How does Prof D. J. H. Garling generalize recursion theorem?
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From textbook A Course in Mathematical Analysis by Prof D. J. H. Garling, I'm not able to understand how he generalizes recursion theorem.
In the first step, he states the theorem:
Here, mapping $s:P→P$ satisfies Peano’s axioms.
In the second step, he reasons:
Finally, he gets the theorem in a more general form:
My confusion is that:
In the original theorem, for each $bara$, there exists a unique mapping $a:P → A$ such that $a_0=bara$ and $a_s(n)=f(a_n)$ $forall nin P$. We substitute $P$ by $Z^+$. As a result, for each $bara$, there exists a unique mapping $a:Z^+ → A$ such that $a_0=bara$ and $a_s(n)=f(a_n)$ $forall nin Z^+$. However, I don't understand how the author derives the mapping $f^n:A → A$ from the mapping $a:Z^+ → A$.
In the second step, the author lets $f^n(bara)=a_n$. This means $f^0(bara)=a_0$, $f^1(bara)=a_1$, $f^2(bara)=a_2$, ..., $f^n(bara)=a_n$. From my understanding, he only defines $f^0, f^1, f^2, f^3,..., f^n$ for only $bara$. I don't understand how the author defines $f^0, f^1, f^2, f^3,..., f^n$ for the other elements in A.
$(f^n(bara)=a_n$ and it is possible that $ran(a)subsetneq A) → ran(f^n)subsetneq A$. How can $f^n$ be defined for all $ain A$?
For the sake of clarity, there is a screenshot of full proof of recursion theorem by the author at Possible typos in the proof of Recursion Theorem.
Many thanks for clarifying my doubt!
elementary-set-theory
asked Jan 23 at 1:19
Le Anh Dung
709419
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show 1 more comment
up vote
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From textbook A Course in Mathematical Analysis by Prof D. J. H. Garling, I'm not able to understand how he generalizes recursion theorem.
In the first step, he states the theorem:
Here, mapping $s:P→P$ satisfies Peano’s axioms.
In the second step, he reasons:
Finally, he gets the theorem in a more general form:
My confusion is that:
In the original theorem, for each $bara$, there exists a unique mapping $a:P → A$ such that $a_0=bara$ and $a_s(n)=f(a_n)$ $forall nin P$. We substitute $P$ by $Z^+$. As a result, for each $bara$, there exists a unique mapping $a:Z^+ → A$ such that $a_0=bara$ and $a_s(n)=f(a_n)$ $forall nin Z^+$. However, I don't understand how the author derives the mapping $f^n:A → A$ from the mapping $a:Z^+ → A$.
In the second step, the author lets $f^n(bara)=a_n$. This means $f^0(bara)=a_0$, $f^1(bara)=a_1$, $f^2(bara)=a_2$, ..., $f^n(bara)=a_n$. From my understanding, he only defines $f^0, f^1, f^2, f^3,..., f^n$ for only $bara$. I don't understand how the author defines $f^0, f^1, f^2, f^3,..., f^n$ for the other elements in A.
$(f^n(bara)=a_n$ and it is possible that $ran(a)subsetneq A) → ran(f^n)subsetneq A$. How can $f^n$ be defined for all $ain A$?
For the sake of clarity, there is a screenshot of full proof of recursion theorem by the author at Possible typos in the proof of Recursion Theorem.
Many thanks for clarifying my doubt!
elementary-set-theory
asked Jan 23 at 1:19
Le Anh Dung
709419
Could you clarify what $P$ is? Re 2, I think $f^n$ is supposed to be provided by Theorem 1.8.2, and "$f^n(bar a)=a_n$" is defining $a_n$, not $f^n$.
– stewbasic
Jan 23 at 2:30
@stewbasic, I have just edited my post for clarity. Here, mapping $s:P→P$ satisfies Peano’s axioms. You said "$f^n(bar a)=a_n$" is defining $a_n$, not $f^n$, so i wonder how $f^n$ is defined properly?
– Le Anh Dung
Jan 23 at 2:36
I thought the paragraph was proving Theorem 1.8.2 $Rightarrow$ Theorem 1.8.1, but it's hard to tell without more context. Does the book provide a separate proof of Theorem 1.8.2?
