Mixing
Relation $(a_n)0$, $a_n-b_n<epsilon$ for all $n$ large enough, defines a total order $<$ on Cauchy...
$begingroup$
Honestly, I figured out the idea to prove this theorem without much difficulty. But I found very hard to formalize it into a rigorous proof and it takes me three days to do it.
Could you please have a check on my attempt? Thank you so much!
A sequence $langle a_n rangle_{n=0}^{infty}$ of rational numbers is a Cauchy sequence if for every rational $epsilon>0$ there exists an integer $N$ such that $|a_m - a_n| < epsilon$ for all $m,n > N$. Let $mathfrak{C}$ be the set of all Cauchy sequences of rational numbers. We define a relation $preccurlyeq$ on $mathfrak{C}$ by $langle a_n rangle_{n=0}^{infty} preccurlyeq langle b_n rangle_{n=0}^{infty}$ if and only if for every rational $epsilon>0$ there exists an integer $N$ such that $a_n - b_n < epsilon$ for all $n > N$.
Theorem: Any two elements of $mathfrak{C}$ are comparable w.r.t $preccurlyeq$.
My attempt:
Assume the contrary that there exist $langle a_n rangle_{n=0}^{infty}$, $langle b_n rangle_{n=0}^{infty} in mathfrak{C}$ that are incomparable. Then $langle a_n rangle_{n=0}^{infty} not preccurlyeq langle b_n rangle_{n=0}^{infty}$ and $langle b_n rangle_{n=0}^{infty} not preccurlyeq langle a_n rangle_{n=0}^{infty}$. As a result, $exists epsilon >0,forall N,exists n>N:a_n - b_n ge epsilon$ and $exists epsilon >0,forall N,exists m>N:b_m - a_m ge epsilon$.
Then $exists epsilon >0,forall N,exists (n>N wedge m>N):a_n - b_n ge epsilon wedge b_m - a_m ge epsilon$. By Axiom of Choice, there exist functions $f:Bbb N to Bbb N$ and $h:Bbb N to Bbb N$ such that $f(N)>N$, $h(N)>N$, $a_{f(N)} - b_{f(N)} ge epsilon$, and $b_{h(N)} - a_{h(N)} ge epsilon$.
Moreover, $langle b_n rangle_{n=0}^{infty}$ is a Cauchy sequence $implies$ $exists N_0,forall m,n > N_0:|b_m - b_n| < epsilon$.
We define functions $f':Bbb N to Bbb N$ and $h':Bbb N to Bbb N$ by $f'(N)=f(N_0)$, $h'(N)=h(N_0)$ for all $N<N_0$ and $f'(N)=f(N)$, $h'(N)=h(N)$ for all $N ge N_0$.
It follows that $f'(N)>N$, $h'(N)>N$, $a_{f'(N)} - b_{f'(N)} ge epsilon$, $b_{h'(N)} - a_{h'(N)} ge epsilon$, and $|b_{h'(N)} - b_{f'(N)}| < epsilon$ for all $N in Bbb N$. As a result,
$begin{align}|a_{f’(N)} - a_{h’(N)}| &= |(a_{f’(N)} - b_{f’(N)}) + (b_{h’(N)} - a_{h’(N)} ) + (b_{f’(N)} - b_{h’(N)})|\ &ge |(a_{f’(N)} - b_{f’(N)}) + (b_{h’(N)} - a_{h’(N)} )| - |b_{f’(N)} - b_{h’(N)}|\ &= |a_{f’(N)} - b_{f’(N)}| + |b_{h’(N)} - a_{h’(N)} | - |b_{f’(N)} - b_{h’(N)}|\ &> epsilon + epsilon - epsilon =epsilonend{align}$
Hence $exists epsilon>0, forall N, exists(f’(N)>N wedge h’(N)>N):a_{f’(N)} - a_{h’(N)}>epsilon$. This contradicts the fact that $langle a_n rangle_{n=0}^{infty}$ is a Cauchy sequence.
real-analysis proof-verification cauchy-sequences
edited Jan 30 at 10:34
Did
249k23227466
asked Jan 30 at 10:22
Le Anh DungLe Anh Dung
1,4511621
$endgroup$
add a comment |
$begingroup$
Honestly, I figured out the idea to prove this theorem without much difficulty. But I found very hard to formalize it into a rigorous proof and it takes me three days to do it.
