Mixing
What does this set-theoretic notation mean in the proof of this theorem?
$begingroup$
From A friendly introduction to mathematical logic,
In the first red box below, what does this notation mean?
And in the second red box, how exactly is this process repeated?
logic notation proof-explanation first-order-logic model-theory
edited Jan 3 at 17:09
Mauro ALLEGRANZA
64.8k448112
asked Jan 3 at 15:14
Oliver GOliver G
1,6171529
$endgroup$
add a comment |
$begingroup$
From A friendly introduction to mathematical logic,
In the first red box below, what does this notation mean?
And in the second red box, how exactly is this process repeated?
logic notation proof-explanation first-order-logic model-theory
edited Jan 3 at 17:09
Mauro ALLEGRANZA
64.8k448112
asked Jan 3 at 15:14
Oliver GOliver G
1,6171529
$endgroup$
add a comment |
$begingroup$
From A friendly introduction to mathematical logic,
In the first red box below, what does this notation mean?
And in the second red box, how exactly is this process repeated?
logic notation proof-explanation first-order-logic model-theory
edited Jan 3 at 17:09
Mauro ALLEGRANZA
64.8k448112
asked Jan 3 at 15:14
Oliver GOliver G
1,6171529
$endgroup$
From A friendly introduction to mathematical logic,
In the first red box below, what does this notation mean?
And in the second red box, how exactly is this process repeated?
logic notation proof-explanation first-order-logic model-theory
logic notation proof-explanation first-order-logic model-theory
edited Jan 3 at 17:09
Mauro ALLEGRANZA
64.8k448112
asked Jan 3 at 15:14
Oliver GOliver G
1,6171529
edited Jan 3 at 17:09
Mauro ALLEGRANZA
64.8k448112
asked Jan 3 at 15:14
Oliver GOliver G
1,6171529
edited Jan 3 at 17:09
Mauro ALLEGRANZA
64.8k448112
edited Jan 3 at 17:09
Mauro ALLEGRANZA
64.8k448112
edited Jan 3 at 17:09
Mauro ALLEGRANZA
64.8k448112
64.8k448112
asked Jan 3 at 15:14
Oliver GOliver G
1,6171529
asked Jan 3 at 15:14
Oliver GOliver G
1,6171529
asked Jan 3 at 15:14
Oliver GOliver G
1,6171529
1,6171529
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
what does [the highlighted] notation mean?
Note 1 : the set $B$ is the uncountable domain of the structure $mathfrak B$ and $A_0$ is a nonempty countable subset of $B$.
The function $s: text {Vars} to A_0$ is the usual variable assignment function that maps variables of the language into objects of the domain of an interpretation: in this case $A_0$.
We have : "for each formula $exists x alpha$ and functions $s$ and $s'$ ... choose an element $a_{alpha, s'}$ in $B$ [i.e. an object of the domain] such that..."
Now take the set ${ a_{alpha, s'} }_{text {all } alpha, s': text{Vars} to A_0}$ of all such elements and set : $A_1 = A_0 cup { a_{alpha, s'} }$.
Then consider "for each formula $exists x alpha$ and functions $s$ and $s'$ ... such that..." and build the new set ${ a_{alpha, s'} }_{text {all } alpha, s': text{Vars} to A_1}$ and set $A_2 = A_1 cup { a_{alpha, s'} }$.
And so on...
At the end of the process, the union of all $A_i$'s will be the domain of the countable substructure needed for the theorem.
how exactly is this process repeated?
Note 2 : in the initial stock of functions we have $s : text {Vars} to A_0$. We use them to enlarge the initial set $A_0$ to a new set $A_1$ with more elements.
Due to the fact that $A_1$ has new elements, in the succesive step the stock of functions $s : text {Vars} to A_1$ will be "larger", due to the fact we start with $A_0 ⊂ B$ but we "throw in" new elements of $B$ that not necessarily are in $A_0$.
The tricky point is the reason for "iteration"...
We have to recall the clauses defining the satisfaction relation $vDash$ : $mathfrak B vDash exists x alpha [s]$ means that there is a variable assignment function $s'$ that agree with $s$ on every free variable of the formula such that $mathfrak B vDash alpha [s'[x|a]]$, for some $a in B$.