– stewbasic
Jan 23 at 2:48
@stewbasic, I have a screenshot of the full proof at math.stackexchange.com/questions/2613220/…, but there are some typos. Please be careful about them!
– Le Anh Dung
Jan 23 at 2:52
I assume this paragraph follows the proof in your screenshot? Then probably I am wrong, the paragraph is meant to show Theorem 1.8.1 $Rightarrow$ Theorem 1.8.2. In that case we'd need to apply Theorem 1.8.1 to each $ain A$. That is, define $f^n=(a(0),a(n))mid a:mathbf Z^+to A,,acirc s=fcirc a$ and use Theorem 1.8.1 to show $f^n$ is a function.
– stewbasic
Jan 23 at 3:04
 |Â
show 1 more comment
up vote
1
down vote
favorite
up vote
1
down vote
favorite
From textbook A Course in Mathematical Analysis by Prof D. J. H. Garling, I'm not able to understand how he generalizes recursion theorem.
In the first step, he states the theorem:
Here, mapping $s:P→P$ satisfies Peano’s axioms.
In the second step, he reasons:
Finally, he gets the theorem in a more general form:
My confusion is that:
In the original theorem, for each $bara$, there exists a unique mapping $a:P → A$ such that $a_0=bara$ and $a_s(n)=f(a_n)$ $forall nin P$. We substitute $P$ by $Z^+$. As a result, for each $bara$, there exists a unique mapping $a:Z^+ → A$ such that $a_0=bara$ and $a_s(n)=f(a_n)$ $forall nin Z^+$. However, I don't understand how the author derives the mapping $f^n:A → A$ from the mapping $a:Z^+ → A$.
In the second step, the author lets $f^n(bara)=a_n$. This means $f^0(bara)=a_0$, $f^1(bara)=a_1$, $f^2(bara)=a_2$, ..., $f^n(bara)=a_n$. From my understanding, he only defines $f^0, f^1, f^2, f^3,..., f^n$ for only $bara$. I don't understand how the author defines $f^0, f^1, f^2, f^3,..., f^n$ for the other elements in A.
$(f^n(bara)=a_n$ and it is possible that $ran(a)subsetneq A) → ran(f^n)subsetneq A$. How can $f^n$ be defined for all $ain A$?
For the sake of clarity, there is a screenshot of full proof of recursion theorem by the author at Possible typos in the proof of Recursion Theorem.
Many thanks for clarifying my doubt!
elementary-set-theory
asked Jan 23 at 1:19
Le Anh Dung
709419
From textbook A Course in Mathematical Analysis by Prof D. J. H. Garling, I'm not able to understand how he generalizes recursion theorem.
In the first step, he states the theorem:
Here, mapping $s:P→P$ satisfies Peano’s axioms.
In the second step, he reasons:
Finally, he gets the theorem in a more general form:
My confusion is that:
In the original theorem, for each $bara$, there exists a unique mapping $a:P → A$ such that $a_0=bara$ and $a_s(n)=f(a_n)$ $forall nin P$. We substitute $P$ by $Z^+$. As a result, for each $bara$, there exists a unique mapping $a:Z^+ → A$ such that $a_0=bara$ and $a_s(n)=f(a_n)$ $forall nin Z^+$. However, I don't understand how the author derives the mapping $f^n:A → A$ from the mapping $a:Z^+ → A$.
In the second step, the author lets $f^n(bara)=a_n$. This means $f^0(bara)=a_0$, $f^1(bara)=a_1$, $f^2(bara)=a_2$, ..., $f^n(bara)=a_n$. From my understanding, he only defines $f^0, f^1, f^2, f^3,..., f^n$ for only $bara$. I don't understand how the author defines $f^0, f^1, f^2, f^3,..., f^n$ for the other elements in A.
$(f^n(bara)=a_n$ and it is possible that $ran(a)subsetneq A) → ran(f^n)subsetneq A$. How can $f^n$ be defined for all $ain A$?
For the sake of clarity, there is a screenshot of full proof of recursion theorem by the author at Possible typos in the proof of Recursion Theorem.
