The structure $(mathbb N, <, f)_{fin mathcal F}$ seems to have no elementary extension
$begingroup$
This is exercise 2.3.1 (2) from Tent/Ziegler, A Course in Model Theory.
Let $mathcal F$ be the set of all functions $mathbb N to mathbb N$. Show that $(mathbb N, <, f)_{fin mathcal F}$ has no countable proper elementary extension.
My approach was to assume we have some proper elementary extension $mathfrak A$ with $mathbb N subsetneq A$. Choose $a in A setminus mathbb N$. As we have a first order sentences true in $mathbb N$ stating that there is no maximal element, either $a$ is between two natural numbers $n,m$ with $m = n+1$, or comes before every natural number. In the first case let $overline n, overline m : mathbb N to mathbb N$ the constant function giving $n$ and $m$. Then in $(mathbb N, <, f)_{fin mathcal F}$ the sentence $exists x forall y ( overline n(y) < x < overline m(y) )$ is false, but in $mathfrak A$ it is true. Hence this case could not arise. Similar if $a$ comes before every natural number, using $overline 0 : mathbb N to mathbb N$ I can give a sentence true in this extension, but not true in $mathbb N$.
So, in consequence there would be no elementary extension at all. But this contradicts the Theorem stated on these slides (see slide number 7), that every infinite model has a proper elementary extension.
So what is wrong with my above reasoning? Why this contradiction?
proof-verification logic first-order-logic model-theory
asked Dec 13 '18 at 17:14
StefanHStefanH
8,11152365
$endgroup$
add a comment |
$begingroup$
This is exercise 2.3.1 (2) from Tent/Ziegler, A Course in Model Theory.
Let $mathcal F$ be the set of all functions $mathbb N to mathbb N$. Show that $(mathbb N, <, f)_{fin mathcal F}$ has no countable proper elementary extension.
My approach was to assume we have some proper elementary extension $mathfrak A$ with $mathbb N subsetneq A$. Choose $a in A setminus mathbb N$. As we have a first order sentences true in $mathbb N$ stating that there is no maximal element, either $a$ is between two natural numbers $n,m$ with $m = n+1$, or comes before every natural number. In the first case let $overline n, overline m : mathbb N to mathbb N$ the constant function giving $n$ and $m$. Then in $(mathbb N, <, f)_{fin mathcal F}$ the sentence $exists x forall y ( overline n(y) < x < overline m(y) )$ is false, but in $mathfrak A$ it is true. Hence this case could not arise. Similar if $a$ comes before every natural number, using $overline 0 : mathbb N to mathbb N$ I can give a sentence true in this extension, but not true in $mathbb N$.
So, in consequence there would be no elementary extension at all. But this contradicts the Theorem stated on these slides (see slide number 7), that every infinite model has a proper elementary extension.
So what is wrong with my above reasoning? Why this contradiction?
proof-verification logic first-order-logic model-theory
asked Dec 13 '18 at 17:14
StefanHStefanH
8,11152365
$endgroup$
2
$begingroup$
"Either $a$ is between two natural numbers $n$, $m$ with $m=n+1$, or comes before every natural number". What?? Exactly the opposite is true! It's true that $mathbb{N}models forall x, exists y, (x<y)$ (there is no maximal element), so $mathfrak{A}$ also satisfies this sentence. But this doesn't rule out having an element of $mathfrak{A}$ which is bigger than all elements of $mathbb{N}$ - it's just that there also have to be even bigger elements of $mathfrak{A}$!
$endgroup$
– Alex Kruckman
Dec 13 '18 at 21:06
add a comment |
$begingroup$
This is exercise 2.3.1 (2) from Tent/Ziegler, A Course in Model Theory.
Let $mathcal F$ be the set of all functions $mathbb N to mathbb N$. Show that $(mathbb N, <, f)_{fin mathcal F}$ has no countable proper elementary extension.
