A sufficient (?) condition under which a linear order is order-isomorphic to a subset of the real line
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Let $(mathbb{X},<)$ be a linear order. Assume that there exists $g:mathbb{X} times mathbb{R} rightarrow mathbb{R}$ such that for each fixed $x<x'$ the function $y mapsto g(x,y)-g(x',y)$ is a strictly increasing bijection of $mathbb{R}$ onto itself.
I am trying to prove (or disprove) that then $(mathbb{X},<)$ is necessarily order-isomorphic to a subset of $mathbb{R}$ with the usual order. That is, there exists a strictly increasing function $f:mathbb{X} rightarrow mathbb{R}$.
Thank you.
order-theory
asked Nov 16 at 17:31
Mikhail
605
add a comment |
up vote
0
down vote
favorite
Let $(mathbb{X},<)$ be a linear order. Assume that there exists $g:mathbb{X} times mathbb{R} rightarrow mathbb{R}$ such that for each fixed $x<x'$ the function $y mapsto g(x,y)-g(x',y)$ is a strictly increasing bijection of $mathbb{R}$ onto itself.
I am trying to prove (or disprove) that then $(mathbb{X},<)$ is necessarily order-isomorphic to a subset of $mathbb{R}$ with the usual order. That is, there exists a strictly increasing function $f:mathbb{X} rightarrow mathbb{R}$.
Thank you.
order-theory
asked Nov 16 at 17:31
Mikhail
605
Do you know that every finite or countably infinite linear ordering is order isomorphic to a subset of the rationals? I haven't given much thought to your actual question, so if this result is well known to you and doesn't seem relevant, then ignore my comment.
– Dave L. Renfro
Nov 16 at 18:15
Is it even straightforward that $|X|le|Bbb R|$?
– Hagen von Eitzen
Nov 16 at 18:17
@DaveL.Renfro, thank you. Yes, I know this fact, but I don't know how to use it to prove (or disprove) the statement of interest.
– Mikhail
Nov 16 at 18:54
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $(mathbb{X},<)$ be a linear order. Assume that there exists $g:mathbb{X} times mathbb{R} rightarrow mathbb{R}$ such that for each fixed $x<x'$ the function $y mapsto g(x,y)-g(x',y)$ is a strictly increasing bijection of $mathbb{R}$ onto itself.
I am trying to prove (or disprove) that then $(mathbb{X},<)$ is necessarily order-isomorphic to a subset of $mathbb{R}$ with the usual order. That is, there exists a strictly increasing function $f:mathbb{X} rightarrow mathbb{R}$.
Thank you.
order-theory
asked Nov 16 at 17:31
Mikhail
605
Let $(mathbb{X},<)$ be a linear order. Assume that there exists $g:mathbb{X} times mathbb{R} rightarrow mathbb{R}$ such that for each fixed $x<x'$ the function $y mapsto g(x,y)-g(x',y)$ is a strictly increasing bijection of $mathbb{R}$ onto itself.
I am trying to prove (or disprove) that then $(mathbb{X},<)$ is necessarily order-isomorphic to a subset of $mathbb{R}$ with the usual order. That is, there exists a strictly increasing function $f:mathbb{X} rightarrow mathbb{R}$.
Thank you.
order-theory
order-theory
asked Nov 16 at 17:31
Mikhail
605
asked Nov 16 at 17:31
Mikhail
605
edited Nov 16 at 18:07
asked Nov 16 at 17:31
Mikhail
605
asked Nov 16 at 17:31
Mikhail
605
asked Nov 16 at 17:31
Mikhail
605
605
Do you know that every finite or countably infinite linear ordering is order isomorphic to a subset of the rationals? I haven't given much thought to your actual question, so if this result is well known to you and doesn't seem relevant, then ignore my comment.
– Dave L. Renfro
Nov 16 at 18:15
Is it even straightforward that $|X|le|Bbb R|$?
– Hagen von Eitzen
Nov 16 at 18:17
@DaveL.Renfro, thank you. Yes, I know this fact, but I don't know how to use it to prove (or disprove) the statement of interest.
– Mikhail
Nov 16 at 18:54
add a comment |
Do you know that every finite or countably infinite linear ordering is order isomorphic to a subset of the rationals? I haven't given much thought to your actual question, so if this result is well known to you and doesn't seem relevant, then ignore my comment.
– Dave L. Renfro
Nov 16 at 18:15
Is it even straightforward that $|X|le|Bbb R|$?
– Hagen von Eitzen
Nov 16 at 18:17
@DaveL.Renfro, thank you. Yes, I know this fact, but I don't know how to use it to prove (or disprove) the statement of interest.
– Mikhail
Nov 16 at 18:54
Do you know that every finite or countably infinite linear ordering is order isomorphic to a subset of the rationals? I haven't given much thought to your actual question, so if this result is well known to you and doesn't seem relevant, then ignore my comment.
– Dave L. Renfro
Nov 16 at 18:15
Do you know that every finite or countably infinite linear ordering is order isomorphic to a subset of the rationals? I haven't given much thought to your actual question, so if this result is well known to you and doesn't seem relevant, then ignore my comment.
