$bigcap_{i}mathrm{co}(A_i)=mathrm{co}(bigcap_{i}A_i)$
$begingroup$
Let $(A_i: i in I)$ be a family of closed sets contained in $[0,1]$ such that for all $i,j in I$ there exists $k in I$ for which $A_i cap A_j=A_k$. Denote by $mathrm{co}(X)$ the convex hull of $X$. Is it true that
$$
bigcap_{i in I}mathrm{co}(A_i)=mathrm{co}left(bigcap_{i in I}A_iright),,,?
$$
Ps. Here a related question. Here, under weaker conditions, it is already shown that the right hand side is always contained in the left hand side. I guess a positive answer also on the converse inclusion.
general-topology compactness convex-hulls
edited Dec 22 '18 at 4:09
Alex Ravsky
42.4k32383
asked Dec 21 '18 at 14:11
Paolo LeonettiPaolo Leonetti
11.5k21550
$endgroup$
add a comment |
$begingroup$
Let $(A_i: i in I)$ be a family of closed sets contained in $[0,1]$ such that for all $i,j in I$ there exists $k in I$ for which $A_i cap A_j=A_k$. Denote by $mathrm{co}(X)$ the convex hull of $X$. Is it true that
$$
bigcap_{i in I}mathrm{co}(A_i)=mathrm{co}left(bigcap_{i in I}A_iright),,,?
$$
Ps. Here a related question. Here, under weaker conditions, it is already shown that the right hand side is always contained in the left hand side. I guess a positive answer also on the converse inclusion.
general-topology compactness convex-hulls
edited Dec 22 '18 at 4:09
Alex Ravsky
42.4k32383
asked Dec 21 '18 at 14:11
Paolo LeonettiPaolo Leonetti
11.5k21550
$endgroup$
$begingroup$
What happens when you try to prove it? Can you show the left is a subset of the right? Can you show the right is a subset of the left? --> In this forum, you should show your work when you pose a question. If you do not, the question may be "put on hold" until you do.
$endgroup$
– GEdgar
Dec 21 '18 at 14:23
1
$begingroup$
I already proved in the related link that RHS is a subset of LHS, under weaker conditions. However, the related question is false. A counterexample will be posted soon; on the other hand, the sets were not bounded. Here, assuming that each $A_i$ is compact you can use, by the Caratheodory theorem, that operators closure and conv commute. Hence the identity simplifies to the above one.
$endgroup$
– Paolo Leonetti
Dec 21 '18 at 14:26
add a comment |
$begingroup$
Let $(A_i: i in I)$ be a family of closed sets contained in $[0,1]$ such that for all $i,j in I$ there exists $k in I$ for which $A_i cap A_j=A_k$. Denote by $mathrm{co}(X)$ the convex hull of $X$. Is it true that
$$
bigcap_{i in I}mathrm{co}(A_i)=mathrm{co}left(bigcap_{i in I}A_iright),,,?
$$
Ps. Here a related question. Here, under weaker conditions, it is already shown that the right hand side is always contained in the left hand side. I guess a positive answer also on the converse inclusion.
general-topology compactness convex-hulls
edited Dec 22 '18 at 4:09
Alex Ravsky
42.4k32383
asked Dec 21 '18 at 14:11
Paolo LeonettiPaolo Leonetti
11.5k21550
$endgroup$
Let $(A_i: i in I)$ be a family of closed sets contained in $[0,1]$ such that for all $i,j in I$ there exists $k in I$ for which $A_i cap A_j=A_k$. Denote by $mathrm{co}(X)$ the convex hull of $X$. Is it true that
$$
bigcap_{i in I}mathrm{co}(A_i)=mathrm{co}left(bigcap_{i in I}A_iright),,,?
$$
Ps. Here a related question. Here, under weaker conditions, it is already shown that the right hand side is always contained in the left hand side. I guess a positive answer also on the converse inclusion.
general-topology compactness convex-hulls
general-topology compactness convex-hulls
edited Dec 22 '18 at 4:09
Alex Ravsky
42.4k32383
asked Dec 21 '18 at 14:11
Paolo LeonettiPaolo Leonetti
11.5k21550
edited Dec 22 '18 at 4:09
Alex Ravsky
42.4k32383
asked Dec 21 '18 at 14:11
Paolo LeonettiPaolo Leonetti
11.5k21550
edited Dec 22 '18 at 4:09
Alex Ravsky
42.4k32383
edited Dec 22 '18 at 4:09
Alex Ravsky
42.4k32383
edited Dec 22 '18 at 4:09
Alex Ravsky
42.4k32383
42.4k32383
asked Dec 21 '18 at 14:11
Paolo LeonettiPaolo Leonetti
11.5k21550
asked Dec 21 '18 at 14:11
Paolo LeonettiPaolo Leonetti
11.5k21550
asked Dec 21 '18 at 14:11
Paolo LeonettiPaolo Leonetti
11.5k21550
11.5k21550
$begingroup$
What happens when you try to prove it? Can you show the left is a subset of the right? Can you show the right is a subset of the left? --> In this forum, you should show your work when you pose a question. If you do not, the question may be "put on hold" until you do.