My approach was to assume we have some proper elementary extension $mathfrak A$ with $mathbb N subsetneq A$. Choose $a in A setminus mathbb N$. As we have a first order sentences true in $mathbb N$ stating that there is no maximal element, either $a$ is between two natural numbers $n,m$ with $m = n+1$, or comes before every natural number. In the first case let $overline n, overline m : mathbb N to mathbb N$ the constant function giving $n$ and $m$. Then in $(mathbb N, <, f)_{fin mathcal F}$ the sentence $exists x forall y ( overline n(y) < x < overline m(y) )$ is false, but in $mathfrak A$ it is true. Hence this case could not arise. Similar if $a$ comes before every natural number, using $overline 0 : mathbb N to mathbb N$ I can give a sentence true in this extension, but not true in $mathbb N$.
So, in consequence there would be no elementary extension at all. But this contradicts the Theorem stated on these slides (see slide number 7), that every infinite model has a proper elementary extension.
So what is wrong with my above reasoning? Why this contradiction?
proof-verification logic first-order-logic model-theory
asked Dec 13 '18 at 17:14
StefanHStefanH
8,11152365
$endgroup$
This is exercise 2.3.1 (2) from Tent/Ziegler, A Course in Model Theory.
Let $mathcal F$ be the set of all functions $mathbb N to mathbb N$. Show that $(mathbb N, <, f)_{fin mathcal F}$ has no countable proper elementary extension.
My approach was to assume we have some proper elementary extension $mathfrak A$ with $mathbb N subsetneq A$. Choose $a in A setminus mathbb N$. As we have a first order sentences true in $mathbb N$ stating that there is no maximal element, either $a$ is between two natural numbers $n,m$ with $m = n+1$, or comes before every natural number. In the first case let $overline n, overline m : mathbb N to mathbb N$ the constant function giving $n$ and $m$. Then in $(mathbb N, <, f)_{fin mathcal F}$ the sentence $exists x forall y ( overline n(y) < x < overline m(y) )$ is false, but in $mathfrak A$ it is true. Hence this case could not arise. Similar if $a$ comes before every natural number, using $overline 0 : mathbb N to mathbb N$ I can give a sentence true in this extension, but not true in $mathbb N$.
So, in consequence there would be no elementary extension at all. But this contradicts the Theorem stated on these slides (see slide number 7), that every infinite model has a proper elementary extension.
So what is wrong with my above reasoning? Why this contradiction?
proof-verification logic first-order-logic model-theory
proof-verification logic first-order-logic model-theory
asked Dec 13 '18 at 17:14
StefanHStefanH
8,11152365
asked Dec 13 '18 at 17:14
StefanHStefanH
8,11152365
asked Dec 13 '18 at 17:14
StefanHStefanH
8,11152365
asked Dec 13 '18 at 17:14
StefanHStefanH
8,11152365
asked Dec 13 '18 at 17:14
StefanHStefanH
8,11152365
8,11152365
2
$begingroup$
"Either $a$ is between two natural numbers $n$, $m$ with $m=n+1$, or comes before every natural number". What?? Exactly the opposite is true! It's true that $mathbb{N}models forall x, exists y, (x<y)$ (there is no maximal element), so $mathfrak{A}$ also satisfies this sentence. But this doesn't rule out having an element of $mathfrak{A}$ which is bigger than all elements of $mathbb{N}$ - it's just that there also have to be even bigger elements of $mathfrak{A}$!
$endgroup$
– Alex Kruckman
Dec 13 '18 at 21:06
add a comment |
2
$begingroup$
"Either $a$ is between two natural numbers $n$, $m$ with $m=n+1$, or comes before every natural number". What?? Exactly the opposite is true! It's true that $mathbb{N}models forall x, exists y, (x<y)$ (there is no maximal element), so $mathfrak{A}$ also satisfies this sentence. But this doesn't rule out having an element of $mathfrak{A}$ which is bigger than all elements of $mathbb{N}$ - it's just that there also have to be even bigger elements of $mathfrak{A}$!
$endgroup$
– Alex Kruckman
Dec 13 '18 at 21:06
2
2
$begingroup$
"Either $a$ is between two natural numbers $n$, $m$ with $m=n+1$, or comes before every natural number". What?? Exactly the opposite is true! It's true that $mathbb{N}models forall x, exists y, (x<y)$ (there is no maximal element), so $mathfrak{A}$ also satisfies this sentence. But this doesn't rule out having an element of $mathfrak{A}$ which is bigger than all elements of $mathbb{N}$ - it's just that there also have to be even bigger elements of $mathfrak{A}$!
