Mixing
Order Topology on a Preorder
$begingroup$
While looking at the definition of the order topology defined on a total order (https://en.wikipedia.org/wiki/Order_topology), I realized I needed a generalization to preorders. So ultimately the question: Is there a generalization of the order topology to preorders. If so, what is it? If not, what are some workarounds?
Some motivation:
My idea is to use “dense” preorders to model cause and effect where a<=b iff a causes b. The topology would make it possible to capture the idea that it is posible certian events are “closer” to directly causing another event.
general-topology relations order-theory order-topology
asked Jan 17 at 14:46
user512716user512716
816
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add a comment |
$begingroup$
While looking at the definition of the order topology defined on a total order (https://en.wikipedia.org/wiki/Order_topology), I realized I needed a generalization to preorders. So ultimately the question: Is there a generalization of the order topology to preorders. If so, what is it? If not, what are some workarounds?
Some motivation:
My idea is to use “dense” preorders to model cause and effect where a<=b iff a causes b. The topology would make it possible to capture the idea that it is posible certian events are “closer” to directly causing another event.
general-topology relations order-theory order-topology
asked Jan 17 at 14:46
user512716user512716
816
$endgroup$
add a comment |
$begingroup$
While looking at the definition of the order topology defined on a total order (https://en.wikipedia.org/wiki/Order_topology), I realized I needed a generalization to preorders. So ultimately the question: Is there a generalization of the order topology to preorders. If so, what is it? If not, what are some workarounds?
Some motivation:
My idea is to use “dense” preorders to model cause and effect where a<=b iff a causes b. The topology would make it possible to capture the idea that it is posible certian events are “closer” to directly causing another event.
general-topology relations order-theory order-topology
asked Jan 17 at 14:46
user512716user512716
816
$endgroup$
While looking at the definition of the order topology defined on a total order (https://en.wikipedia.org/wiki/Order_topology), I realized I needed a generalization to preorders. So ultimately the question: Is there a generalization of the order topology to preorders. If so, what is it? If not, what are some workarounds?
Some motivation:
My idea is to use “dense” preorders to model cause and effect where a<=b iff a causes b. The topology would make it possible to capture the idea that it is posible certian events are “closer” to directly causing another event.
general-topology relations order-theory order-topology
general-topology relations order-theory order-topology
asked Jan 17 at 14:46
user512716user512716
816
asked Jan 17 at 14:46
user512716user512716
816
asked Jan 17 at 14:46
user512716user512716
816
asked Jan 17 at 14:46
user512716user512716
816
asked Jan 17 at 14:46
user512716user512716
816
816
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
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Define for each $x in X$, the lower and upper sets $L(x) = {y in X: y < x}$ and $U(x) = {y in X: y > x}$. By definition all such sets together form a subbase for the order topology when $(X,<)$ is a linear order.
So if you want to generalise to a partial order $(X,le)$, just define $x < y$ as $ x le y$ and $x neq y$ and use the same subbase.
answered Jan 17 at 18:52
Henno BrandsmaHenno Brandsma
111k348118
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add a comment |
$begingroup$
There is the Alexandrov topology on a pre-ordered set.
It is not a generalization of the order topology you link to, but it might be what you're looking for
answered Jan 17 at 15:28
amrsaamrsa
3,6702618
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add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
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active
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active
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$begingroup$
Define for each $x in X$, the lower and upper sets $L(x) = {y in X: y < x}$ and $U(x) = {y in X: y > x}$. By definition all such sets together form a subbase for the order topology when $(X,<)$ is a linear order.
So if you want to generalise to a partial order $(X,le)$, just define $x < y$ as $ x le y$ and $x neq y$ and use the same subbase.
answered Jan 17 at 18:52
Henno BrandsmaHenno Brandsma
111k348118
$endgroup$
add a comment |
$begingroup$
Define for each $x in X$, the lower and upper sets $L(x) = {y in X: y < x}$ and $U(x) = {y in X: y > x}$. By definition all such sets together form a subbase for the order topology when $(X,<)$ is a linear order.
So if you want to generalise to a partial order $(X,le)$, just define $x < y$ as $ x le y$ and $x neq y$ and use the same subbase.
answered Jan 17 at 18:52
Henno BrandsmaHenno Brandsma
111k348118
$endgroup$
add a comment |
$begingroup$
Define for each $x in X$, the lower and upper sets $L(x) = {y in X: y < x}$ and $U(x) = {y in X: y > x}$. By definition all such sets together form a subbase for the order topology when $(X,<)$ is a linear order.
So if you want to generalise to a partial order $(X,le)$, just define $x < y$ as $ x le y$ and $x neq y$ and use the same subbase.
answered Jan 17 at 18:52
Henno BrandsmaHenno Brandsma
111k348118
$endgroup$
Define for each $x in X$, the lower and upper sets $L(x) = {y in X: y < x}$ and $U(x) = {y in X: y > x}$. By definition all such sets together form a subbase for the order topology when $(X,<)$ is a linear order.
So if you want to generalise to a partial order $(X,le)$, just define $x < y$ as $ x le y$ and $x neq y$ and use the same subbase.
answered Jan 17 at 18:52
Henno BrandsmaHenno Brandsma
111k348118
answered Jan 17 at 18:52
Henno BrandsmaHenno Brandsma
111k348118
answered Jan 17 at 18:52
Henno BrandsmaHenno Brandsma
111k348118
answered Jan 17 at 18:52
Henno BrandsmaHenno Brandsma
111k348118
111k348118
add a comment |
add a comment |
$begingroup$
There is the Alexandrov topology on a pre-ordered set.
It is not a generalization of the order topology you link to, but it might be what you're looking for
answered Jan 17 at 15:28
amrsaamrsa
3,6702618
$endgroup$
add a comment |
$begingroup$
There is the Alexandrov topology on a pre-ordered set.
It is not a generalization of the order topology you link to, but it might be what you're looking for
answered Jan 17 at 15:28
amrsaamrsa
3,6702618
$endgroup$
add a comment |
$begingroup$
There is the Alexandrov topology on a pre-ordered set.
It is not a generalization of the order topology you link to, but it might be what you're looking for
answered Jan 17 at 15:28
amrsaamrsa
3,6702618
$endgroup$
There is the Alexandrov topology on a pre-ordered set.
It is not a generalization of the order topology you link to, but it might be what you're looking for
answered Jan 17 at 15:28
amrsaamrsa
3,6702618
answered Jan 17 at 15:28
amrsaamrsa
3,6702618
answered Jan 17 at 15:28
amrsaamrsa
3,6702618
answered Jan 17 at 15:28
amrsaamrsa
3,6702618
3,6702618
add a comment |
add a comment |
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