Mixing
Product Order of Total Orders
2
$begingroup$
Given a product order $langle X_1 times X_2 times ... times X_p, preceqrangle$ of ordered sets $langle X_1, preceq^ast rangle$, $langle X_2, preceq^" rangle$, ...
Is there a possibility to deduce whether these ordered sets ($langle X_1, preceq^ast rangle$, $langle X_2, preceq^" rangle$, ...) are total orders from the properties of the product order?
order-theory
asked Jan 25 at 18:06
andrestlessandrestless
134
$endgroup$
add a comment |
2
$begingroup$
Given a product order $langle X_1 times X_2 times ... times X_p, preceqrangle$ of ordered sets $langle X_1, preceq^ast rangle$, $langle X_2, preceq^" rangle$, ...
Is there a possibility to deduce whether these ordered sets ($langle X_1, preceq^ast rangle$, $langle X_2, preceq^" rangle$, ...) are total orders from the properties of the product order?
order-theory
asked Jan 25 at 18:06
andrestlessandrestless
134
$endgroup$
add a comment |
2
2
2
$begingroup$
Given a product order $langle X_1 times X_2 times ... times X_p, preceqrangle$ of ordered sets $langle X_1, preceq^ast rangle$, $langle X_2, preceq^" rangle$, ...
Is there a possibility to deduce whether these ordered sets ($langle X_1, preceq^ast rangle$, $langle X_2, preceq^" rangle$, ...) are total orders from the properties of the product order?
order-theory
asked Jan 25 at 18:06
andrestlessandrestless
134
$endgroup$
Given a product order $langle X_1 times X_2 times ... times X_p, preceqrangle$ of ordered sets $langle X_1, preceq^ast rangle$, $langle X_2, preceq^" rangle$, ...
Is there a possibility to deduce whether these ordered sets ($langle X_1, preceq^ast rangle$, $langle X_2, preceq^" rangle$, ...) are total orders from the properties of the product order?
order-theory
order-theory
asked Jan 25 at 18:06
andrestlessandrestless
134
asked Jan 25 at 18:06
andrestlessandrestless
134
asked Jan 25 at 18:06
andrestlessandrestless
134
asked Jan 25 at 18:06
andrestlessandrestless
134
asked Jan 25 at 18:06
andrestlessandrestless
134
134
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
1
$begingroup$
Let $A$ and $B$ be ordered sets with $b$ in $B$, give $Atimes B$ the product order and let $p:Atimes B to A$, $(x,y) mapsto x$ be the first projection.
Show $p(Atimes{b})$ is order isomorphic to $A$.
Thus knowing the product order, one knows the order of the components.
edited Jan 26 at 10:49
amrsa
3,7852618
answered Jan 26 at 4:38
William ElliotWilliam Elliot
8,7572820
$endgroup$
$begingroup$
Thanks for the answer. But I'm looking for some sort of property which tells me right away if the ordered sets are totally ordered sets. If I have the totally ordered sets $X_1, X_2, ..., X_p$ and give the product order $X_1 times X_2 times ... times X_p$. Is it fair to say if the width of the product order is $p$ then the ordered sets $X_1, X_2, ..., X_p$ are all totally ordered?
$endgroup$
– andrestless
Jan 26 at 16:55
$begingroup$
@andrestless. Does the sum of the widths equal the width of a not empty product?
$endgroup$
– William Elliot
Jan 26 at 23:20
$begingroup$
@andrestless Suppose $X$ is the product of two two-element chains, and $Y$ is a one-element chain. In this situation, $X times Y$ has width $2$, and you would conclude $X$ and $Y$ were chains, which is not the case. On the other hand, if $X$ and $Y$ are three-element chains, then $Xtimes Y$ has width $3$, and you would conclude that at least one of $X$ and $Y$ were not a chain, which they were. So that condition of yours is neither necessary nor sufficient.
$endgroup$
– amrsa
Jan 27 at 11:37
$begingroup$
@amrsa Why is the width of the first X one?
$endgroup$
– William Elliot
Jan 28 at 0:31
$begingroup$
@amrsa. Prove the width of {0,1,2}×{0,1,2} is three.
$endgroup$
– William Elliot
Jan 28 at 0:34
|
show 4 more comments
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
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1
$begingroup$
Let $A$ and $B$ be ordered sets with $b$ in $B$, give $Atimes B$ the product order and let $p:Atimes B to A$, $(x,y) mapsto x$ be the first projection.
Show $p(Atimes{b})$ is order isomorphic to $A$.
Thus knowing the product order, one knows the order of the components.
edited Jan 26 at 10:49
amrsa
3,7852618
answered Jan 26 at 4:38
William ElliotWilliam Elliot
8,7572820
$endgroup$
$begingroup$
Thanks for the answer. But I'm looking for some sort of property which tells me right away if the ordered sets are totally ordered sets. If I have the totally ordered sets $X_1, X_2, ..., X_p$ and give the product order $X_1 times X_2 times ... times X_p$. Is it fair to say if the width of the product order is $p$ then the ordered sets $X_1, X_2, ..., X_p$ are all totally ordered?