$endgroup$
– GEdgar
Dec 21 '18 at 14:23
1
$begingroup$
I already proved in the related link that RHS is a subset of LHS, under weaker conditions. However, the related question is false. A counterexample will be posted soon; on the other hand, the sets were not bounded. Here, assuming that each $A_i$ is compact you can use, by the Caratheodory theorem, that operators closure and conv commute. Hence the identity simplifies to the above one.
$endgroup$
– Paolo Leonetti
Dec 21 '18 at 14:26
add a comment |
$begingroup$
What happens when you try to prove it? Can you show the left is a subset of the right? Can you show the right is a subset of the left? --> In this forum, you should show your work when you pose a question. If you do not, the question may be "put on hold" until you do.
$endgroup$
– GEdgar
Dec 21 '18 at 14:23
1
$begingroup$
I already proved in the related link that RHS is a subset of LHS, under weaker conditions. However, the related question is false. A counterexample will be posted soon; on the other hand, the sets were not bounded. Here, assuming that each $A_i$ is compact you can use, by the Caratheodory theorem, that operators closure and conv commute. Hence the identity simplifies to the above one.
$endgroup$
– Paolo Leonetti
Dec 21 '18 at 14:26
$begingroup$
What happens when you try to prove it? Can you show the left is a subset of the right? Can you show the right is a subset of the left? --> In this forum, you should show your work when you pose a question. If you do not, the question may be "put on hold" until you do.
$endgroup$
– GEdgar
Dec 21 '18 at 14:23
$begingroup$
What happens when you try to prove it? Can you show the left is a subset of the right? Can you show the right is a subset of the left? --> In this forum, you should show your work when you pose a question. If you do not, the question may be "put on hold" until you do.
$endgroup$
– GEdgar
Dec 21 '18 at 14:23
1
1
$begingroup$
I already proved in the related link that RHS is a subset of LHS, under weaker conditions. However, the related question is false. A counterexample will be posted soon; on the other hand, the sets were not bounded. Here, assuming that each $A_i$ is compact you can use, by the Caratheodory theorem, that operators closure and conv commute. Hence the identity simplifies to the above one.
$endgroup$
– Paolo Leonetti
Dec 21 '18 at 14:26
$begingroup$
I already proved in the related link that RHS is a subset of LHS, under weaker conditions. However, the related question is false. A counterexample will be posted soon; on the other hand, the sets were not bounded. Here, assuming that each $A_i$ is compact you can use, by the Caratheodory theorem, that operators closure and conv commute. Hence the identity simplifies to the above one.
$endgroup$
– Paolo Leonetti
Dec 21 '18 at 14:26
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The converse inclusion holds for any family $(A_i: i in I)$ of compact subsets of a (Hausdorff) locally convex topological vector space $V$ (over $Bbb R$). Indeed, let $ain Vsetminus mathrm{co}left(bigcap_{i in I}A_iright)$. By Hahn–Banach separation theorem there exists a continuous linear map $lambda : Vto Bbb R$ and $t inBbb R$ such that $lambda(a) < t < lambda (b)$ for all $bin mathrm{co}left(bigcap_{i in I}A_iright)$. Therefore $U=lambda^{-1}(t,infty)$ is a convex open subset of $V$ containing $bigcap_{i in I}A_i$ but not containing $a$. By so-called generalized Shura-Bura’s lemma, there exists a finite subset $J$ of $I$ such that $bigcap_{i in J}A_isubset U$ (indeed, otherwise $(A_isetminus U: i in I)$ is a centered family of compact sets with empty intersection). Since $bigcap_{i in J}A_i=A_j$ for some $jin I$, we have $mathrm{co} A_jsubset Usubset Vsetminus{a}$.
answered Dec 22 '18 at 4:08
Alex RavskyAlex Ravsky
42.4k32383
$endgroup$
add a comment |
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$begingroup$
The converse inclusion holds for any family $(A_i: i in I)$ of compact subsets of a (Hausdorff) locally convex topological vector space $V$ (over $Bbb R$). Indeed, let $ain Vsetminus mathrm{co}left(bigcap_{i in I}A_iright)$. By Hahn–Banach separation theorem there exists a continuous linear map $lambda : Vto Bbb R$ and $t inBbb R$ such that $lambda(a) < t < lambda (b)$ for all $bin mathrm{co}left(bigcap_{i in I}A_iright)$. Therefore $U=lambda^{-1}(t,infty)$ is a convex open subset of $V$ containing $bigcap_{i in I}A_i$ but not containing $a$. By so-called generalized Shura-Bura’s lemma, there exists a finite subset $J$ of $I$ such that $bigcap_{i in J}A_isubset U$ (indeed, otherwise $(A_isetminus U: i in I)$ is a centered family of compact sets with empty intersection). Since $bigcap_{i in J}A_i=A_j$ for some $jin I$, we have $mathrm{co} A_jsubset Usubset Vsetminus{a}$.