Could you please have a check on my attempt? Thank you so much!
A sequence $langle a_n rangle_{n=0}^{infty}$ of rational numbers is a Cauchy sequence if for every rational $epsilon>0$ there exists an integer $N$ such that $|a_m - a_n| < epsilon$ for all $m,n > N$. Let $mathfrak{C}$ be the set of all Cauchy sequences of rational numbers. We define a relation $preccurlyeq$ on $mathfrak{C}$ by $langle a_n rangle_{n=0}^{infty} preccurlyeq langle b_n rangle_{n=0}^{infty}$ if and only if for every rational $epsilon>0$ there exists an integer $N$ such that $a_n - b_n < epsilon$ for all $n > N$.
Theorem: Any two elements of $mathfrak{C}$ are comparable w.r.t $preccurlyeq$.
My attempt:
Assume the contrary that there exist $langle a_n rangle_{n=0}^{infty}$, $langle b_n rangle_{n=0}^{infty} in mathfrak{C}$ that are incomparable. Then $langle a_n rangle_{n=0}^{infty} not preccurlyeq langle b_n rangle_{n=0}^{infty}$ and $langle b_n rangle_{n=0}^{infty} not preccurlyeq langle a_n rangle_{n=0}^{infty}$. As a result, $exists epsilon >0,forall N,exists n>N:a_n - b_n ge epsilon$ and $exists epsilon >0,forall N,exists m>N:b_m - a_m ge epsilon$.
Then $exists epsilon >0,forall N,exists (n>N wedge m>N):a_n - b_n ge epsilon wedge b_m - a_m ge epsilon$. By Axiom of Choice, there exist functions $f:Bbb N to Bbb N$ and $h:Bbb N to Bbb N$ such that $f(N)>N$, $h(N)>N$, $a_{f(N)} - b_{f(N)} ge epsilon$, and $b_{h(N)} - a_{h(N)} ge epsilon$.
Moreover, $langle b_n rangle_{n=0}^{infty}$ is a Cauchy sequence $implies$ $exists N_0,forall m,n > N_0:|b_m - b_n| < epsilon$.
We define functions $f':Bbb N to Bbb N$ and $h':Bbb N to Bbb N$ by $f'(N)=f(N_0)$, $h'(N)=h(N_0)$ for all $N<N_0$ and $f'(N)=f(N)$, $h'(N)=h(N)$ for all $N ge N_0$.
It follows that $f'(N)>N$, $h'(N)>N$, $a_{f'(N)} - b_{f'(N)} ge epsilon$, $b_{h'(N)} - a_{h'(N)} ge epsilon$, and $|b_{h'(N)} - b_{f'(N)}| < epsilon$ for all $N in Bbb N$. As a result,
$begin{align}|a_{f’(N)} - a_{h’(N)}| &= |(a_{f’(N)} - b_{f’(N)}) + (b_{h’(N)} - a_{h’(N)} ) + (b_{f’(N)} - b_{h’(N)})|\ &ge |(a_{f’(N)} - b_{f’(N)}) + (b_{h’(N)} - a_{h’(N)} )| - |b_{f’(N)} - b_{h’(N)}|\ &= |a_{f’(N)} - b_{f’(N)}| + |b_{h’(N)} - a_{h’(N)} | - |b_{f’(N)} - b_{h’(N)}|\ &> epsilon + epsilon - epsilon =epsilonend{align}$
Hence $exists epsilon>0, forall N, exists(f’(N)>N wedge h’(N)>N):a_{f’(N)} - a_{h’(N)}>epsilon$. This contradicts the fact that $langle a_n rangle_{n=0}^{infty}$ is a Cauchy sequence.
real-analysis proof-verification cauchy-sequences
edited Jan 30 at 10:34
Did
249k23227466
asked Jan 30 at 10:22
Le Anh DungLe Anh Dung
1,4511621
$endgroup$
add a comment |
$begingroup$
Honestly, I figured out the idea to prove this theorem without much difficulty. But I found very hard to formalize it into a rigorous proof and it takes me three days to do it.
Could you please have a check on my attempt? Thank you so much!