$endgroup$
– andrestless
Jan 26 at 16:55
$begingroup$
@andrestless. Does the sum of the widths equal the width of a not empty product?
$endgroup$
– William Elliot
Jan 26 at 23:20
$begingroup$
@andrestless Suppose $X$ is the product of two two-element chains, and $Y$ is a one-element chain. In this situation, $X times Y$ has width $2$, and you would conclude $X$ and $Y$ were chains, which is not the case. On the other hand, if $X$ and $Y$ are three-element chains, then $Xtimes Y$ has width $3$, and you would conclude that at least one of $X$ and $Y$ were not a chain, which they were. So that condition of yours is neither necessary nor sufficient.
$endgroup$
– amrsa
Jan 27 at 11:37
$begingroup$
@amrsa Why is the width of the first X one?
$endgroup$
– William Elliot
Jan 28 at 0:31
$begingroup$
@amrsa. Prove the width of {0,1,2}×{0,1,2} is three.
$endgroup$
– William Elliot
Jan 28 at 0:34
|
show 4 more comments
1
$begingroup$
Let $A$ and $B$ be ordered sets with $b$ in $B$, give $Atimes B$ the product order and let $p:Atimes B to A$, $(x,y) mapsto x$ be the first projection.
Show $p(Atimes{b})$ is order isomorphic to $A$.
Thus knowing the product order, one knows the order of the components.
edited Jan 26 at 10:49
amrsa
3,7852618
answered Jan 26 at 4:38
William ElliotWilliam Elliot
8,7572820
$endgroup$
$begingroup$
Thanks for the answer. But I'm looking for some sort of property which tells me right away if the ordered sets are totally ordered sets. If I have the totally ordered sets $X_1, X_2, ..., X_p$ and give the product order $X_1 times X_2 times ... times X_p$. Is it fair to say if the width of the product order is $p$ then the ordered sets $X_1, X_2, ..., X_p$ are all totally ordered?
$endgroup$
– andrestless
Jan 26 at 16:55
$begingroup$
@andrestless. Does the sum of the widths equal the width of a not empty product?
$endgroup$
– William Elliot
Jan 26 at 23:20
$begingroup$
@andrestless Suppose $X$ is the product of two two-element chains, and $Y$ is a one-element chain. In this situation, $X times Y$ has width $2$, and you would conclude $X$ and $Y$ were chains, which is not the case. On the other hand, if $X$ and $Y$ are three-element chains, then $Xtimes Y$ has width $3$, and you would conclude that at least one of $X$ and $Y$ were not a chain, which they were. So that condition of yours is neither necessary nor sufficient.
$endgroup$
– amrsa
Jan 27 at 11:37
$begingroup$
@amrsa Why is the width of the first X one?
$endgroup$
– William Elliot
Jan 28 at 0:31
$begingroup$
@amrsa. Prove the width of {0,1,2}×{0,1,2} is three.
$endgroup$
– William Elliot
Jan 28 at 0:34
|
show 4 more comments
1
1
1
$begingroup$
Let $A$ and $B$ be ordered sets with $b$ in $B$, give $Atimes B$ the product order and let $p:Atimes B to A$, $(x,y) mapsto x$ be the first projection.
Show $p(Atimes{b})$ is order isomorphic to $A$.
Thus knowing the product order, one knows the order of the components.
edited Jan 26 at 10:49
amrsa
3,7852618
answered Jan 26 at 4:38
William ElliotWilliam Elliot
8,7572820
$endgroup$
Let $A$ and $B$ be ordered sets with $b$ in $B$, give $Atimes B$ the product order and let $p:Atimes B to A$, $(x,y) mapsto x$ be the first projection.
Show $p(Atimes{b})$ is order isomorphic to $A$.
Thus knowing the product order, one knows the order of the components.
edited Jan 26 at 10:49
amrsa
3,7852618
answered Jan 26 at 4:38
William ElliotWilliam Elliot
8,7572820
edited Jan 26 at 10:49
amrsa
3,7852618
edited Jan 26 at 10:49
amrsa
3,7852618
edited Jan 26 at 10:49
amrsa
3,7852618
3,7852618
answered Jan 26 at 4:38
William ElliotWilliam Elliot
8,7572820
answered Jan 26 at 4:38
William ElliotWilliam Elliot
8,7572820
answered Jan 26 at 4:38
William ElliotWilliam Elliot
8,7572820
8,7572820
$begingroup$
Thanks for the answer. But I'm looking for some sort of property which tells me right away if the ordered sets are totally ordered sets. If I have the totally ordered sets $X_1, X_2, ..., X_p$ and give the product order $X_1 times X_2 times ... times X_p$. Is it fair to say if the width of the product order is $p$ then the ordered sets $X_1, X_2, ..., X_p$ are all totally ordered?
$endgroup$
– andrestless
Jan 26 at 16:55
$begingroup$
@andrestless. Does the sum of the widths equal the width of a not empty product?
