$B in mathcal{B}(M) Leftrightarrow (rB+x) in mathcal{B}(N)$
$begingroup$
Let $M$ be a $d$-dimensional manifold in $mathbb{R}^p$ and let $r>0, x in mathbb{R}^p$. Then $N:=rM+x$ is also a $d$-dimensional manifold in $mathbb{R^p}$.
I want to show that for $B subseteq M mathcal{}\B in mathcal{B}(M) Leftrightarrow (rB+x) in mathcal{B}(N)$
where $mathcal{B}$ is the Borel $sigma$ algebra.
How can I show this? I thought about using the fast that $T(B)=rB+x$ is a homeomorphism somehow but I don't know if that works
real-analysis analysis measure-theory
asked Dec 2 '18 at 11:32
conradconrad
757
$endgroup$
add a comment |
$begingroup$
Let $M$ be a $d$-dimensional manifold in $mathbb{R}^p$ and let $r>0, x in mathbb{R}^p$. Then $N:=rM+x$ is also a $d$-dimensional manifold in $mathbb{R^p}$.
I want to show that for $B subseteq M mathcal{}\B in mathcal{B}(M) Leftrightarrow (rB+x) in mathcal{B}(N)$
where $mathcal{B}$ is the Borel $sigma$ algebra.
How can I show this? I thought about using the fast that $T(B)=rB+x$ is a homeomorphism somehow but I don't know if that works
real-analysis analysis measure-theory
asked Dec 2 '18 at 11:32
conradconrad
757
$endgroup$
add a comment |
$begingroup$
Let $M$ be a $d$-dimensional manifold in $mathbb{R}^p$ and let $r>0, x in mathbb{R}^p$. Then $N:=rM+x$ is also a $d$-dimensional manifold in $mathbb{R^p}$.
I want to show that for $B subseteq M mathcal{}\B in mathcal{B}(M) Leftrightarrow (rB+x) in mathcal{B}(N)$
where $mathcal{B}$ is the Borel $sigma$ algebra.
How can I show this? I thought about using the fast that $T(B)=rB+x$ is a homeomorphism somehow but I don't know if that works
real-analysis analysis measure-theory
asked Dec 2 '18 at 11:32
conradconrad
757
$endgroup$
Let $M$ be a $d$-dimensional manifold in $mathbb{R}^p$ and let $r>0, x in mathbb{R}^p$. Then $N:=rM+x$ is also a $d$-dimensional manifold in $mathbb{R^p}$.
I want to show that for $B subseteq M mathcal{}\B in mathcal{B}(M) Leftrightarrow (rB+x) in mathcal{B}(N)$
where $mathcal{B}$ is the Borel $sigma$ algebra.
How can I show this? I thought about using the fast that $T(B)=rB+x$ is a homeomorphism somehow but I don't know if that works
real-analysis analysis measure-theory
real-analysis analysis measure-theory
asked Dec 2 '18 at 11:32
conradconrad
757
asked Dec 2 '18 at 11:32
conradconrad
757
asked Dec 2 '18 at 11:32
conradconrad
757
asked Dec 2 '18 at 11:32
conradconrad
757
asked Dec 2 '18 at 11:32
conradconrad
757
757
add a comment |
add a comment |
1 Answer
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Any homeomorphism $T$ between topological spaces $X$ and $Y$ preserves Borel sets in the sense $B$ is Borel in $X$ iff $T(X)$ is Borel in $Y$. This is easy to see from the definition of Borel sigma algebra as the one generated by open sets. In this case $Ty=ry+x$ defines a homeomorphism.
answered Dec 2 '18 at 11:40
Kavi Rama MurthyKavi Rama Murthy
60.6k42161
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add a comment |
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1 Answer
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$begingroup$
Any homeomorphism $T$ between topological spaces $X$ and $Y$ preserves Borel sets in the sense $B$ is Borel in $X$ iff $T(X)$ is Borel in $Y$. This is easy to see from the definition of Borel sigma algebra as the one generated by open sets. In this case $Ty=ry+x$ defines a homeomorphism.
answered Dec 2 '18 at 11:40
Kavi Rama MurthyKavi Rama Murthy
60.6k42161
$endgroup$
add a comment |
$begingroup$
Any homeomorphism $T$ between topological spaces $X$ and $Y$ preserves Borel sets in the sense $B$ is Borel in $X$ iff $T(X)$ is Borel in $Y$. This is easy to see from the definition of Borel sigma algebra as the one generated by open sets. In this case $Ty=ry+x$ defines a homeomorphism.
answered Dec 2 '18 at 11:40
Kavi Rama MurthyKavi Rama Murthy
60.6k42161
$endgroup$
add a comment |
$begingroup$
Any homeomorphism $T$ between topological spaces $X$ and $Y$ preserves Borel sets in the sense $B$ is Borel in $X$ iff $T(X)$ is Borel in $Y$. This is easy to see from the definition of Borel sigma algebra as the one generated by open sets. In this case $Ty=ry+x$ defines a homeomorphism.
answered Dec 2 '18 at 11:40
Kavi Rama MurthyKavi Rama Murthy
60.6k42161
$endgroup$
Any homeomorphism $T$ between topological spaces $X$ and $Y$ preserves Borel sets in the sense $B$ is Borel in $X$ iff $T(X)$ is Borel in $Y$. This is easy to see from the definition of Borel sigma algebra as the one generated by open sets. In this case $Ty=ry+x$ defines a homeomorphism.
answered Dec 2 '18 at 11:40
Kavi Rama MurthyKavi Rama Murthy
60.6k42161
answered Dec 2 '18 at 11:40
Kavi Rama MurthyKavi Rama Murthy
60.6k42161
answered Dec 2 '18 at 11:40
Kavi Rama MurthyKavi Rama Murthy
60.6k42161
answered Dec 2 '18 at 11:40
Kavi Rama MurthyKavi Rama Murthy
60.6k42161
60.6k42161
add a comment |
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