We start with $exists x alpha$ and $s : text {Vars} to A_0$ such that $mathfrak B vDash exists x alpha [s]$.
But it may happen that $mathfrak B vDash exists x alpha [s]$ does not hold because there is no suitable "$x$-variant" $s' : text {Vars} to A_0$.
In the following steps, when we have enlarged $A_0$, this may change and a new $s$ pops-up such that $mathfrak B vDash exists x alpha [s]$.
This is why at each step we have to review" every existential formulas.
Note 3 : Why consider only existential formulas ?
Because obviously every universal one that is true in the larger structure will holds also in every subset of the structure (if every ball in my box are black and I pick up at random five balls from it, for sure they will be black. It is not the same if in the box there is one white ball: I'm not certain that in the five chosen balls there is the white one.)
The issue is that with the existential formula like $exists x alpha$, we have no certainty that the "witness" (if any) is included into $A_0$.
The process described above amount "simply" to this : starting from the countable $A_0$, throw in countable new elements [because thera are countable many iterations] that are enough to satisfy all the existential formulas.
answered Jan 3 at 15:33
Mauro ALLEGRANZAMauro ALLEGRANZA
64.8k448112
$endgroup$
$begingroup$
By "the set of all such elements", do you mean the set of all elements for the particular combination of $(exists x a, s, s')$? And the following $A_k$ are the set of all elements for the next combination?
$endgroup$
– Oliver G
Jan 3 at 15:48
$begingroup$
@OliverG - for each formula and each function $s$ ve have an object $a_{alpha, s'}$. Then, considering all possible pairs formula-function we have a set ${ a_{alpha, s'} }$ of objects. With it we form the new set $A_1 = A_0 cup { a_{alpha, s'} }$.
$endgroup$
– Mauro ALLEGRANZA
Jan 3 at 15:52
$begingroup$
If the set contains every $b$ for each combination $(exists x a, s, s')$, how is $A_2$ formed? What process is being repeated?
$endgroup$
– Oliver G
Jan 3 at 15:57
$begingroup$
What is the specific process of creating $A_1$? I don't understand what is being done in order to create it.
$endgroup$
– Oliver G
Jan 3 at 16:06
add a comment |
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1 Answer
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$begingroup$
what does [the highlighted] notation mean?
Note 1 : the set $B$ is the uncountable domain of the structure $mathfrak B$ and $A_0$ is a nonempty countable subset of $B$.
The function $s: text {Vars} to A_0$ is the usual variable assignment function that maps variables of the language into objects of the domain of an interpretation: in this case $A_0$.
We have : "for each formula $exists x alpha$ and functions $s$ and $s'$ ... choose an element $a_{alpha, s'}$ in $B$ [i.e. an object of the domain] such that..."
Now take the set ${ a_{alpha, s'} }_{text {all } alpha, s': text{Vars} to A_0}$ of all such elements and set : $A_1 = A_0 cup { a_{alpha, s'} }$.
Then consider "for each formula $exists x alpha$ and functions $s$ and $s'$ ... such that..." and build the new set ${ a_{alpha, s'} }_{text {all } alpha, s': text{Vars} to A_1}$ and set $A_2 = A_1 cup { a_{alpha, s'} }$.
And so on...
At the end of the process, the union of all $A_i$'s will be the domain of the countable substructure needed for the theorem.
how exactly is this process repeated?
Note 2 : in the initial stock of functions we have $s : text {Vars} to A_0$. We use them to enlarge the initial set $A_0$ to a new set $A_1$ with more elements.
Due to the fact that $A_1$ has new elements, in the succesive step the stock of functions $s : text {Vars} to A_1$ will be "larger", due to the fact we start with $A_0 ⊂ B$ but we "throw in" new elements of $B$ that not necessarily are in $A_0$.
The tricky point is the reason for "iteration"...
We have to recall the clauses defining the satisfaction relation $vDash$ : $mathfrak B vDash exists x alpha [s]$ means that there is a variable assignment function $s'$ that agree with $s$ on every free variable of the formula such that $mathfrak B vDash alpha [s'[x|a]]$, for some $a in B$.
