The theory of dense linear orders without end-points is not $2^omega$-categorical
$begingroup$
It seems best to prove this by counter example. Both $mathbb{R}$ and $I := mathbb{R} backslash mathbb{Q}$ under the usual order $<$ are models of the theory of dense linear orders without end-points and I think they are not isomorphic (if I understand the definition correctly, that means there is no order preserving bijection between $mathbb{R}$ and $I$). I couldn't manage to prove this.
My thoughts so far: suppose there exists such an isomorphism $beta: mathbb{R} to I$, then no irrational elements can be mapped to $beta(mathbb{Q})$, which messes up the order maybe because $mathbb{Q}$ is dense in $mathbb{R}$?
order-theory model-theory
asked Dec 30 '18 at 22:54
Pel de PindaPel de Pinda
802213
$endgroup$
add a comment |
$begingroup$
It seems best to prove this by counter example. Both $mathbb{R}$ and $I := mathbb{R} backslash mathbb{Q}$ under the usual order $<$ are models of the theory of dense linear orders without end-points and I think they are not isomorphic (if I understand the definition correctly, that means there is no order preserving bijection between $mathbb{R}$ and $I$). I couldn't manage to prove this.
My thoughts so far: suppose there exists such an isomorphism $beta: mathbb{R} to I$, then no irrational elements can be mapped to $beta(mathbb{Q})$, which messes up the order maybe because $mathbb{Q}$ is dense in $mathbb{R}$?
order-theory model-theory
asked Dec 30 '18 at 22:54
Pel de PindaPel de Pinda
802213
$endgroup$
2
$begingroup$
Hint: Which of those two orders are Dedekind complete?
$endgroup$
– Alessandro Codenotti
Dec 30 '18 at 23:01
add a comment |
$begingroup$
It seems best to prove this by counter example. Both $mathbb{R}$ and $I := mathbb{R} backslash mathbb{Q}$ under the usual order $<$ are models of the theory of dense linear orders without end-points and I think they are not isomorphic (if I understand the definition correctly, that means there is no order preserving bijection between $mathbb{R}$ and $I$). I couldn't manage to prove this.
My thoughts so far: suppose there exists such an isomorphism $beta: mathbb{R} to I$, then no irrational elements can be mapped to $beta(mathbb{Q})$, which messes up the order maybe because $mathbb{Q}$ is dense in $mathbb{R}$?
order-theory model-theory
asked Dec 30 '18 at 22:54
Pel de PindaPel de Pinda
802213
$endgroup$
It seems best to prove this by counter example. Both $mathbb{R}$ and $I := mathbb{R} backslash mathbb{Q}$ under the usual order $<$ are models of the theory of dense linear orders without end-points and I think they are not isomorphic (if I understand the definition correctly, that means there is no order preserving bijection between $mathbb{R}$ and $I$). I couldn't manage to prove this.
My thoughts so far: suppose there exists such an isomorphism $beta: mathbb{R} to I$, then no irrational elements can be mapped to $beta(mathbb{Q})$, which messes up the order maybe because $mathbb{Q}$ is dense in $mathbb{R}$?
order-theory model-theory
order-theory model-theory
asked Dec 30 '18 at 22:54
Pel de PindaPel de Pinda
802213
asked Dec 30 '18 at 22:54
Pel de PindaPel de Pinda
802213
asked Dec 30 '18 at 22:54
Pel de PindaPel de Pinda
802213
asked Dec 30 '18 at 22:54
Pel de PindaPel de Pinda
802213
asked Dec 30 '18 at 22:54
Pel de PindaPel de Pinda
802213
802213
2
$begingroup$
Hint: Which of those two orders are Dedekind complete?
$endgroup$
– Alessandro Codenotti
Dec 30 '18 at 23:01
add a comment |
2
$begingroup$
Hint: Which of those two orders are Dedekind complete?
$endgroup$
– Alessandro Codenotti
Dec 30 '18 at 23:01
2
2
$begingroup$
Hint: Which of those two orders are Dedekind complete?
$endgroup$
– Alessandro Codenotti
Dec 30 '18 at 23:01
$begingroup$
Hint: Which of those two orders are Dedekind complete?