$endgroup$
– William Elliot
Jan 26 at 23:20
$begingroup$
@andrestless Suppose $X$ is the product of two two-element chains, and $Y$ is a one-element chain. In this situation, $X times Y$ has width $2$, and you would conclude $X$ and $Y$ were chains, which is not the case. On the other hand, if $X$ and $Y$ are three-element chains, then $Xtimes Y$ has width $3$, and you would conclude that at least one of $X$ and $Y$ were not a chain, which they were. So that condition of yours is neither necessary nor sufficient.
$endgroup$
– amrsa
Jan 27 at 11:37
$begingroup$
@amrsa Why is the width of the first X one?
$endgroup$
– William Elliot
Jan 28 at 0:31
$begingroup$
@amrsa. Prove the width of {0,1,2}×{0,1,2} is three.
$endgroup$
– William Elliot
Jan 28 at 0:34
|
show 4 more comments
$begingroup$
Thanks for the answer. But I'm looking for some sort of property which tells me right away if the ordered sets are totally ordered sets. If I have the totally ordered sets $X_1, X_2, ..., X_p$ and give the product order $X_1 times X_2 times ... times X_p$. Is it fair to say if the width of the product order is $p$ then the ordered sets $X_1, X_2, ..., X_p$ are all totally ordered?
$endgroup$
– andrestless
Jan 26 at 16:55
$begingroup$
@andrestless. Does the sum of the widths equal the width of a not empty product?
$endgroup$
– William Elliot
Jan 26 at 23:20
$begingroup$
@andrestless Suppose $X$ is the product of two two-element chains, and $Y$ is a one-element chain. In this situation, $X times Y$ has width $2$, and you would conclude $X$ and $Y$ were chains, which is not the case. On the other hand, if $X$ and $Y$ are three-element chains, then $Xtimes Y$ has width $3$, and you would conclude that at least one of $X$ and $Y$ were not a chain, which they were. So that condition of yours is neither necessary nor sufficient.
$endgroup$
– amrsa
Jan 27 at 11:37
$begingroup$
@amrsa Why is the width of the first X one?
$endgroup$
– William Elliot
Jan 28 at 0:31
$begingroup$
@amrsa. Prove the width of {0,1,2}×{0,1,2} is three.
$endgroup$
– William Elliot
Jan 28 at 0:34
$begingroup$
Thanks for the answer. But I'm looking for some sort of property which tells me right away if the ordered sets are totally ordered sets. If I have the totally ordered sets $X_1, X_2, ..., X_p$ and give the product order $X_1 times X_2 times ... times X_p$. Is it fair to say if the width of the product order is $p$ then the ordered sets $X_1, X_2, ..., X_p$ are all totally ordered?
$endgroup$
– andrestless
Jan 26 at 16:55
$begingroup$
Thanks for the answer. But I'm looking for some sort of property which tells me right away if the ordered sets are totally ordered sets. If I have the totally ordered sets $X_1, X_2, ..., X_p$ and give the product order $X_1 times X_2 times ... times X_p$. Is it fair to say if the width of the product order is $p$ then the ordered sets $X_1, X_2, ..., X_p$ are all totally ordered?
$endgroup$
– andrestless
Jan 26 at 16:55
$begingroup$
@andrestless. Does the sum of the widths equal the width of a not empty product?
$endgroup$
– William Elliot
Jan 26 at 23:20
$begingroup$
@andrestless. Does the sum of the widths equal the width of a not empty product?
$endgroup$
– William Elliot
Jan 26 at 23:20
$begingroup$
@andrestless Suppose $X$ is the product of two two-element chains, and $Y$ is a one-element chain. In this situation, $X times Y$ has width $2$, and you would conclude $X$ and $Y$ were chains, which is not the case. On the other hand, if $X$ and $Y$ are three-element chains, then $Xtimes Y$ has width $3$, and you would conclude that at least one of $X$ and $Y$ were not a chain, which they were. So that condition of yours is neither necessary nor sufficient.
$endgroup$
– amrsa
Jan 27 at 11:37
$begingroup$
@andrestless Suppose $X$ is the product of two two-element chains, and $Y$ is a one-element chain. In this situation, $X times Y$ has width $2$, and you would conclude $X$ and $Y$ were chains, which is not the case. On the other hand, if $X$ and $Y$ are three-element chains, then $Xtimes Y$ has width $3$, and you would conclude that at least one of $X$ and $Y$ were not a chain, which they were. So that condition of yours is neither necessary nor sufficient.
$endgroup$
– amrsa
Jan 27 at 11:37
$begingroup$
@amrsa Why is the width of the first X one?
$endgroup$
– William Elliot
Jan 28 at 0:31
$begingroup$
@amrsa Why is the width of the first X one?
$endgroup$
– William Elliot
Jan 28 at 0:31
$begingroup$
@amrsa. Prove the width of {0,1,2}×{0,1,2} is three.
$endgroup$
– William Elliot
Jan 28 at 0:34
$begingroup$
@amrsa. Prove the width of {0,1,2}×{0,1,2} is three.
$endgroup$
– William Elliot
Jan 28 at 0:34
|
show 4 more comments
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