We start with $exists x alpha$ and $s : text {Vars} to A_0$ such that $mathfrak B vDash exists x alpha [s]$.
But it may happen that $mathfrak B vDash exists x alpha [s]$ does not hold because there is no suitable "$x$-variant" $s' : text {Vars} to A_0$.
In the following steps, when we have enlarged $A_0$, this may change and a new $s$ pops-up such that $mathfrak B vDash exists x alpha [s]$.
This is why at each step we have to review" every existential formulas.
Note 3 : Why consider only existential formulas ?
Because obviously every universal one that is true in the larger structure will holds also in every subset of the structure (if every ball in my box are black and I pick up at random five balls from it, for sure they will be black. It is not the same if in the box there is one white ball: I'm not certain that in the five chosen balls there is the white one.)
The issue is that with the existential formula like $exists x alpha$, we have no certainty that the "witness" (if any) is included into $A_0$.
The process described above amount "simply" to this : starting from the countable $A_0$, throw in countable new elements [because thera are countable many iterations] that are enough to satisfy all the existential formulas.
answered Jan 3 at 15:33
Mauro ALLEGRANZAMauro ALLEGRANZA
64.8k448112
$endgroup$
$begingroup$
By "the set of all such elements", do you mean the set of all elements for the particular combination of $(exists x a, s, s')$? And the following $A_k$ are the set of all elements for the next combination?
$endgroup$
– Oliver G
Jan 3 at 15:48
$begingroup$
@OliverG - for each formula and each function $s$ ve have an object $a_{alpha, s'}$. Then, considering all possible pairs formula-function we have a set ${ a_{alpha, s'} }$ of objects. With it we form the new set $A_1 = A_0 cup { a_{alpha, s'} }$.
$endgroup$
– Mauro ALLEGRANZA
Jan 3 at 15:52
$begingroup$
If the set contains every $b$ for each combination $(exists x a, s, s')$, how is $A_2$ formed? What process is being repeated?
$endgroup$
– Oliver G
Jan 3 at 15:57
$begingroup$
What is the specific process of creating $A_1$? I don't understand what is being done in order to create it.
$endgroup$
– Oliver G
Jan 3 at 16:06
add a comment |
$begingroup$
what does [the highlighted] notation mean?
Note 1 : the set $B$ is the uncountable domain of the structure $mathfrak B$ and $A_0$ is a nonempty countable subset of $B$.
The function $s: text {Vars} to A_0$ is the usual variable assignment function that maps variables of the language into objects of the domain of an interpretation: in this case $A_0$.
We have : "for each formula $exists x alpha$ and functions $s$ and $s'$ ... choose an element $a_{alpha, s'}$ in $B$ [i.e. an object of the domain] such that..."
Now take the set ${ a_{alpha, s'} }_{text {all } alpha, s': text{Vars} to A_0}$ of all such elements and set : $A_1 = A_0 cup { a_{alpha, s'} }$.
Then consider "for each formula $exists x alpha$ and functions $s$ and $s'$ ... such that..." and build the new set ${ a_{alpha, s'} }_{text {all } alpha, s': text{Vars} to A_1}$ and set $A_2 = A_1 cup { a_{alpha, s'} }$.
And so on...
At the end of the process, the union of all $A_i$'s will be the domain of the countable substructure needed for the theorem.
how exactly is this process repeated?
Note 2 : in the initial stock of functions we have $s : text {Vars} to A_0$. We use them to enlarge the initial set $A_0$ to a new set $A_1$ with more elements.
Due to the fact that $A_1$ has new elements, in the succesive step the stock of functions $s : text {Vars} to A_1$ will be "larger", due to the fact we start with $A_0 ⊂ B$ but we "throw in" new elements of $B$ that not necessarily are in $A_0$.
The tricky point is the reason for "iteration"...
We have to recall the clauses defining the satisfaction relation $vDash$ : $mathfrak B vDash exists x alpha [s]$ means that there is a variable assignment function $s'$ that agree with $s$ on every free variable of the formula such that $mathfrak B vDash alpha [s'[x|a]]$, for some $a in B$.
We start with $exists x alpha$ and $s : text {Vars} to A_0$ such that $mathfrak B vDash exists x alpha [s]$.