$endgroup$
– Alex Kruckman
Dec 13 '18 at 21:06
$begingroup$
"Either $a$ is between two natural numbers $n$, $m$ with $m=n+1$, or comes before every natural number". What?? Exactly the opposite is true! It's true that $mathbb{N}models forall x, exists y, (x<y)$ (there is no maximal element), so $mathfrak{A}$ also satisfies this sentence. But this doesn't rule out having an element of $mathfrak{A}$ which is bigger than all elements of $mathbb{N}$ - it's just that there also have to be even bigger elements of $mathfrak{A}$!
$endgroup$
– Alex Kruckman
Dec 13 '18 at 21:06
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
There seems to be some confusion here. Let $mathcal{M} = (mathbb{N}, < f)_{f in mathcal{F}}$. We want to show that if $mathcal{M} prec mathcal{N}$, then $|mathcal{N}| > aleph_0$ (by upward Löwenheim–Skolem/Ultrapower construction, we know such $mathcal{N}$ exists). Any element in $a in mathcal{N} backslash mathcal{M}$ will be greater than any natural number (in above, you say that it must be less than any natural number or inbetween two of them).
Now, one has to show that $|mathcal{N}| > aleph_0$. Choose $a in mathcal{N} backslash mathcal{M}$. We know that $M models forall x exists y f(x) = y$ for every $f in mathcal{F}$. Since $M prec mathcal{N}$, we know that for every $f in mathcal{F}$, $f(a) in mathcal{N}$. Now, it's your job to show that $|mathcal{N}| > aleph_0$.
(As a hint, you need to somehow incorporate the ordering.)
answered Dec 13 '18 at 20:32
KyleKyle
5,28321334
$endgroup$
$begingroup$
The set of monotone functions is uncountable, and for a monotone function $f$ I would have $f(a) in mathcal N setminus mathcal M$. If I can show that all these values $f(a)$ are distinct for different $f$ this would give the result, but I guess this does not hold. So do you have one hint more how to proceed?
$endgroup$
– StefanH
Dec 14 '18 at 13:37
1
$begingroup$
@StefanH (1) Find an uncountable subfamily $mathcal F_0subseteqmathcal F$ with the following property: for every two different $f,qinmathcal F_0$, there is $n_0inmathbb N$ such that $f(n)neq g(n)$ holds for all $n>n_0$. (2) Conclude that for different $f,ginmathcal F_0$, $f(a)neq g(a)$.
$endgroup$
– SMM
Dec 14 '18 at 16:17
add a comment |
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$begingroup$
There seems to be some confusion here. Let $mathcal{M} = (mathbb{N}, < f)_{f in mathcal{F}}$. We want to show that if $mathcal{M} prec mathcal{N}$, then $|mathcal{N}| > aleph_0$ (by upward Löwenheim–Skolem/Ultrapower construction, we know such $mathcal{N}$ exists). Any element in $a in mathcal{N} backslash mathcal{M}$ will be greater than any natural number (in above, you say that it must be less than any natural number or inbetween two of them).
Now, one has to show that $|mathcal{N}| > aleph_0$. Choose $a in mathcal{N} backslash mathcal{M}$. We know that $M models forall x exists y f(x) = y$ for every $f in mathcal{F}$. Since $M prec mathcal{N}$, we know that for every $f in mathcal{F}$, $f(a) in mathcal{N}$. Now, it's your job to show that $|mathcal{N}| > aleph_0$.
(As a hint, you need to somehow incorporate the ordering.)
answered Dec 13 '18 at 20:32
KyleKyle
5,28321334
$endgroup$
$begingroup$
The set of monotone functions is uncountable, and for a monotone function $f$ I would have $f(a) in mathcal N setminus mathcal M$. If I can show that all these values $f(a)$ are distinct for different $f$ this would give the result, but I guess this does not hold. So do you have one hint more how to proceed?
