A Banach space with a cone's order












0














I'm reading a paper on a Banach space with an order induced by a cone.



Let $(mathbb{X}, || cdot||)$ be a real Banach space.



We define a subset $P$ of $mathbb{X}$ by $P := { x in mathbb{X} : xgeq 0}$.



My question is what does $ xgeq 0$ mean, knowing that $mathbb{X}$ is quite abstract and could be any real Banach space.






real-analysis banach-spaces order-theory







share|cite|improve this question
























  • $ge$ is also an abstract partial ordering on $X$ (satisfying certain basic conditions). The point is that it determines and is determined by its positive cone $P$ defined as above.
    – Berci
    Nov 24 at 15:41










  • @Berci can you explain more
    – Motaka
    Nov 24 at 16:09
















0














I'm reading a paper on a Banach space with an order induced by a cone.



Let $(mathbb{X}, || cdot||)$ be a real Banach space.



We define a subset $P$ of $mathbb{X}$ by $P := { x in mathbb{X} : xgeq 0}$.



My question is what does $ xgeq 0$ mean, knowing that $mathbb{X}$ is quite abstract and could be any real Banach space.






real-analysis banach-spaces order-theory







share|cite|improve this question
























  • $ge$ is also an abstract partial ordering on $X$ (satisfying certain basic conditions). The point is that it determines and is determined by its positive cone $P$ defined as above.
    – Berci
    Nov 24 at 15:41










  • @Berci can you explain more
    – Motaka
    Nov 24 at 16:09














0












0








0







I'm reading a paper on a Banach space with an order induced by a cone.



Let $(mathbb{X}, || cdot||)$ be a real Banach space.



We define a subset $P$ of $mathbb{X}$ by $P := { x in mathbb{X} : xgeq 0}$.



My question is what does $ xgeq 0$ mean, knowing that $mathbb{X}$ is quite abstract and could be any real Banach space.






real-analysis banach-spaces order-theory







share|cite|improve this question















I'm reading a paper on a Banach space with an order induced by a cone.



Let $(mathbb{X}, || cdot||)$ be a real Banach space.



We define a subset $P$ of $mathbb{X}$ by $P := { x in mathbb{X} : xgeq 0}$.



My question is what does $ xgeq 0$ mean, knowing that $mathbb{X}$ is quite abstract and could be any real Banach space.






real-analysis banach-spaces order-theory




real-analysis banach-spaces order-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 24 at 16:07









Dinisaur

1009




1009










asked Nov 24 at 15:37









Motaka

229111




229111












  • $ge$ is also an abstract partial ordering on $X$ (satisfying certain basic conditions). The point is that it determines and is determined by its positive cone $P$ defined as above.
    – Berci
    Nov 24 at 15:41










  • @Berci can you explain more
    – Motaka
    Nov 24 at 16:09


















  • $ge$ is also an abstract partial ordering on $X$ (satisfying certain basic conditions). The point is that it determines and is determined by its positive cone $P$ defined as above.
    – Berci
    Nov 24 at 15:41










  • @Berci can you explain more
    – Motaka
    Nov 24 at 16:09
















$ge$ is also an abstract partial ordering on $X$ (satisfying certain basic conditions). The point is that it determines and is determined by its positive cone $P$ defined as above.
– Berci
Nov 24 at 15:41




$ge$ is also an abstract partial ordering on $X$ (satisfying certain basic conditions). The point is that it determines and is determined by its positive cone $P$ defined as above.
– Berci
Nov 24 at 15:41












@Berci can you explain more
– Motaka
Nov 24 at 16:09




@Berci can you explain more
– Motaka
Nov 24 at 16:09










1 Answer
1






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oldest

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1














Given an open cone $K$ in a Banach space (i.e. a subset closed with respect to summation and multiplication by non negative factors $lambdageq 0$, and with zero intersection with it's opposite $-Kcap K = {0}$) you can define the order relation by $a>b$ iff $a-bin K$.



In an analogous way, given a closed cone $C$ you can define the order relation by $ageq b$ iff $a-bin C$.



So if you assume the cone to be closed then you'll simply have $P=C$, otherwhise $P$ will be the closure of $K$.



To rephrase it to answer your question: $xgeq 0$ means for $x$ to be in the closure of the cone you use to define your partial order on $mathbb{X}$.