But it may happen that $mathfrak B vDash exists x alpha [s]$ does not hold because there is no suitable "$x$-variant" $s' : text {Vars} to A_0$.
In the following steps, when we have enlarged $A_0$, this may change and a new $s$ pops-up such that $mathfrak B vDash exists x alpha [s]$.
This is why at each step we have to review" every existential formulas.
Note 3 : Why consider only existential formulas ?
Because obviously every universal one that is true in the larger structure will holds also in every subset of the structure (if every ball in my box are black and I pick up at random five balls from it, for sure they will be black. It is not the same if in the box there is one white ball: I'm not certain that in the five chosen balls there is the white one.)
The issue is that with the existential formula like $exists x alpha$, we have no certainty that the "witness" (if any) is included into $A_0$.
The process described above amount "simply" to this : starting from the countable $A_0$, throw in countable new elements [because thera are countable many iterations] that are enough to satisfy all the existential formulas.
answered Jan 3 at 15:33
Mauro ALLEGRANZAMauro ALLEGRANZA
64.8k448112
$endgroup$
$begingroup$
By "the set of all such elements", do you mean the set of all elements for the particular combination of $(exists x a, s, s')$? And the following $A_k$ are the set of all elements for the next combination?
$endgroup$
– Oliver G
Jan 3 at 15:48
$begingroup$
@OliverG - for each formula and each function $s$ ve have an object $a_{alpha, s'}$. Then, considering all possible pairs formula-function we have a set ${ a_{alpha, s'} }$ of objects. With it we form the new set $A_1 = A_0 cup { a_{alpha, s'} }$.
$endgroup$
– Mauro ALLEGRANZA
Jan 3 at 15:52
$begingroup$
If the set contains every $b$ for each combination $(exists x a, s, s')$, how is $A_2$ formed? What process is being repeated?
$endgroup$
– Oliver G
Jan 3 at 15:57
$begingroup$
What is the specific process of creating $A_1$? I don't understand what is being done in order to create it.
$endgroup$
– Oliver G
Jan 3 at 16:06
add a comment |
$begingroup$
what does [the highlighted] notation mean?
Note 1 : the set $B$ is the uncountable domain of the structure $mathfrak B$ and $A_0$ is a nonempty countable subset of $B$.
The function $s: text {Vars} to A_0$ is the usual variable assignment function that maps variables of the language into objects of the domain of an interpretation: in this case $A_0$.
We have : "for each formula $exists x alpha$ and functions $s$ and $s'$ ... choose an element $a_{alpha, s'}$ in $B$ [i.e. an object of the domain] such that..."
Now take the set ${ a_{alpha, s'} }_{text {all } alpha, s': text{Vars} to A_0}$ of all such elements and set : $A_1 = A_0 cup { a_{alpha, s'} }$.
Then consider "for each formula $exists x alpha$ and functions $s$ and $s'$ ... such that..." and build the new set ${ a_{alpha, s'} }_{text {all } alpha, s': text{Vars} to A_1}$ and set $A_2 = A_1 cup { a_{alpha, s'} }$.
And so on...
At the end of the process, the union of all $A_i$'s will be the domain of the countable substructure needed for the theorem.
how exactly is this process repeated?
Note 2 : in the initial stock of functions we have $s : text {Vars} to A_0$. We use them to enlarge the initial set $A_0$ to a new set $A_1$ with more elements.
Due to the fact that $A_1$ has new elements, in the succesive step the stock of functions $s : text {Vars} to A_1$ will be "larger", due to the fact we start with $A_0 ⊂ B$ but we "throw in" new elements of $B$ that not necessarily are in $A_0$.
The tricky point is the reason for "iteration"...
We have to recall the clauses defining the satisfaction relation $vDash$ : $mathfrak B vDash exists x alpha [s]$ means that there is a variable assignment function $s'$ that agree with $s$ on every free variable of the formula such that $mathfrak B vDash alpha [s'[x|a]]$, for some $a in B$.
We start with $exists x alpha$ and $s : text {Vars} to A_0$ such that $mathfrak B vDash exists x alpha [s]$.
But it may happen that $mathfrak B vDash exists x alpha [s]$ does not hold because there is no suitable "$x$-variant" $s' : text {Vars} to A_0$.
