence follows because $sigma(mathcal F)=bigcap Sigma_{mathcal F}$ and thus $T^{-1}(sigma(mathcal F))= bigcap T^{-1}Sigma_{mathcal F}=bigcapSigma_{T^{-1}mathcal F} = sigma(T^{-1}mathcal F)$.
measure-theory elementary-set-theory
$endgroup$
Question. Let $X$ be a set and $T:Xto X$ be a surjective map. Let $mathcal F$ be a collection of subsets of $X$. Is it true that
$$sigma(T^{-1}mathcal F) = T^{-1}(sigma(mathcal F))$$
(Here $sigma(mathcal F)$ denote the $sigma$-algebra generated by $mathcal F$, $T^{-1}(mathcal F)={T^{-1}F:Fin mathcal F}$ etc).
It is clear that the LHS above is contained in the RHS. This is because the RHS is a $sigma$-algebra which contains $T^{-1}mathcal F$.
I am able to prove the reverse containment under the additional assumption that $T$ be bijective.
To do this, let us write $Sigma_{mathcal C}$ to denote the collection of all the $sigma$-algebras on $X$ which contain a given collection of subsets $mathcal C$ of $X$. Then we have a natural map $Sigma_{mathcal C}to Sigma_{T^{-1}mathcal C}$ which takes $mathcal Ain Sigma_{mathcal C}$ to $T^{-1}mathcal A$ and another natural map $Sigma_{T^{-1}mathcal C}to Sigma_{mathcal C}$ which is the inverse of the previous map.
The equivalence follows because $sigma(mathcal F)=bigcap Sigma_{mathcal F}$ and thus $T^{-1}(sigma(mathcal F))= bigcap T^{-1}Sigma_{mathcal F}=bigcapSigma_{T^{-1}mathcal F} = sigma(T^{-1}mathcal F)$.
measure-theory elementary-set-theory
measure-theory elementary-set-theory
asked Dec 25 '18 at 7:48
caffeinemachinecaffeinemachine
6,63221355
6,63221355
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1 Answer
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$begingroup$
Consider ${Ain sigma (mathcal F): T^{-1}(A) in sigma (T^{-1} (mathcal F))}$. Verify that this is a sigma algebra which contains $mathcal F$. Hence it contains $sigma (mathcal F)$ which proves the equality without any extra assumptions on $T$.
$endgroup$
$begingroup$
+1 very nice...
$endgroup$
– mathworker21
Dec 25 '18 at 8:06
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So we don't even need $T$ to be surjective? That is very nice!
$endgroup$
– caffeinemachine
Dec 25 '18 at 8:13
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@caffeinemachine The moral here is: inverse images behave very well w.r.t. set theoretic operations, hence w.r.t. sigma algebras.
$endgroup$
– Kavi Rama Murthy
Dec 25 '18 at 8:18
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@KaviRamaMurthy Yes Prof. Murthy. In fact this is what prompted me to conjecture this in the first place. Somehow I missed considering the set you considered. In hindsight it is natural. I learnt an important lesson from your solution(s). Thank you.
$endgroup$
– caffeinemachine
Dec 25 '18 at 8:29
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Consider ${Ain sigma (mathcal F): T^{-1}(A) in sigma (T^{-1} (mathcal F))}$. Verify that this is a sigma algebra which contains $mathcal F$. Hence it contains $sigma (mathcal F)$ which proves the equality without any extra assumptions on $T$.
$endgroup$
$begingroup$
+1 very nice...
$endgroup$
– mathworker21
Dec 25 '18 at 8:06
$begingroup$
So we don't even need $T$ to be surjective? That is very nice!
$endgroup$
– caffeinemachine
Dec 25 '18 at 8:13
$begingroup$
@caffeinemachine The moral here is: inverse images behave very well w.r.t. set theoretic operations, hence w.r.t. sigma algebras.
$endgroup$
– Kavi Rama Murthy
Dec 25 '18 at 8:18
$begingroup$
@KaviRamaMurthy Yes Prof. Murthy. In fact this is what prompted me to conjecture this in the first place. Somehow I missed considering the set you considered. In hindsight it is natural. I learnt an important lesson from your solution(s). Thank you.
$endgroup$
– caffeinemachine
Dec 25 '18 at 8:29
add a comment |
$begingroup$
Consider ${Ain sigma (mathcal F): T^{-1}(A) in sigma (T^{-1} (mathcal F))}$. Verify that this is a sigma algebra which contains $mathcal F$. Hence it contains $sigma (mathcal F)$ which proves the equality without any extra assumptions on $T$.
$endgroup$
$begingroup$
+1 very nice...
$endgroup$
– mathworker21
Dec 25 '18 at 8:06
$begingroup$
So we don't even need $T$ to be surjective? That is very nice!