Many thanks for clarifying my doubt!
elementary-set-theory
elementary-set-theory
asked Jan 23 at 1:19
Le Anh Dung
709419
asked Jan 23 at 1:19
Le Anh Dung
709419
edited Aug 31 at 0:52
asked Jan 23 at 1:19
Le Anh Dung
709419
asked Jan 23 at 1:19
Le Anh Dung
709419
asked Jan 23 at 1:19
Le Anh Dung
709419
709419
Could you clarify what $P$ is? Re 2, I think $f^n$ is supposed to be provided by Theorem 1.8.2, and "$f^n(bar a)=a_n$" is defining $a_n$, not $f^n$.
– stewbasic
Jan 23 at 2:30
@stewbasic, I have just edited my post for clarity. Here, mapping $s:P→P$ satisfies Peano’s axioms. You said "$f^n(bar a)=a_n$" is defining $a_n$, not $f^n$, so i wonder how $f^n$ is defined properly?
– Le Anh Dung
Jan 23 at 2:36
I thought the paragraph was proving Theorem 1.8.2 $Rightarrow$ Theorem 1.8.1, but it's hard to tell without more context. Does the book provide a separate proof of Theorem 1.8.2?
– stewbasic
Jan 23 at 2:48
@stewbasic, I have a screenshot of the full proof at math.stackexchange.com/questions/2613220/…, but there are some typos. Please be careful about them!
– Le Anh Dung
Jan 23 at 2:52
I assume this paragraph follows the proof in your screenshot? Then probably I am wrong, the paragraph is meant to show Theorem 1.8.1 $Rightarrow$ Theorem 1.8.2. In that case we'd need to apply Theorem 1.8.1 to each $ain A$. That is, define $f^n=(a(0),a(n))mid a:mathbf Z^+to A,,acirc s=fcirc a$ and use Theorem 1.8.1 to show $f^n$ is a function.
– stewbasic
Jan 23 at 3:04
 |Â
show 1 more comment
Could you clarify what $P$ is? Re 2, I think $f^n$ is supposed to be provided by Theorem 1.8.2, and "$f^n(bar a)=a_n$" is defining $a_n$, not $f^n$.
– stewbasic
Jan 23 at 2:30
@stewbasic, I have just edited my post for clarity. Here, mapping $s:P→P$ satisfies Peano’s axioms. You said "$f^n(bar a)=a_n$" is defining $a_n$, not $f^n$, so i wonder how $f^n$ is defined properly?
– Le Anh Dung
Jan 23 at 2:36
I thought the paragraph was proving Theorem 1.8.2 $Rightarrow$ Theorem 1.8.1, but it's hard to tell without more context. Does the book provide a separate proof of Theorem 1.8.2?
– stewbasic
Jan 23 at 2:48
@stewbasic, I have a screenshot of the full proof at math.stackexchange.com/questions/2613220/…, but there are some typos. Please be careful about them!
– Le Anh Dung
Jan 23 at 2:52
I assume this paragraph follows the proof in your screenshot? Then probably I am wrong, the paragraph is meant to show Theorem 1.8.1 $Rightarrow$ Theorem 1.8.2. In that case we'd need to apply Theorem 1.8.1 to each $ain A$. That is, define $f^n=(a(0),a(n))mid a:mathbf Z^+to A,,acirc s=fcirc a$ and use Theorem 1.8.1 to show $f^n$ is a function.
– stewbasic
Jan 23 at 3:04
Could you clarify what $P$ is? Re 2, I think $f^n$ is supposed to be provided by Theorem 1.8.2, and "$f^n(bar a)=a_n$" is defining $a_n$, not $f^n$.
– stewbasic
Jan 23 at 2:30
Could you clarify what $P$ is? Re 2, I think $f^n$ is supposed to be provided by Theorem 1.8.2, and "$f^n(bar a)=a_n$" is defining $a_n$, not $f^n$.
– stewbasic
Jan 23 at 2:30
@stewbasic, I have just edited my post for clarity. Here, mapping $s:P→P$ satisfies Peano’s axioms. You said "$f^n(bar a)=a_n$" is defining $a_n$, not $f^n$, so i wonder how $f^n$ is defined properly?
– Le Anh Dung
Jan 23 at 2:36
@stewbasic, I have just edited my post for clarity. Here, mapping $s:P→P$ satisfies Peano’s axioms. You said "$f^n(bar a)=a_n$" is defining $a_n$, not $f^n$, so i wonder how $f^n$ is defined properly?