$endgroup$
– StefanH
Dec 14 '18 at 13:37
1
$begingroup$
@StefanH (1) Find an uncountable subfamily $mathcal F_0subseteqmathcal F$ with the following property: for every two different $f,qinmathcal F_0$, there is $n_0inmathbb N$ such that $f(n)neq g(n)$ holds for all $n>n_0$. (2) Conclude that for different $f,ginmathcal F_0$, $f(a)neq g(a)$.
$endgroup$
– SMM
Dec 14 '18 at 16:17
add a comment |
$begingroup$
There seems to be some confusion here. Let $mathcal{M} = (mathbb{N}, < f)_{f in mathcal{F}}$. We want to show that if $mathcal{M} prec mathcal{N}$, then $|mathcal{N}| > aleph_0$ (by upward Löwenheim–Skolem/Ultrapower construction, we know such $mathcal{N}$ exists). Any element in $a in mathcal{N} backslash mathcal{M}$ will be greater than any natural number (in above, you say that it must be less than any natural number or inbetween two of them).
Now, one has to show that $|mathcal{N}| > aleph_0$. Choose $a in mathcal{N} backslash mathcal{M}$. We know that $M models forall x exists y f(x) = y$ for every $f in mathcal{F}$. Since $M prec mathcal{N}$, we know that for every $f in mathcal{F}$, $f(a) in mathcal{N}$. Now, it's your job to show that $|mathcal{N}| > aleph_0$.
(As a hint, you need to somehow incorporate the ordering.)
answered Dec 13 '18 at 20:32
KyleKyle
5,28321334
$endgroup$
$begingroup$
The set of monotone functions is uncountable, and for a monotone function $f$ I would have $f(a) in mathcal N setminus mathcal M$. If I can show that all these values $f(a)$ are distinct for different $f$ this would give the result, but I guess this does not hold. So do you have one hint more how to proceed?
$endgroup$
– StefanH
Dec 14 '18 at 13:37
1
$begingroup$
@StefanH (1) Find an uncountable subfamily $mathcal F_0subseteqmathcal F$ with the following property: for every two different $f,qinmathcal F_0$, there is $n_0inmathbb N$ such that $f(n)neq g(n)$ holds for all $n>n_0$. (2) Conclude that for different $f,ginmathcal F_0$, $f(a)neq g(a)$.
$endgroup$
– SMM
Dec 14 '18 at 16:17
add a comment |
$begingroup$
There seems to be some confusion here. Let $mathcal{M} = (mathbb{N}, < f)_{f in mathcal{F}}$. We want to show that if $mathcal{M} prec mathcal{N}$, then $|mathcal{N}| > aleph_0$ (by upward Löwenheim–Skolem/Ultrapower construction, we know such $mathcal{N}$ exists). Any element in $a in mathcal{N} backslash mathcal{M}$ will be greater than any natural number (in above, you say that it must be less than any natural number or inbetween two of them).
Now, one has to show that $|mathcal{N}| > aleph_0$. Choose $a in mathcal{N} backslash mathcal{M}$. We know that $M models forall x exists y f(x) = y$ for every $f in mathcal{F}$. Since $M prec mathcal{N}$, we know that for every $f in mathcal{F}$, $f(a) in mathcal{N}$. Now, it's your job to show that $|mathcal{N}| > aleph_0$.
(As a hint, you need to somehow incorporate the ordering.)
answered Dec 13 '18 at 20:32
KyleKyle
5,28321334
$endgroup$
There seems to be some confusion here. Let $mathcal{M} = (mathbb{N}, < f)_{f in mathcal{F}}$. We want to show that if $mathcal{M} prec mathcal{N}$, then $|mathcal{N}| > aleph_0$ (by upward Löwenheim–Skolem/Ultrapower construction, we know such $mathcal{N}$ exists). Any element in $a in mathcal{N} backslash mathcal{M}$ will be greater than any natural number (in above, you say that it must be less than any natural number or inbetween two of them).
Now, one has to show that $|mathcal{N}| > aleph_0$. Choose $a in mathcal{N} backslash mathcal{M}$. We know that $M models forall x exists y f(x) = y$ for every $f in mathcal{F}$. Since $M prec mathcal{N}$, we know that for every $f in mathcal{F}$, $f(a) in mathcal{N}$. Now, it's your job to show that $|mathcal{N}| > aleph_0$.