$endgroup$
– caffeinemachine
Dec 25 '18 at 8:13
$begingroup$
@caffeinemachine The moral here is: inverse images behave very well w.r.t. set theoretic operations, hence w.r.t. sigma algebras.
$endgroup$
– Kavi Rama Murthy
Dec 25 '18 at 8:18
$begingroup$
@KaviRamaMurthy Yes Prof. Murthy. In fact this is what prompted me to conjecture this in the first place. Somehow I missed considering the set you considered. In hindsight it is natural. I learnt an important lesson from your solution(s). Thank you.
$endgroup$
– caffeinemachine
Dec 25 '18 at 8:29
add a comment |
$begingroup$
Consider ${Ain sigma (mathcal F): T^{-1}(A) in sigma (T^{-1} (mathcal F))}$. Verify that this is a sigma algebra which contains $mathcal F$. Hence it contains $sigma (mathcal F)$ which proves the equality without any extra assumptions on $T$.
$endgroup$
Consider ${Ain sigma (mathcal F): T^{-1}(A) in sigma (T^{-1} (mathcal F))}$. Verify that this is a sigma algebra which contains $mathcal F$. Hence it contains $sigma (mathcal F)$ which proves the equality without any extra assumptions on $T$.
edited Dec 25 '18 at 8:07
answered Dec 25 '18 at 7:55
Kavi Rama MurthyKavi Rama Murthy
66.9k53067
66.9k53067
$begingroup$
+1 very nice...
$endgroup$
– mathworker21
Dec 25 '18 at 8:06
$begingroup$
So we don't even need $T$ to be surjective? That is very nice!
$endgroup$
– caffeinemachine
Dec 25 '18 at 8:13
$begingroup$
@caffeinemachine The moral here is: inverse images behave very well w.r.t. set theoretic operations, hence w.r.t. sigma algebras.
$endgroup$
– Kavi Rama Murthy
Dec 25 '18 at 8:18
$begingroup$
@KaviRamaMurthy Yes Prof. Murthy. In fact this is what prompted me to conjecture this in the first place. Somehow I missed considering the set you considered. In hindsight it is natural. I learnt an important lesson from your solution(s). Thank you.
$endgroup$
– caffeinemachine
Dec 25 '18 at 8:29
add a comment |
$begingroup$
+1 very nice...
$endgroup$
– mathworker21
Dec 25 '18 at 8:06
$begingroup$
So we don't even need $T$ to be surjective? That is very nice!
$endgroup$
– caffeinemachine
Dec 25 '18 at 8:13
$begingroup$
@caffeinemachine The moral here is: inverse images behave very well w.r.t. set theoretic operations, hence w.r.t. sigma algebras.
$endgroup$
– Kavi Rama Murthy
Dec 25 '18 at 8:18
$begingroup$
@KaviRamaMurthy Yes Prof. Murthy. In fact this is what prompted me to conjecture this in the first place. Somehow I missed considering the set you considered. In hindsight it is natural. I learnt an important lesson from your solution(s). Thank you.
$endgroup$
– caffeinemachine
Dec 25 '18 at 8:29
$begingroup$
+1 very nice...
$endgroup$
– mathworker21
Dec 25 '18 at 8:06
$begingroup$
+1 very nice...
$endgroup$
– mathworker21
Dec 25 '18 at 8:06
$begingroup$
So we don't even need $T$ to be surjective? That is very nice!
$endgroup$
– caffeinemachine
Dec 25 '18 at 8:13
$begingroup$
So we don't even need $T$ to be surjective? That is very nice!
$endgroup$
– caffeinemachine
Dec 25 '18 at 8:13
$begingroup$
@caffeinemachine The moral here is: inverse images behave very well w.r.t. set theoretic operations, hence w.r.t. sigma algebras.
$endgroup$
– Kavi Rama Murthy
Dec 25 '18 at 8:18
$begingroup$
@caffeinemachine The moral here is: inverse images behave very well w.r.t. set theoretic operations, hence w.r.t. sigma algebras.
$endgroup$
– Kavi Rama Murthy
Dec 25 '18 at 8:18
$begingroup$
@KaviRamaMurthy Yes Prof. Murthy. In fact this is what prompted me to conjecture this in the first place. Somehow I missed considering the set you considered. In hindsight it is natural. I learnt an important lesson from your solution(s). Thank you.
$endgroup$
– caffeinemachine
Dec 25 '18 at 8:29
$begingroup$
@KaviRamaMurthy Yes Prof. Murthy. In fact this is what prompted me to conjecture this in the first place. Somehow I missed considering the set you considered. In hindsight it is natural. I learnt an important lesson from your solution(s). Thank you.
$endgroup$
– caffeinemachine
Dec 25 '18 at 8:29
add a comment |
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