– Le Anh Dung
Jan 23 at 2:36
I thought the paragraph was proving Theorem 1.8.2 $Rightarrow$ Theorem 1.8.1, but it's hard to tell without more context. Does the book provide a separate proof of Theorem 1.8.2?
– stewbasic
Jan 23 at 2:48
I thought the paragraph was proving Theorem 1.8.2 $Rightarrow$ Theorem 1.8.1, but it's hard to tell without more context. Does the book provide a separate proof of Theorem 1.8.2?
– stewbasic
Jan 23 at 2:48
@stewbasic, I have a screenshot of the full proof at math.stackexchange.com/questions/2613220/…, but there are some typos. Please be careful about them!
– Le Anh Dung
Jan 23 at 2:52
@stewbasic, I have a screenshot of the full proof at math.stackexchange.com/questions/2613220/…, but there are some typos. Please be careful about them!
– Le Anh Dung
Jan 23 at 2:52
I assume this paragraph follows the proof in your screenshot? Then probably I am wrong, the paragraph is meant to show Theorem 1.8.1 $Rightarrow$ Theorem 1.8.2. In that case we'd need to apply Theorem 1.8.1 to each $ain A$. That is, define $f^n=(a(0),a(n))mid a:mathbf Z^+to A,,acirc s=fcirc a$ and use Theorem 1.8.1 to show $f^n$ is a function.
– stewbasic
Jan 23 at 3:04
I assume this paragraph follows the proof in your screenshot? Then probably I am wrong, the paragraph is meant to show Theorem 1.8.1 $Rightarrow$ Theorem 1.8.2. In that case we'd need to apply Theorem 1.8.1 to each $ain A$. That is, define $f^n=(a(0),a(n))mid a:mathbf Z^+to A,,acirc s=fcirc a$ and use Theorem 1.8.1 to show $f^n$ is a function.
– stewbasic
Jan 23 at 3:04
 |Â
show 1 more comment
1 Answer
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Intuitively, to construct $f^n$ we must specify $f^n(bar a)$ for each $bar ain A$. So for each $bar ain A$, apply Theorem 1.8.1 to obtain a sequence $a$ (this depends on $bar a$), and let $f^n(bar a)=a(n)$.
More formally we could define $f^n$ as
$$
pin Atimes A:textthere exists g:mathbf Z^+to Atext with gcirc s=fcirc g,,g(0)=p_1,,g(n)=p_2.
$$
I claim that for all $ain A$, $|f^ncap(atimes A)|=1$. Indeed fix $ain A$. By Theorem 1.8.1, there exists $g:mathbf Z^+to A$ with $gcirc s=fcirc g$ and $g(0)=a$. Then $(a,g(n))in f^n$. Conversely if $pin f^ncap(atimes A)$ then there exists $g':mathbf Z^+to A$ with $g'circ s=fcirc g$, $g'(0)=p_1=a$ and $g'(n)=p_2$. By the uniqueness in Theorem 1.8.1, $g'=g$, so $p_2=g(n)$. Thus $f^ncap(atimes A)=(a,g(n))$. This shows $f^n$ is a function.
answered Jan 23 at 3:27
stewbasic
5,6481826
By my understanding of your answer, I get: 1. Mapping $f^n$ depends on $ain A$. This means for each $ain A$, there exists a unique mapping $f^n$. 2. For each $ain A$, $f^n=(a,p)in Atimes A:exists g:mathbf Z^+to Atext s.t gcirc s=fcirc g,,p=g(n).$ Is it right?
– Le Anh Dung
Jan 25 at 17:49
I don't know what $|f^ncap(atimes A)|=1$ means? Please elaborate!
– Le Anh Dung
Jan 25 at 18:02
No, $f^n$ does not depend on $a$. $f^n(a)$ depends on $a$. I defined $f^n$ as a subset of $Atimes A$, so $f^ncap(atimes A)$ is also a subset of $Atimes A$, and I prove it has one element. In other words, for each $ain A$ there is a unique $bin A$ such that $(a,b)in f^n$; this ensures $f^n$ is a function.
– stewbasic
Jan 26 at 0:46
I'm now clear. Many thanks for you!