answered Dec 22 '18 at 4:08
Alex RavskyAlex Ravsky
42.4k32383
$endgroup$
add a comment |
$begingroup$
The converse inclusion holds for any family $(A_i: i in I)$ of compact subsets of a (Hausdorff) locally convex topological vector space $V$ (over $Bbb R$). Indeed, let $ain Vsetminus mathrm{co}left(bigcap_{i in I}A_iright)$. By Hahn–Banach separation theorem there exists a continuous linear map $lambda : Vto Bbb R$ and $t inBbb R$ such that $lambda(a) < t < lambda (b)$ for all $bin mathrm{co}left(bigcap_{i in I}A_iright)$. Therefore $U=lambda^{-1}(t,infty)$ is a convex open subset of $V$ containing $bigcap_{i in I}A_i$ but not containing $a$. By so-called generalized Shura-Bura’s lemma, there exists a finite subset $J$ of $I$ such that $bigcap_{i in J}A_isubset U$ (indeed, otherwise $(A_isetminus U: i in I)$ is a centered family of compact sets with empty intersection). Since $bigcap_{i in J}A_i=A_j$ for some $jin I$, we have $mathrm{co} A_jsubset Usubset Vsetminus{a}$.
answered Dec 22 '18 at 4:08
Alex RavskyAlex Ravsky
42.4k32383
$endgroup$
add a comment |
$begingroup$
The converse inclusion holds for any family $(A_i: i in I)$ of compact subsets of a (Hausdorff) locally convex topological vector space $V$ (over $Bbb R$). Indeed, let $ain Vsetminus mathrm{co}left(bigcap_{i in I}A_iright)$. By Hahn–Banach separation theorem there exists a continuous linear map $lambda : Vto Bbb R$ and $t inBbb R$ such that $lambda(a) < t < lambda (b)$ for all $bin mathrm{co}left(bigcap_{i in I}A_iright)$. Therefore $U=lambda^{-1}(t,infty)$ is a convex open subset of $V$ containing $bigcap_{i in I}A_i$ but not containing $a$. By so-called generalized Shura-Bura’s lemma, there exists a finite subset $J$ of $I$ such that $bigcap_{i in J}A_isubset U$ (indeed, otherwise $(A_isetminus U: i in I)$ is a centered family of compact sets with empty intersection). Since $bigcap_{i in J}A_i=A_j$ for some $jin I$, we have $mathrm{co} A_jsubset Usubset Vsetminus{a}$.
answered Dec 22 '18 at 4:08
Alex RavskyAlex Ravsky
42.4k32383
$endgroup$
The converse inclusion holds for any family $(A_i: i in I)$ of compact subsets of a (Hausdorff) locally convex topological vector space $V$ (over $Bbb R$). Indeed, let $ain Vsetminus mathrm{co}left(bigcap_{i in I}A_iright)$. By Hahn–Banach separation theorem there exists a continuous linear map $lambda : Vto Bbb R$ and $t inBbb R$ such that $lambda(a) < t < lambda (b)$ for all $bin mathrm{co}left(bigcap_{i in I}A_iright)$. Therefore $U=lambda^{-1}(t,infty)$ is a convex open subset of $V$ containing $bigcap_{i in I}A_i$ but not containing $a$. By so-called generalized Shura-Bura’s lemma, there exists a finite subset $J$ of $I$ such that $bigcap_{i in J}A_isubset U$ (indeed, otherwise $(A_isetminus U: i in I)$ is a centered family of compact sets with empty intersection). Since $bigcap_{i in J}A_i=A_j$ for some $jin I$, we have $mathrm{co} A_jsubset Usubset Vsetminus{a}$.
answered Dec 22 '18 at 4:08
Alex RavskyAlex Ravsky
42.4k32383
answered Dec 22 '18 at 4:08
Alex RavskyAlex Ravsky
42.4k32383
answered Dec 22 '18 at 4:08
Alex RavskyAlex Ravsky
42.4k32383
answered Dec 22 '18 at 4:08
Alex RavskyAlex Ravsky
42.4k32383
42.4k32383
add a comment |
add a comment |
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$begingroup$
What happens when you try to prove it? Can you show the left is a subset of the right? Can you show the right is a subset of the left? --> In this forum, you should show your work when you pose a question. If you do not, the question may be "put on hold" until you do.
$endgroup$
– GEdgar
Dec 21 '18 at 14:23
1
$begingroup$
I already proved in the related link that RHS is a subset of LHS, under weaker conditions. However, the related question is false. A counterexample will be posted soon; on the other hand, the sets were not bounded. Here, assuming that each $A_i$ is compact you can use, by the Caratheodory theorem, that operators closure and conv commute. Hence the identity simplifies to the above one.
$endgroup$
– Paolo Leonetti
Dec 21 '18 at 14:26