In the following steps, when we have enlarged $A_0$, this may change and a new $s$ pops-up such that $mathfrak B vDash exists x alpha [s]$.
This is why at each step we have to review" every existential formulas.
Note 3 : Why consider only existential formulas ?
Because obviously every universal one that is true in the larger structure will holds also in every subset of the structure (if every ball in my box are black and I pick up at random five balls from it, for sure they will be black. It is not the same if in the box there is one white ball: I'm not certain that in the five chosen balls there is the white one.)
The issue is that with the existential formula like $exists x alpha$, we have no certainty that the "witness" (if any) is included into $A_0$.
The process described above amount "simply" to this : starting from the countable $A_0$, throw in countable new elements [because thera are countable many iterations] that are enough to satisfy all the existential formulas.
answered Jan 3 at 15:33
Mauro ALLEGRANZAMauro ALLEGRANZA
64.8k448112
$endgroup$
what does [the highlighted] notation mean?
Note 1 : the set $B$ is the uncountable domain of the structure $mathfrak B$ and $A_0$ is a nonempty countable subset of $B$.
The function $s: text {Vars} to A_0$ is the usual variable assignment function that maps variables of the language into objects of the domain of an interpretation: in this case $A_0$.
We have : "for each formula $exists x alpha$ and functions $s$ and $s'$ ... choose an element $a_{alpha, s'}$ in $B$ [i.e. an object of the domain] such that..."
Now take the set ${ a_{alpha, s'} }_{text {all } alpha, s': text{Vars} to A_0}$ of all such elements and set : $A_1 = A_0 cup { a_{alpha, s'} }$.
Then consider "for each formula $exists x alpha$ and functions $s$ and $s'$ ... such that..." and build the new set ${ a_{alpha, s'} }_{text {all } alpha, s': text{Vars} to A_1}$ and set $A_2 = A_1 cup { a_{alpha, s'} }$.
And so on...
At the end of the process, the union of all $A_i$'s will be the domain of the countable substructure needed for the theorem.
how exactly is this process repeated?
Note 2 : in the initial stock of functions we have $s : text {Vars} to A_0$. We use them to enlarge the initial set $A_0$ to a new set $A_1$ with more elements.
Due to the fact that $A_1$ has new elements, in the succesive step the stock of functions $s : text {Vars} to A_1$ will be "larger", due to the fact we start with $A_0 ⊂ B$ but we "throw in" new elements of $B$ that not necessarily are in $A_0$.
The tricky point is the reason for "iteration"...
We have to recall the clauses defining the satisfaction relation $vDash$ : $mathfrak B vDash exists x alpha [s]$ means that there is a variable assignment function $s'$ that agree with $s$ on every free variable of the formula such that $mathfrak B vDash alpha [s'[x|a]]$, for some $a in B$.
We start with $exists x alpha$ and $s : text {Vars} to A_0$ such that $mathfrak B vDash exists x alpha [s]$.
But it may happen that $mathfrak B vDash exists x alpha [s]$ does not hold because there is no suitable "$x$-variant" $s' : text {Vars} to A_0$.
In the following steps, when we have enlarged $A_0$, this may change and a new $s$ pops-up such that $mathfrak B vDash exists x alpha [s]$.
This is why at each step we have to review" every existential formulas.
Note 3 : Why consider only existential formulas ?
Because obviously every universal one that is true in the larger structure will holds also in every subset of the structure (if every ball in my box are black and I pick up at random five balls from it, for sure they will be black. It is not the same if in the box there is one white ball: I'm not certain that in the five chosen balls there is the white one.)
The issue is that with the existential formula like $exists x alpha$, we have no certainty that the "witness" (if any) is included into $A_0$.
The process described above amount "simply" to this : starting from the countable $A_0$, throw in countable new elements [because thera are countable many iterations] that are enough to satisfy all the existential formulas.
answered Jan 3 at 15:33
Mauro ALLEGRANZAMauro ALLEGRANZA
64.8k448112
edited Jan 3 at 17:09
answered Jan 3 at 15:33
Mauro ALLEGRANZAMauro ALLEGRANZA
64.8k448112
answered Jan 3 at 15:33
Mauro ALLEGRANZAMauro ALLEGRANZA
64.8k448112
answered Jan 3 at 15:33
Mauro ALLEGRANZAMauro ALLEGRANZA
64.8k448112
64.8k448112
$begingroup$
By "the set of all such elements", do you mean the set of all elements for the particular combination of $(exists x a, s, s')$? And the following $A_k$ are the set of all elements for the next combination?