– Le Anh Dung
Jan 26 at 9:20
add a comment |Â
1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Intuitively, to construct $f^n$ we must specify $f^n(bar a)$ for each $bar ain A$. So for each $bar ain A$, apply Theorem 1.8.1 to obtain a sequence $a$ (this depends on $bar a$), and let $f^n(bar a)=a(n)$.
More formally we could define $f^n$ as
$$
pin Atimes A:textthere exists g:mathbf Z^+to Atext with gcirc s=fcirc g,,g(0)=p_1,,g(n)=p_2.
$$
I claim that for all $ain A$, $|f^ncap(atimes A)|=1$. Indeed fix $ain A$. By Theorem 1.8.1, there exists $g:mathbf Z^+to A$ with $gcirc s=fcirc g$ and $g(0)=a$. Then $(a,g(n))in f^n$. Conversely if $pin f^ncap(atimes A)$ then there exists $g':mathbf Z^+to A$ with $g'circ s=fcirc g$, $g'(0)=p_1=a$ and $g'(n)=p_2$. By the uniqueness in Theorem 1.8.1, $g'=g$, so $p_2=g(n)$. Thus $f^ncap(atimes A)=(a,g(n))$. This shows $f^n$ is a function.
answered Jan 23 at 3:27
stewbasic
5,6481826
By my understanding of your answer, I get: 1. Mapping $f^n$ depends on $ain A$. This means for each $ain A$, there exists a unique mapping $f^n$. 2. For each $ain A$, $f^n=(a,p)in Atimes A:exists g:mathbf Z^+to Atext s.t gcirc s=fcirc g,,p=g(n).$ Is it right?
– Le Anh Dung
Jan 25 at 17:49
I don't know what $|f^ncap(atimes A)|=1$ means? Please elaborate!
– Le Anh Dung
Jan 25 at 18:02
No, $f^n$ does not depend on $a$. $f^n(a)$ depends on $a$. I defined $f^n$ as a subset of $Atimes A$, so $f^ncap(atimes A)$ is also a subset of $Atimes A$, and I prove it has one element. In other words, for each $ain A$ there is a unique $bin A$ such that $(a,b)in f^n$; this ensures $f^n$ is a function.
– stewbasic
Jan 26 at 0:46
I'm now clear. Many thanks for you!
– Le Anh Dung
Jan 26 at 9:20
add a comment |Â
up vote
1
down vote
accepted
Intuitively, to construct $f^n$ we must specify $f^n(bar a)$ for each $bar ain A$. So for each $bar ain A$, apply Theorem 1.8.1 to obtain a sequence $a$ (this depends on $bar a$), and let $f^n(bar a)=a(n)$.
More formally we could define $f^n$ as
$$
pin Atimes A:textthere exists g:mathbf Z^+to Atext with gcirc s=fcirc g,,g(0)=p_1,,g(n)=p_2.
$$
I claim that for all $ain A$, $|f^ncap(atimes A)|=1$. Indeed fix $ain A$. By Theorem 1.8.1, there exists $g:mathbf Z^+to A$ with $gcirc s=fcirc g$ and $g(0)=a$. Then $(a,g(n))in f^n$. Conversely if $pin f^ncap(atimes A)$ then there exists $g':mathbf Z^+to A$ with $g'circ s=fcirc g$, $g'(0)=p_1=a$ and $g'(n)=p_2$. By the uniqueness in Theorem 1.8.1, $g'=g$, so $p_2=g(n)$. Thus $f^ncap(atimes A)=(a,g(n))$. This shows $f^n$ is a function.
answered Jan 23 at 3:27
stewbasic
5,6481826
By my understanding of your answer, I get: 1. Mapping $f^n$ depends on $ain A$. This means for each $ain A$, there exists a unique mapping $f^n$. 2. For each $ain A$, $f^n=(a,p)in Atimes A:exists g:mathbf Z^+to Atext s.t gcirc s=fcirc g,,p=g(n).$ Is it right?
– Le Anh Dung
Jan 25 at 17:49
I don't know what $|f^ncap(atimes A)|=1$ means? Please elaborate!
– Le Anh Dung
Jan 25 at 18:02
No, $f^n$ does not depend on $a$. $f^n(a)$ depends on $a$. I defined $f^n$ as a subset of $Atimes A$, so $f^ncap(atimes A)$ is also a subset of $Atimes A$, and I prove it has one element. In other words, for each $ain A$ there is a unique $bin A$ such that $(a,b)in f^n$; this ensures $f^n$ is a function.