A sequence $langle a_n rangle_{n=0}^{infty}$ of rational numbers is a Cauchy sequence if for every rational $epsilon>0$ there exists an integer $N$ such that $|a_m - a_n| < epsilon$ for all $m,n > N$. Let $mathfrak{C}$ be the set of all Cauchy sequences of rational numbers. We define a relation $preccurlyeq$ on $mathfrak{C}$ by $langle a_n rangle_{n=0}^{infty} preccurlyeq langle b_n rangle_{n=0}^{infty}$ if and only if for every rational $epsilon>0$ there exists an integer $N$ such that $a_n - b_n < epsilon$ for all $n > N$.
Theorem: Any two elements of $mathfrak{C}$ are comparable w.r.t $preccurlyeq$.
My attempt:
Assume the contrary that there exist $langle a_n rangle_{n=0}^{infty}$, $langle b_n rangle_{n=0}^{infty} in mathfrak{C}$ that are incomparable. Then $langle a_n rangle_{n=0}^{infty} not preccurlyeq langle b_n rangle_{n=0}^{infty}$ and $langle b_n rangle_{n=0}^{infty} not preccurlyeq langle a_n rangle_{n=0}^{infty}$. As a result, $exists epsilon >0,forall N,exists n>N:a_n - b_n ge epsilon$ and $exists epsilon >0,forall N,exists m>N:b_m - a_m ge epsilon$.
Then $exists epsilon >0,forall N,exists (n>N wedge m>N):a_n - b_n ge epsilon wedge b_m - a_m ge epsilon$. By Axiom of Choice, there exist functions $f:Bbb N to Bbb N$ and $h:Bbb N to Bbb N$ such that $f(N)>N$, $h(N)>N$, $a_{f(N)} - b_{f(N)} ge epsilon$, and $b_{h(N)} - a_{h(N)} ge epsilon$.
Moreover, $langle b_n rangle_{n=0}^{infty}$ is a Cauchy sequence $implies$ $exists N_0,forall m,n > N_0:|b_m - b_n| < epsilon$.
We define functions $f':Bbb N to Bbb N$ and $h':Bbb N to Bbb N$ by $f'(N)=f(N_0)$, $h'(N)=h(N_0)$ for all $N<N_0$ and $f'(N)=f(N)$, $h'(N)=h(N)$ for all $N ge N_0$.
It follows that $f'(N)>N$, $h'(N)>N$, $a_{f'(N)} - b_{f'(N)} ge epsilon$, $b_{h'(N)} - a_{h'(N)} ge epsilon$, and $|b_{h'(N)} - b_{f'(N)}| < epsilon$ for all $N in Bbb N$. As a result,
$begin{align}|a_{f’(N)} - a_{h’(N)}| &= |(a_{f’(N)} - b_{f’(N)}) + (b_{h’(N)} - a_{h’(N)} ) + (b_{f’(N)} - b_{h’(N)})|\ &ge |(a_{f’(N)} - b_{f’(N)}) + (b_{h’(N)} - a_{h’(N)} )| - |b_{f’(N)} - b_{h’(N)}|\ &= |a_{f’(N)} - b_{f’(N)}| + |b_{h’(N)} - a_{h’(N)} | - |b_{f’(N)} - b_{h’(N)}|\ &> epsilon + epsilon - epsilon =epsilonend{align}$
Hence $exists epsilon>0, forall N, exists(f’(N)>N wedge h’(N)>N):a_{f’(N)} - a_{h’(N)}>epsilon$. This contradicts the fact that $langle a_n rangle_{n=0}^{infty}$ is a Cauchy sequence.
real-analysis proof-verification cauchy-sequences
edited Jan 30 at 10:34
Did
249k23227466
asked Jan 30 at 10:22
Le Anh DungLe Anh Dung
1,4511621
$endgroup$
Honestly, I figured out the idea to prove this theorem without much difficulty. But I found very hard to formalize it into a rigorous proof and it takes me three days to do it.
Could you please have a check on my attempt? Thank you so much!