(As a hint, you need to somehow incorporate the ordering.)
answered Dec 13 '18 at 20:32
KyleKyle
5,28321334
edited Dec 14 '18 at 1:58
answered Dec 13 '18 at 20:32
KyleKyle
5,28321334
answered Dec 13 '18 at 20:32
KyleKyle
5,28321334
answered Dec 13 '18 at 20:32
KyleKyle
5,28321334
5,28321334
$begingroup$
The set of monotone functions is uncountable, and for a monotone function $f$ I would have $f(a) in mathcal N setminus mathcal M$. If I can show that all these values $f(a)$ are distinct for different $f$ this would give the result, but I guess this does not hold. So do you have one hint more how to proceed?
$endgroup$
– StefanH
Dec 14 '18 at 13:37
1
$begingroup$
@StefanH (1) Find an uncountable subfamily $mathcal F_0subseteqmathcal F$ with the following property: for every two different $f,qinmathcal F_0$, there is $n_0inmathbb N$ such that $f(n)neq g(n)$ holds for all $n>n_0$. (2) Conclude that for different $f,ginmathcal F_0$, $f(a)neq g(a)$.
$endgroup$
– SMM
Dec 14 '18 at 16:17
add a comment |
$begingroup$
The set of monotone functions is uncountable, and for a monotone function $f$ I would have $f(a) in mathcal N setminus mathcal M$. If I can show that all these values $f(a)$ are distinct for different $f$ this would give the result, but I guess this does not hold. So do you have one hint more how to proceed?
$endgroup$
– StefanH
Dec 14 '18 at 13:37
1
$begingroup$
@StefanH (1) Find an uncountable subfamily $mathcal F_0subseteqmathcal F$ with the following property: for every two different $f,qinmathcal F_0$, there is $n_0inmathbb N$ such that $f(n)neq g(n)$ holds for all $n>n_0$. (2) Conclude that for different $f,ginmathcal F_0$, $f(a)neq g(a)$.
$endgroup$
– SMM
Dec 14 '18 at 16:17
$begingroup$
The set of monotone functions is uncountable, and for a monotone function $f$ I would have $f(a) in mathcal N setminus mathcal M$. If I can show that all these values $f(a)$ are distinct for different $f$ this would give the result, but I guess this does not hold. So do you have one hint more how to proceed?
$endgroup$
– StefanH
Dec 14 '18 at 13:37
$begingroup$
The set of monotone functions is uncountable, and for a monotone function $f$ I would have $f(a) in mathcal N setminus mathcal M$. If I can show that all these values $f(a)$ are distinct for different $f$ this would give the result, but I guess this does not hold. So do you have one hint more how to proceed?
$endgroup$
– StefanH
Dec 14 '18 at 13:37
1
1
$begingroup$
@StefanH (1) Find an uncountable subfamily $mathcal F_0subseteqmathcal F$ with the following property: for every two different $f,qinmathcal F_0$, there is $n_0inmathbb N$ such that $f(n)neq g(n)$ holds for all $n>n_0$. (2) Conclude that for different $f,ginmathcal F_0$, $f(a)neq g(a)$.
$endgroup$
– SMM
Dec 14 '18 at 16:17
$begingroup$
@StefanH (1) Find an uncountable subfamily $mathcal F_0subseteqmathcal F$ with the following property: for every two different $f,qinmathcal F_0$, there is $n_0inmathbb N$ such that $f(n)neq g(n)$ holds for all $n>n_0$. (2) Conclude that for different $f,ginmathcal F_0$, $f(a)neq g(a)$.
$endgroup$
– SMM
Dec 14 '18 at 16:17
add a comment |
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$begingroup$
"Either $a$ is between two natural numbers $n$, $m$ with $m=n+1$, or comes before every natural number". What?? Exactly the opposite is true! It's true that $mathbb{N}models forall x, exists y, (x<y)$ (there is no maximal element), so $mathfrak{A}$ also satisfies this sentence. But this doesn't rule out having an element of $mathfrak{A}$ which is bigger than all elements of $mathbb{N}$ - it's just that there also have to be even bigger elements of $mathfrak{A}$!
$endgroup$
– Alex Kruckman
Dec 13 '18 at 21:06