– stewbasic
Jan 26 at 0:46
I'm now clear. Many thanks for you!
– Le Anh Dung
Jan 26 at 9:20
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Intuitively, to construct $f^n$ we must specify $f^n(bar a)$ for each $bar ain A$. So for each $bar ain A$, apply Theorem 1.8.1 to obtain a sequence $a$ (this depends on $bar a$), and let $f^n(bar a)=a(n)$.
More formally we could define $f^n$ as
$$
pin Atimes A:textthere exists g:mathbf Z^+to Atext with gcirc s=fcirc g,,g(0)=p_1,,g(n)=p_2.
$$
I claim that for all $ain A$, $|f^ncap(atimes A)|=1$. Indeed fix $ain A$. By Theorem 1.8.1, there exists $g:mathbf Z^+to A$ with $gcirc s=fcirc g$ and $g(0)=a$. Then $(a,g(n))in f^n$. Conversely if $pin f^ncap(atimes A)$ then there exists $g':mathbf Z^+to A$ with $g'circ s=fcirc g$, $g'(0)=p_1=a$ and $g'(n)=p_2$. By the uniqueness in Theorem 1.8.1, $g'=g$, so $p_2=g(n)$. Thus $f^ncap(atimes A)=(a,g(n))$. This shows $f^n$ is a function.
answered Jan 23 at 3:27
stewbasic
5,6481826
Intuitively, to construct $f^n$ we must specify $f^n(bar a)$ for each $bar ain A$. So for each $bar ain A$, apply Theorem 1.8.1 to obtain a sequence $a$ (this depends on $bar a$), and let $f^n(bar a)=a(n)$.
More formally we could define $f^n$ as
$$
pin Atimes A:textthere exists g:mathbf Z^+to Atext with gcirc s=fcirc g,,g(0)=p_1,,g(n)=p_2.
$$
I claim that for all $ain A$, $|f^ncap(atimes A)|=1$. Indeed fix $ain A$. By Theorem 1.8.1, there exists $g:mathbf Z^+to A$ with $gcirc s=fcirc g$ and $g(0)=a$. Then $(a,g(n))in f^n$. Conversely if $pin f^ncap(atimes A)$ then there exists $g':mathbf Z^+to A$ with $g'circ s=fcirc g$, $g'(0)=p_1=a$ and $g'(n)=p_2$. By the uniqueness in Theorem 1.8.1, $g'=g$, so $p_2=g(n)$. Thus $f^ncap(atimes A)=(a,g(n))$. This shows $f^n$ is a function.
answered Jan 23 at 3:27
stewbasic
5,6481826
answered Jan 23 at 3:27
stewbasic
5,6481826
answered Jan 23 at 3:27
stewbasic
5,6481826
answered Jan 23 at 3:27
stewbasic
5,6481826
5,6481826
By my understanding of your answer, I get: 1. Mapping $f^n$ depends on $ain A$. This means for each $ain A$, there exists a unique mapping $f^n$. 2. For each $ain A$, $f^n=(a,p)in Atimes A:exists g:mathbf Z^+to Atext s.t gcirc s=fcirc g,,p=g(n).$ Is it right?
– Le Anh Dung
Jan 25 at 17:49
I don't know what $|f^ncap(atimes A)|=1$ means? Please elaborate!
– Le Anh Dung
Jan 25 at 18:02
No, $f^n$ does not depend on $a$. $f^n(a)$ depends on $a$. I defined $f^n$ as a subset of $Atimes A$, so $f^ncap(atimes A)$ is also a subset of $Atimes A$, and I prove it has one element. In other words, for each $ain A$ there is a unique $bin A$ such that $(a,b)in f^n$; this ensures $f^n$ is a function.
– stewbasic
Jan 26 at 0:46
I'm now clear. Many thanks for you!
– Le Anh Dung
Jan 26 at 9:20
add a comment |Â
By my understanding of your answer, I get: 1. Mapping $f^n$ depends on $ain A$. This means for each $ain A$, there exists a unique mapping $f^n$. 2. For each $ain A$, $f^n=(a,p)in Atimes A:exists g:mathbf Z^+to Atext s.t gcirc s=fcirc g,,p=g(n).$ Is it right?