$endgroup$
– Alessandro Codenotti
Dec 30 '18 at 23:01
add a comment |
2 Answers
2
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In the irrationals, take a sequence decreasing to $0$, then look at where those points map in the reals. Those real points will be a bounded decreasing sequence, therefore they will have a real limit $x$. Where does $x$ map in the irrationals ... call it $y$. Then $y<0$, but then there are irrationals between $y$ and $0$ and that should give you a contradiction since $x$ was the limit of the decreasing real sequence.
answered Dec 30 '18 at 23:06
NedNed
1,993910
$endgroup$
add a comment |
$begingroup$
(I decided to put it as an answer in the end)
A different $I$ that I believe is easier to work with is $I=Bbb Q ∪[0,1]$, assume that there is isomorphism $φ:Ito Bbb R$, then $φ(1.5)<φ(2)$, because it is order preserving, but $[1.5,2]$ in $I$ is countable but $[φ(1.5),φ(2)]$ in $Bbb R$ is uncountable, which leads to contradiction.
Also, although it is not $2^omega$-categorical, all models of cardinality $kappa$ that satisfy those properties are elementary equivalent: let $cal M,N$ be 2 models of that theory of size $kappa$, then, by (downward) Löwenheim–Skolem theorem, there exists $cal M',N'$ elementary substructure of $cal M,N$ respectively of size $omega$. Because the theory is $omega$-categorical, $cal M',N'$ are isomorphic, which implies that they are elementary equivalent, so $cal Mequiv M'equiv N'equiv N$
answered Dec 31 '18 at 7:01
HoloHolo
5,60321030
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
In the irrationals, take a sequence decreasing to $0$, then look at where those points map in the reals. Those real points will be a bounded decreasing sequence, therefore they will have a real limit $x$. Where does $x$ map in the irrationals ... call it $y$. Then $y<0$, but then there are irrationals between $y$ and $0$ and that should give you a contradiction since $x$ was the limit of the decreasing real sequence.
answered Dec 30 '18 at 23:06
NedNed
1,993910
$endgroup$
add a comment |
$begingroup$
In the irrationals, take a sequence decreasing to $0$, then look at where those points map in the reals. Those real points will be a bounded decreasing sequence, therefore they will have a real limit $x$. Where does $x$ map in the irrationals ... call it $y$. Then $y<0$, but then there are irrationals between $y$ and $0$ and that should give you a contradiction since $x$ was the limit of the decreasing real sequence.
answered Dec 30 '18 at 23:06
NedNed
1,993910
$endgroup$
add a comment |
$begingroup$
In the irrationals, take a sequence decreasing to $0$, then look at where those points map in the reals. Those real points will be a bounded decreasing sequence, therefore they will have a real limit $x$. Where does $x$ map in the irrationals ... call it $y$. Then $y<0$, but then there are irrationals between $y$ and $0$ and that should give you a contradiction since $x$ was the limit of the decreasing real sequence.
answered Dec 30 '18 at 23:06
NedNed
1,993910
$endgroup$
In the irrationals, take a sequence decreasing to $0$, then look at where those points map in the reals. Those real points will be a bounded decreasing sequence, therefore they will have a real limit $x$. Where does $x$ map in the irrationals ... call it $y$. Then $y<0$, but then there are irrationals between $y$ and $0$ and that should give you a contradiction since $x$ was the limit of the decreasing real sequence.
answered Dec 30 '18 at 23:06
NedNed
1,993910
answered Dec 30 '18 at 23:06
NedNed
1,993910
answered Dec 30 '18 at 23:06
NedNed
1,993910
answered Dec 30 '18 at 23:06
NedNed
1,993910
1,993910
add a comment |
add a comment |
$begingroup$
(I decided to put it as an answer in the end)
A different $I$ that I believe is easier to work with is $I=Bbb Q ∪[0,1]$, assume that there is isomorphism $φ:Ito Bbb R$, then $φ(1.5)<φ(2)$, because it is order preserving, but $[1.5,2]$ in $I$ is countable but $[φ(1.5),φ(2)]$ in $Bbb R$ is uncountable, which leads to contradiction.