A sequence $langle a_n rangle_{n=0}^{infty}$ of rational numbers is a Cauchy sequence if for every rational $epsilon>0$ there exists an integer $N$ such that $|a_m - a_n| < epsilon$ for all $m,n > N$. Let $mathfrak{C}$ be the set of all Cauchy sequences of rational numbers. We define a relation $preccurlyeq$ on $mathfrak{C}$ by $langle a_n rangle_{n=0}^{infty} preccurlyeq langle b_n rangle_{n=0}^{infty}$ if and only if for every rational $epsilon>0$ there exists an integer $N$ such that $a_n - b_n < epsilon$ for all $n > N$.
Theorem: Any two elements of $mathfrak{C}$ are comparable w.r.t $preccurlyeq$.
My attempt:
Assume the contrary that there exist $langle a_n rangle_{n=0}^{infty}$, $langle b_n rangle_{n=0}^{infty} in mathfrak{C}$ that are incomparable. Then $langle a_n rangle_{n=0}^{infty} not preccurlyeq langle b_n rangle_{n=0}^{infty}$ and $langle b_n rangle_{n=0}^{infty} not preccurlyeq langle a_n rangle_{n=0}^{infty}$. As a result, $exists epsilon >0,forall N,exists n>N:a_n - b_n ge epsilon$ and $exists epsilon >0,forall N,exists m>N:b_m - a_m ge epsilon$.
Then $exists epsilon >0,forall N,exists (n>N wedge m>N):a_n - b_n ge epsilon wedge b_m - a_m ge epsilon$. By Axiom of Choice, there exist functions $f:Bbb N to Bbb N$ and $h:Bbb N to Bbb N$ such that $f(N)>N$, $h(N)>N$, $a_{f(N)} - b_{f(N)} ge epsilon$, and $b_{h(N)} - a_{h(N)} ge epsilon$.
Moreover, $langle b_n rangle_{n=0}^{infty}$ is a Cauchy sequence $implies$ $exists N_0,forall m,n > N_0:|b_m - b_n| < epsilon$.
We define functions $f':Bbb N to Bbb N$ and $h':Bbb N to Bbb N$ by $f'(N)=f(N_0)$, $h'(N)=h(N_0)$ for all $N<N_0$ and $f'(N)=f(N)$, $h'(N)=h(N)$ for all $N ge N_0$.
It follows that $f'(N)>N$, $h'(N)>N$, $a_{f'(N)} - b_{f'(N)} ge epsilon$, $b_{h'(N)} - a_{h'(N)} ge epsilon$, and $|b_{h'(N)} - b_{f'(N)}| < epsilon$ for all $N in Bbb N$. As a result,
$begin{align}|a_{f’(N)} - a_{h’(N)}| &= |(a_{f’(N)} - b_{f’(N)}) + (b_{h’(N)} - a_{h’(N)} ) + (b_{f’(N)} - b_{h’(N)})|\ &ge |(a_{f’(N)} - b_{f’(N)}) + (b_{h’(N)} - a_{h’(N)} )| - |b_{f’(N)} - b_{h’(N)}|\ &= |a_{f’(N)} - b_{f’(N)}| + |b_{h’(N)} - a_{h’(N)} | - |b_{f’(N)} - b_{h’(N)}|\ &> epsilon + epsilon - epsilon =epsilonend{align}$
Hence $exists epsilon>0, forall N, exists(f’(N)>N wedge h’(N)>N):a_{f’(N)} - a_{h’(N)}>epsilon$. This contradicts the fact that $langle a_n rangle_{n=0}^{infty}$ is a Cauchy sequence.
real-analysis proof-verification cauchy-sequences
real-analysis proof-verification cauchy-sequences
edited Jan 30 at 10:34
Did
249k23227466
asked Jan 30 at 10:22
Le Anh DungLe Anh Dung
1,4511621
edited Jan 30 at 10:34
Did
249k23227466
asked Jan 30 at 10:22
Le Anh DungLe Anh Dung
1,4511621
edited Jan 30 at 10:34
Did
249k23227466
edited Jan 30 at 10:34
Did
249k23227466
edited Jan 30 at 10:34
Did
249k23227466
249k23227466
asked Jan 30 at 10:22
Le Anh DungLe Anh Dung
1,4511621
asked Jan 30 at 10:22
Le Anh DungLe Anh Dung
1,4511621
asked Jan 30 at 10:22
Le Anh DungLe Anh Dung
1,4511621
1,4511621
add a comment |
add a comment |
1 Answer
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I see only a (minor) problem with your attempt: at the first paragraph, those $varepsilon$'s should be two distinct numbers a priori. But that's not a serious problem.