– Le Anh Dung
Jan 25 at 17:49
I don't know what $|f^ncap(atimes A)|=1$ means? Please elaborate!
– Le Anh Dung
Jan 25 at 18:02
No, $f^n$ does not depend on $a$. $f^n(a)$ depends on $a$. I defined $f^n$ as a subset of $Atimes A$, so $f^ncap(atimes A)$ is also a subset of $Atimes A$, and I prove it has one element. In other words, for each $ain A$ there is a unique $bin A$ such that $(a,b)in f^n$; this ensures $f^n$ is a function.
– stewbasic
Jan 26 at 0:46
I'm now clear. Many thanks for you!
– Le Anh Dung
Jan 26 at 9:20
By my understanding of your answer, I get: 1. Mapping $f^n$ depends on $ain A$. This means for each $ain A$, there exists a unique mapping $f^n$. 2. For each $ain A$, $f^n=(a,p)in Atimes A:exists g:mathbf Z^+to Atext s.t gcirc s=fcirc g,,p=g(n).$ Is it right?
– Le Anh Dung
Jan 25 at 17:49
By my understanding of your answer, I get: 1. Mapping $f^n$ depends on $ain A$. This means for each $ain A$, there exists a unique mapping $f^n$. 2. For each $ain A$, $f^n=(a,p)in Atimes A:exists g:mathbf Z^+to Atext s.t gcirc s=fcirc g,,p=g(n).$ Is it right?
– Le Anh Dung
Jan 25 at 17:49
I don't know what $|f^ncap(atimes A)|=1$ means? Please elaborate!
– Le Anh Dung
Jan 25 at 18:02
I don't know what $|f^ncap(atimes A)|=1$ means? Please elaborate!
– Le Anh Dung
Jan 25 at 18:02
No, $f^n$ does not depend on $a$. $f^n(a)$ depends on $a$. I defined $f^n$ as a subset of $Atimes A$, so $f^ncap(atimes A)$ is also a subset of $Atimes A$, and I prove it has one element. In other words, for each $ain A$ there is a unique $bin A$ such that $(a,b)in f^n$; this ensures $f^n$ is a function.
– stewbasic
Jan 26 at 0:46
No, $f^n$ does not depend on $a$. $f^n(a)$ depends on $a$. I defined $f^n$ as a subset of $Atimes A$, so $f^ncap(atimes A)$ is also a subset of $Atimes A$, and I prove it has one element. In other words, for each $ain A$ there is a unique $bin A$ such that $(a,b)in f^n$; this ensures $f^n$ is a function.
– stewbasic
Jan 26 at 0:46
I'm now clear. Many thanks for you!
– Le Anh Dung
Jan 26 at 9:20
I'm now clear. Many thanks for you!
– Le Anh Dung
Jan 26 at 9:20
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Could you clarify what $P$ is? Re 2, I think $f^n$ is supposed to be provided by Theorem 1.8.2, and "$f^n(bar a)=a_n$" is defining $a_n$, not $f^n$.
– stewbasic
Jan 23 at 2:30
@stewbasic, I have just edited my post for clarity. Here, mapping $s:P→P$ satisfies Peano’s axioms. You said "$f^n(bar a)=a_n$" is defining $a_n$, not $f^n$, so i wonder how $f^n$ is defined properly?
– Le Anh Dung
Jan 23 at 2:36
I thought the paragraph was proving Theorem 1.8.2 $Rightarrow$ Theorem 1.8.1, but it's hard to tell without more context. Does the book provide a separate proof of Theorem 1.8.2?
– stewbasic
Jan 23 at 2:48
@stewbasic, I have a screenshot of the full proof at math.stackexchange.com/questions/2613220/…, but there are some typos. Please be careful about them!
– Le Anh Dung
Jan 23 at 2:52
I assume this paragraph follows the proof in your screenshot? Then probably I am wrong, the paragraph is meant to show Theorem 1.8.1 $Rightarrow$ Theorem 1.8.2. In that case we'd need to apply Theorem 1.8.1 to each $ain A$. That is, define $f^n=(a(0),a(n))mid a:mathbf Z^+to A,,acirc s=fcirc a$ and use Theorem 1.8.1 to show $f^n$ is a function.
– stewbasic
Jan 23 at 3:04