– Dave L. Renfro
Nov 16 at 18:15
Is it even straightforward that $|X|le|Bbb R|$?
– Hagen von Eitzen
Nov 16 at 18:17
Is it even straightforward that $|X|le|Bbb R|$?
– Hagen von Eitzen
Nov 16 at 18:17
@DaveL.Renfro, thank you. Yes, I know this fact, but I don't know how to use it to prove (or disprove) the statement of interest.
– Mikhail
Nov 16 at 18:54
@DaveL.Renfro, thank you. Yes, I know this fact, but I don't know how to use it to prove (or disprove) the statement of interest.
– Mikhail
Nov 16 at 18:54
add a comment |
1 Answer
1
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oldest
votes
up vote
2
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accepted
Define $FcolonBbb Xto Bbb R$ as
$$F(x)=g(x,0)-g(x,1).$$
Claim. Then $x<x'$ implies $F(x)<F(x')$.
Proof.
Let $h(y)=g(x,y)-g(x',y)$. We are given that $hcolonBbb RtoBbb R$ is strictly increasing.
Now we have $$F(x')-F(x)=g(x',0)-g(x',1)-g(x,0)+g(x,1)=h(1)-h(0)>0$$
as desired. $square$
answered Nov 16 at 20:50
Hagen von Eitzen
275k21266494
Great! Thank you very much!
– Mikhail
Nov 16 at 21:00
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Define $FcolonBbb Xto Bbb R$ as
$$F(x)=g(x,0)-g(x,1).$$
Claim. Then $x<x'$ implies $F(x)<F(x')$.
Proof.
Let $h(y)=g(x,y)-g(x',y)$. We are given that $hcolonBbb RtoBbb R$ is strictly increasing.
Now we have $$F(x')-F(x)=g(x',0)-g(x',1)-g(x,0)+g(x,1)=h(1)-h(0)>0$$
as desired. $square$
answered Nov 16 at 20:50
Hagen von Eitzen
275k21266494
Great! Thank you very much!
– Mikhail
Nov 16 at 21:00
add a comment |
up vote
2
down vote
accepted
Define $FcolonBbb Xto Bbb R$ as
$$F(x)=g(x,0)-g(x,1).$$
Claim. Then $x<x'$ implies $F(x)<F(x')$.
Proof.
Let $h(y)=g(x,y)-g(x',y)$. We are given that $hcolonBbb RtoBbb R$ is strictly increasing.
Now we have $$F(x')-F(x)=g(x',0)-g(x',1)-g(x,0)+g(x,1)=h(1)-h(0)>0$$
as desired. $square$
answered Nov 16 at 20:50
Hagen von Eitzen
275k21266494
Great! Thank you very much!
– Mikhail
Nov 16 at 21:00
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Define $FcolonBbb Xto Bbb R$ as
$$F(x)=g(x,0)-g(x,1).$$
Claim. Then $x<x'$ implies $F(x)<F(x')$.
Proof.
Let $h(y)=g(x,y)-g(x',y)$. We are given that $hcolonBbb RtoBbb R$ is strictly increasing.
Now we have $$F(x')-F(x)=g(x',0)-g(x',1)-g(x,0)+g(x,1)=h(1)-h(0)>0$$
as desired. $square$
answered Nov 16 at 20:50
Hagen von Eitzen
275k21266494
Define $FcolonBbb Xto Bbb R$ as
$$F(x)=g(x,0)-g(x,1).$$
Claim. Then $x<x'$ implies $F(x)<F(x')$.
Proof.
Let $h(y)=g(x,y)-g(x',y)$. We are given that $hcolonBbb RtoBbb R$ is strictly increasing.
Now we have $$F(x')-F(x)=g(x',0)-g(x',1)-g(x,0)+g(x,1)=h(1)-h(0)>0$$
as desired. $square$
answered Nov 16 at 20:50
Hagen von Eitzen
275k21266494
answered Nov 16 at 20:50
Hagen von Eitzen
275k21266494
answered Nov 16 at 20:50
Hagen von Eitzen
275k21266494
answered Nov 16 at 20:50
Hagen von Eitzen
275k21266494
275k21266494
Great! Thank you very much!
– Mikhail
Nov 16 at 21:00
add a comment |
Great! Thank you very much!
– Mikhail
Nov 16 at 21:00
Great! Thank you very much!
– Mikhail
Nov 16 at 21:00
Great! Thank you very much!
– Mikhail
Nov 16 at 21:00
add a comment |
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Do you know that every finite or countably infinite linear ordering is order isomorphic to a subset of the rationals? I haven't given much thought to your actual question, so if this result is well known to you and doesn't seem relevant, then ignore my comment.
– Dave L. Renfro
Nov 16 at 18:15
Is it even straightforward that $|X|le|Bbb R|$?
– Hagen von Eitzen
Nov 16 at 18:17
@DaveL.Renfro, thank you. Yes, I know this fact, but I don't know how to use it to prove (or disprove) the statement of interest.
– Mikhail
Nov 16 at 18:54