On the other hand, the axiom of choice is not needed for the proof, but using it is not an error.
answered Jan 30 at 10:50
José Carlos SantosJosé Carlos Santos
172k22132239
$endgroup$
$begingroup$
Thank you so much for your verification! Please confirm if my understanding is correct. 1. Although @Did edited my title into ... a total order, I think $preccurlyeq$ is NOT an order relation. This is because $preccurlyeq$ is NOT antisymmetric, i.e there exist $langle a_n rangle_{n=0}^{infty} neq langle b_n rangle_{n=0}^{infty}$ such that $langle a_n rangle_{n=0}^{infty} preccurlyeq langle b_n rangle_{n=0}^{infty}$ and $langle b_n rangle_{n=0}^{infty} preccurlyeq langle a_n rangle_{n=0}^{infty}$.[...]
$endgroup$
– Le Anh Dung
Jan 30 at 11:03
$begingroup$
[...] 2. To avoid the use of Axiom of Choice, we can define $f:Bbb N to Bbb N$ by $f(N)=min {ninBbb N mid n>N wedge a_n - b_n ge epsilon }$ for all $NinBbb N$.
$endgroup$
– Le Anh Dung
Jan 30 at 11:03
$begingroup$
Right: it is not an order relation. While thinking about your attempt, I only cared about the specific claim that you were trying to prove.
$endgroup$
– José Carlos Santos
Jan 30 at 11:04
$begingroup$
Please have a check on my Claim 2. too!
$endgroup$
– Le Anh Dung
Jan 30 at 11:05
$begingroup$
It looks correct to me.
$endgroup$
– José Carlos Santos
Jan 30 at 11:28
|
show 1 more comment
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1 Answer
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1 Answer
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$begingroup$
I see only a (minor) problem with your attempt: at the first paragraph, those $varepsilon$'s should be two distinct numbers a priori. But that's not a serious problem.
On the other hand, the axiom of choice is not needed for the proof, but using it is not an error.
answered Jan 30 at 10:50
José Carlos SantosJosé Carlos Santos
172k22132239
$endgroup$
$begingroup$
Thank you so much for your verification! Please confirm if my understanding is correct. 1. Although @Did edited my title into ... a total order, I think $preccurlyeq$ is NOT an order relation. This is because $preccurlyeq$ is NOT antisymmetric, i.e there exist $langle a_n rangle_{n=0}^{infty} neq langle b_n rangle_{n=0}^{infty}$ such that $langle a_n rangle_{n=0}^{infty} preccurlyeq langle b_n rangle_{n=0}^{infty}$ and $langle b_n rangle_{n=0}^{infty} preccurlyeq langle a_n rangle_{n=0}^{infty}$.[...]
$endgroup$
– Le Anh Dung
Jan 30 at 11:03
$begingroup$
[...] 2. To avoid the use of Axiom of Choice, we can define $f:Bbb N to Bbb N$ by $f(N)=min {ninBbb N mid n>N wedge a_n - b_n ge epsilon }$ for all $NinBbb N$.
$endgroup$
– Le Anh Dung
Jan 30 at 11:03
$begingroup$
Right: it is not an order relation. While thinking about your attempt, I only cared about the specific claim that you were trying to prove.
$endgroup$
– José Carlos Santos
Jan 30 at 11:04
$begingroup$
Please have a check on my Claim 2. too!
$endgroup$
– Le Anh Dung
Jan 30 at 11:05
$begingroup$
It looks correct to me.
$endgroup$
– José Carlos Santos
Jan 30 at 11:28
|
show 1 more comment
$begingroup$
I see only a (minor) problem with your attempt: at the first paragraph, those $varepsilon$'s should be two distinct numbers a priori. But that's not a serious problem.