Also, although it is not $2^omega$-categorical, all models of cardinality $kappa$ that satisfy those properties are elementary equivalent: let $cal M,N$ be 2 models of that theory of size $kappa$, then, by (downward) Löwenheim–Skolem theorem, there exists $cal M',N'$ elementary substructure of $cal M,N$ respectively of size $omega$. Because the theory is $omega$-categorical, $cal M',N'$ are isomorphic, which implies that they are elementary equivalent, so $cal Mequiv M'equiv N'equiv N$
answered Dec 31 '18 at 7:01
HoloHolo
5,60321030
$endgroup$
add a comment |
$begingroup$
(I decided to put it as an answer in the end)
A different $I$ that I believe is easier to work with is $I=Bbb Q ∪[0,1]$, assume that there is isomorphism $φ:Ito Bbb R$, then $φ(1.5)<φ(2)$, because it is order preserving, but $[1.5,2]$ in $I$ is countable but $[φ(1.5),φ(2)]$ in $Bbb R$ is uncountable, which leads to contradiction.
Also, although it is not $2^omega$-categorical, all models of cardinality $kappa$ that satisfy those properties are elementary equivalent: let $cal M,N$ be 2 models of that theory of size $kappa$, then, by (downward) Löwenheim–Skolem theorem, there exists $cal M',N'$ elementary substructure of $cal M,N$ respectively of size $omega$. Because the theory is $omega$-categorical, $cal M',N'$ are isomorphic, which implies that they are elementary equivalent, so $cal Mequiv M'equiv N'equiv N$
answered Dec 31 '18 at 7:01
HoloHolo
5,60321030
$endgroup$
add a comment |
$begingroup$
(I decided to put it as an answer in the end)
A different $I$ that I believe is easier to work with is $I=Bbb Q ∪[0,1]$, assume that there is isomorphism $φ:Ito Bbb R$, then $φ(1.5)<φ(2)$, because it is order preserving, but $[1.5,2]$ in $I$ is countable but $[φ(1.5),φ(2)]$ in $Bbb R$ is uncountable, which leads to contradiction.
Also, although it is not $2^omega$-categorical, all models of cardinality $kappa$ that satisfy those properties are elementary equivalent: let $cal M,N$ be 2 models of that theory of size $kappa$, then, by (downward) Löwenheim–Skolem theorem, there exists $cal M',N'$ elementary substructure of $cal M,N$ respectively of size $omega$. Because the theory is $omega$-categorical, $cal M',N'$ are isomorphic, which implies that they are elementary equivalent, so $cal Mequiv M'equiv N'equiv N$
answered Dec 31 '18 at 7:01
HoloHolo
5,60321030
$endgroup$
(I decided to put it as an answer in the end)
A different $I$ that I believe is easier to work with is $I=Bbb Q ∪[0,1]$, assume that there is isomorphism $φ:Ito Bbb R$, then $φ(1.5)<φ(2)$, because it is order preserving, but $[1.5,2]$ in $I$ is countable but $[φ(1.5),φ(2)]$ in $Bbb R$ is uncountable, which leads to contradiction.
Also, although it is not $2^omega$-categorical, all models of cardinality $kappa$ that satisfy those properties are elementary equivalent: let $cal M,N$ be 2 models of that theory of size $kappa$, then, by (downward) Löwenheim–Skolem theorem, there exists $cal M',N'$ elementary substructure of $cal M,N$ respectively of size $omega$. Because the theory is $omega$-categorical, $cal M',N'$ are isomorphic, which implies that they are elementary equivalent, so $cal Mequiv M'equiv N'equiv N$
answered Dec 31 '18 at 7:01
HoloHolo
5,60321030
answered Dec 31 '18 at 7:01
HoloHolo
5,60321030
answered Dec 31 '18 at 7:01
HoloHolo
5,60321030
answered Dec 31 '18 at 7:01
HoloHolo
5,60321030
5,60321030
add a comment |
add a comment |
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2
$begingroup$
Hint: Which of those two orders are Dedekind complete?
$endgroup$
– Alessandro Codenotti
Dec 30 '18 at 23:01