Edit:



Note that we'd like to define a continuous partal order (i.e. such that if $ageq b_lambda$ for each element of a net ${b_lambda}_{lambdainLambda}$ converging to $b$, then $ageq b$), and in order to have this property, you need your defining cone to be closed.






share|cite|improve this answer























  • Thank you, but yourthird paragraph is a little fuzzy, since you have assumed at the beginning that $P$ and $C$ are closed
    – Motaka
    Nov 24 at 16:37










  • I didn't assume P to be closed, P is the subset of X as you defined it. That subset turns out to be the closure of the cone you used to define your order relation, which can be either open or close. So if the order defining cone is closed (in which case i called it C) then we'll just have that the subsets are equal, so P=C. If on the other hand the order defining cone is open (in which case I called it K) then the subset P will be the topological closure of K. Is it clearer now?
    – Dinisaur
    Nov 24 at 16:43












  • Yes thank you a lot
    – Motaka
    Nov 24 at 17:30






  • 1




    You're welcome, I hope it answer your question :)
    – Dinisaur
    Nov 24 at 18:31











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1 Answer
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1 Answer
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1














Given an open cone $K$ in a Banach space (i.e. a subset closed with respect to summation and multiplication by non negative factors $lambdageq 0$, and with zero intersection with it's opposite $-Kcap K = {0}$) you can define the order relation by $a>b$ iff $a-bin K$.



In an analogous way, given a closed cone $C$ you can define the order relation by $ageq b$ iff $a-bin C$.



So if you assume the cone to be closed then you'll simply have $P=C$, otherwhise $P$ will be the closure of $K$.



To rephrase it to answer your question: $xgeq 0$ means for $x$ to be in the closure of the cone you use to define your partial order on $mathbb{X}$.



Edit:



Note that we'd like to define a continuous partal order (i.e. such that if $ageq b_lambda$ for each element of a net ${b_lambda}_{lambdainLambda}$ converging to $b$, then $ageq b$), and in order to have this property, you need your defining cone to be closed.






share|cite|improve this answer























  • Thank you, but yourthird paragraph is a little fuzzy, since you have assumed at the beginning that $P$ and $C$ are closed
    – Motaka
    Nov 24 at 16:37










  • I didn't assume P to be closed, P is the subset of X as you defined it. That subset turns out to be the closure of the cone you used to define your order relation, which can be either open or close. So if the order defining cone is closed (in which case i called it C) then we'll just have that the subsets are equal, so P=C. If on the other hand the order defining cone is open (in which case I called it K) then the subset P will be the topological closure of K. Is it clearer now?
    – Dinisaur
    Nov 24 at 16:43












  • Yes thank you a lot
    – Motaka
    Nov 24 at 17:30






  • 1




    You're welcome, I hope it answer your question :)
    – Dinisaur
    Nov 24 at 18:31
















1














Given an open cone $K$ in a Banach space (i.e. a subset closed with respect to summation and multiplication by non negative factors $lambdageq 0$, and with zero intersection with it's opposite $-Kcap K = {0}$) you can define the order relation by $a>b$ iff $a-bin K$.



In an analogous way, given a closed cone $C$ you can define the order relation by $ageq b$ iff $a-bin C$.



So if you assume the cone to be closed then you'll simply have $P=C$, otherwhise $P$ will be the closure of $K$.



To rephrase it to answer your question: $xgeq 0$ means for $x$ to be in the closure of the cone you use to define your partial order on $mathbb{X}$.



Edit:



Note that we'd like to define a continuous partal order (i.e. such that if $ageq b_lambda$ for each element of a net ${b_lambda}_{lambdainLambda}$ converging to $b$, then $ageq b$), and in order to have this property, you need your defining cone to be closed.






share|cite|improve this answer























  • Thank you, but yourthird paragraph is a little fuzzy, since you have assumed at the beginning that $P$ and $C$ are closed
    – Motaka
    Nov 24 at 16:37










  • I didn't assume P to be closed, P is the subset of X as you defined it. That subset turns out to be the closure of the cone you used to define your order relation, which can be either open or close. So if the order defining cone is closed (in which case i called it C) then we'll just have that the subsets are equal, so P=C. If on the other hand the order defining cone is open (in which case I called it K) then the subset P will be the topological closure of K. Is it clearer now?
    – Dinisaur
    Nov 24 at 16:43












  • Yes thank you a lot
    – Motaka
    Nov 24 at 17:30






  • 1




    You're welcome, I hope it answer your question :)
    – Dinisaur
    Nov 24 at 18:31














1












1








1






Given an open cone $K$ in a Banach space (i.e. a subset closed with respect to summation and multiplication by non negative factors $lambdageq 0$, and with zero intersection with it's opposite $-Kcap K = {0}$) you can define the order relation by $a>b$ iff $a-bin K$.



In an analogous way, given a closed cone $C$ you can define the order relation by $ageq b$ iff $a-bin C$.