$endgroup$
– Oliver G
Jan 3 at 15:48
$begingroup$
@OliverG - for each formula and each function $s$ ve have an object $a_{alpha, s'}$. Then, considering all possible pairs formula-function we have a set ${ a_{alpha, s'} }$ of objects. With it we form the new set $A_1 = A_0 cup { a_{alpha, s'} }$.
$endgroup$
– Mauro ALLEGRANZA
Jan 3 at 15:52
$begingroup$
If the set contains every $b$ for each combination $(exists x a, s, s')$, how is $A_2$ formed? What process is being repeated?
$endgroup$
– Oliver G
Jan 3 at 15:57
$begingroup$
What is the specific process of creating $A_1$? I don't understand what is being done in order to create it.
$endgroup$
– Oliver G
Jan 3 at 16:06
add a comment |
$begingroup$
By "the set of all such elements", do you mean the set of all elements for the particular combination of $(exists x a, s, s')$? And the following $A_k$ are the set of all elements for the next combination?
$endgroup$
– Oliver G
Jan 3 at 15:48
$begingroup$
@OliverG - for each formula and each function $s$ ve have an object $a_{alpha, s'}$. Then, considering all possible pairs formula-function we have a set ${ a_{alpha, s'} }$ of objects. With it we form the new set $A_1 = A_0 cup { a_{alpha, s'} }$.
$endgroup$
– Mauro ALLEGRANZA
Jan 3 at 15:52
$begingroup$
If the set contains every $b$ for each combination $(exists x a, s, s')$, how is $A_2$ formed? What process is being repeated?
$endgroup$
– Oliver G
Jan 3 at 15:57
$begingroup$
What is the specific process of creating $A_1$? I don't understand what is being done in order to create it.
$endgroup$
– Oliver G
Jan 3 at 16:06
$begingroup$
By "the set of all such elements", do you mean the set of all elements for the particular combination of $(exists x a, s, s')$? And the following $A_k$ are the set of all elements for the next combination?
$endgroup$
– Oliver G
Jan 3 at 15:48
$begingroup$
By "the set of all such elements", do you mean the set of all elements for the particular combination of $(exists x a, s, s')$? And the following $A_k$ are the set of all elements for the next combination?
$endgroup$
– Oliver G
Jan 3 at 15:48
$begingroup$
@OliverG - for each formula and each function $s$ ve have an object $a_{alpha, s'}$. Then, considering all possible pairs formula-function we have a set ${ a_{alpha, s'} }$ of objects. With it we form the new set $A_1 = A_0 cup { a_{alpha, s'} }$.
$endgroup$
– Mauro ALLEGRANZA
Jan 3 at 15:52
$begingroup$
@OliverG - for each formula and each function $s$ ve have an object $a_{alpha, s'}$. Then, considering all possible pairs formula-function we have a set ${ a_{alpha, s'} }$ of objects. With it we form the new set $A_1 = A_0 cup { a_{alpha, s'} }$.
$endgroup$
– Mauro ALLEGRANZA
Jan 3 at 15:52
$begingroup$
If the set contains every $b$ for each combination $(exists x a, s, s')$, how is $A_2$ formed? What process is being repeated?
$endgroup$
– Oliver G
Jan 3 at 15:57
$begingroup$
If the set contains every $b$ for each combination $(exists x a, s, s')$, how is $A_2$ formed? What process is being repeated?
$endgroup$
– Oliver G
Jan 3 at 15:57
$begingroup$
What is the specific process of creating $A_1$? I don't understand what is being done in order to create it.
$endgroup$
– Oliver G
Jan 3 at 16:06
$begingroup$
What is the specific process of creating $A_1$? I don't understand what is being done in order to create it.
$endgroup$
– Oliver G
Jan 3 at 16:06
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