So if you assume the cone to be closed then you'll simply have $P=C$, otherwhise $P$ will be the closure of $K$.



To rephrase it to answer your question: $xgeq 0$ means for $x$ to be in the closure of the cone you use to define your partial order on $mathbb{X}$.



Edit:



Note that we'd like to define a continuous partal order (i.e. such that if $ageq b_lambda$ for each element of a net ${b_lambda}_{lambdainLambda}$ converging to $b$, then $ageq b$), and in order to have this property, you need your defining cone to be closed.






share|cite|improve this answer














Given an open cone $K$ in a Banach space (i.e. a subset closed with respect to summation and multiplication by non negative factors $lambdageq 0$, and with zero intersection with it's opposite $-Kcap K = {0}$) you can define the order relation by $a>b$ iff $a-bin K$.



In an analogous way, given a closed cone $C$ you can define the order relation by $ageq b$ iff $a-bin C$.



So if you assume the cone to be closed then you'll simply have $P=C$, otherwhise $P$ will be the closure of $K$.



To rephrase it to answer your question: $xgeq 0$ means for $x$ to be in the closure of the cone you use to define your partial order on $mathbb{X}$.



Edit:



Note that we'd like to define a continuous partal order (i.e. such that if $ageq b_lambda$ for each element of a net ${b_lambda}_{lambdainLambda}$ converging to $b$, then $ageq b$), and in order to have this property, you need your defining cone to be closed.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 24 at 18:38

























answered Nov 24 at 16:22









Dinisaur

1009




1009












  • Thank you, but yourthird paragraph is a little fuzzy, since you have assumed at the beginning that $P$ and $C$ are closed
    – Motaka
    Nov 24 at 16:37










  • I didn't assume P to be closed, P is the subset of X as you defined it. That subset turns out to be the closure of the cone you used to define your order relation, which can be either open or close. So if the order defining cone is closed (in which case i called it C) then we'll just have that the subsets are equal, so P=C. If on the other hand the order defining cone is open (in which case I called it K) then the subset P will be the topological closure of K. Is it clearer now?
    – Dinisaur
    Nov 24 at 16:43












  • Yes thank you a lot
    – Motaka
    Nov 24 at 17:30






  • 1




    You're welcome, I hope it answer your question :)
    – Dinisaur
    Nov 24 at 18:31


















  • Thank you, but yourthird paragraph is a little fuzzy, since you have assumed at the beginning that $P$ and $C$ are closed
    – Motaka
    Nov 24 at 16:37










  • I didn't assume P to be closed, P is the subset of X as you defined it. That subset turns out to be the closure of the cone you used to define your order relation, which can be either open or close. So if the order defining cone is closed (in which case i called it C) then we'll just have that the subsets are equal, so P=C. If on the other hand the order defining cone is open (in which case I called it K) then the subset P will be the topological closure of K. Is it clearer now?
    – Dinisaur
    Nov 24 at 16:43












  • Yes thank you a lot
    – Motaka
    Nov 24 at 17:30






  • 1




    You're welcome, I hope it answer your question :)
    – Dinisaur
    Nov 24 at 18:31
















Thank you, but yourthird paragraph is a little fuzzy, since you have assumed at the beginning that $P$ and $C$ are closed
– Motaka
Nov 24 at 16:37




Thank you, but yourthird paragraph is a little fuzzy, since you have assumed at the beginning that $P$ and $C$ are closed
– Motaka
Nov 24 at 16:37












I didn't assume P to be closed, P is the subset of X as you defined it. That subset turns out to be the closure of the cone you used to define your order relation, which can be either open or close. So if the order defining cone is closed (in which case i called it C) then we'll just have that the subsets are equal, so P=C. If on the other hand the order defining cone is open (in which case I called it K) then the subset P will be the topological closure of K. Is it clearer now?
– Dinisaur
Nov 24 at 16:43






I didn't assume P to be closed, P is the subset of X as you defined it. That subset turns out to be the closure of the cone you used to define your order relation, which can be either open or close. So if the order defining cone is closed (in which case i called it C) then we'll just have that the subsets are equal, so P=C. If on the other hand the order defining cone is open (in which case I called it K) then the subset P will be the topological closure of K. Is it clearer now?
– Dinisaur
Nov 24 at 16:43














Yes thank you a lot
– Motaka
Nov 24 at 17:30




Yes thank you a lot
– Motaka
Nov 24 at 17:30




1




1




You're welcome, I hope it answer your question :)
– Dinisaur
Nov 24 at 18:31




You're welcome, I hope it answer your question :)
– Dinisaur
Nov 24 at 18:31


















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