On the other hand, the axiom of choice is not needed for the proof, but using it is not an error.
answered Jan 30 at 10:50
José Carlos SantosJosé Carlos Santos
172k22132239
$endgroup$
$begingroup$
Thank you so much for your verification! Please confirm if my understanding is correct. 1. Although @Did edited my title into ... a total order, I think $preccurlyeq$ is NOT an order relation. This is because $preccurlyeq$ is NOT antisymmetric, i.e there exist $langle a_n rangle_{n=0}^{infty} neq langle b_n rangle_{n=0}^{infty}$ such that $langle a_n rangle_{n=0}^{infty} preccurlyeq langle b_n rangle_{n=0}^{infty}$ and $langle b_n rangle_{n=0}^{infty} preccurlyeq langle a_n rangle_{n=0}^{infty}$.[...]
$endgroup$
– Le Anh Dung
Jan 30 at 11:03
$begingroup$
[...] 2. To avoid the use of Axiom of Choice, we can define $f:Bbb N to Bbb N$ by $f(N)=min {ninBbb N mid n>N wedge a_n - b_n ge epsilon }$ for all $NinBbb N$.
$endgroup$
– Le Anh Dung
Jan 30 at 11:03
$begingroup$
Right: it is not an order relation. While thinking about your attempt, I only cared about the specific claim that you were trying to prove.
$endgroup$
– José Carlos Santos
Jan 30 at 11:04
$begingroup$
Please have a check on my Claim 2. too!
$endgroup$
– Le Anh Dung
Jan 30 at 11:05
$begingroup$
It looks correct to me.
$endgroup$
– José Carlos Santos
Jan 30 at 11:28
|
show 1 more comment
$begingroup$
I see only a (minor) problem with your attempt: at the first paragraph, those $varepsilon$'s should be two distinct numbers a priori. But that's not a serious problem.
On the other hand, the axiom of choice is not needed for the proof, but using it is not an error.
answered Jan 30 at 10:50
José Carlos SantosJosé Carlos Santos
172k22132239
$endgroup$
I see only a (minor) problem with your attempt: at the first paragraph, those $varepsilon$'s should be two distinct numbers a priori. But that's not a serious problem.
On the other hand, the axiom of choice is not needed for the proof, but using it is not an error.
answered Jan 30 at 10:50
José Carlos SantosJosé Carlos Santos
172k22132239
answered Jan 30 at 10:50
José Carlos SantosJosé Carlos Santos
172k22132239
answered Jan 30 at 10:50
José Carlos SantosJosé Carlos Santos
172k22132239
answered Jan 30 at 10:50
José Carlos SantosJosé Carlos Santos
172k22132239
172k22132239
$begingroup$
Thank you so much for your verification! Please confirm if my understanding is correct. 1. Although @Did edited my title into ... a total order, I think $preccurlyeq$ is NOT an order relation. This is because $preccurlyeq$ is NOT antisymmetric, i.e there exist $langle a_n rangle_{n=0}^{infty} neq langle b_n rangle_{n=0}^{infty}$ such that $langle a_n rangle_{n=0}^{infty} preccurlyeq langle b_n rangle_{n=0}^{infty}$ and $langle b_n rangle_{n=0}^{infty} preccurlyeq langle a_n rangle_{n=0}^{infty}$.[...]
$endgroup$
– Le Anh Dung
Jan 30 at 11:03
$begingroup$
[...] 2. To avoid the use of Axiom of Choice, we can define $f:Bbb N to Bbb N$ by $f(N)=min {ninBbb N mid n>N wedge a_n - b_n ge epsilon }$ for all $NinBbb N$.
$endgroup$
– Le Anh Dung
Jan 30 at 11:03
$begingroup$
Right: it is not an order relation. While thinking about your attempt, I only cared about the specific claim that you were trying to prove.
$endgroup$
– José Carlos Santos
Jan 30 at 11:04
$begingroup$
Please have a check on my Claim 2. too!
$endgroup$
– Le Anh Dung
Jan 30 at 11:05
$begingroup$
It looks correct to me.
$endgroup$
– José Carlos Santos
Jan 30 at 11:28
|
show 1 more comment
$begingroup$
Thank you so much for your verification! Please confirm if my understanding is correct. 1. Although @Did edited my title into ... a total order, I think $preccurlyeq$ is NOT an order relation. This is because $preccurlyeq$ is NOT antisymmetric, i.e there exist $langle a_n rangle_{n=0}^{infty} neq langle b_n rangle_{n=0}^{infty}$ such that $langle a_n rangle_{n=0}^{infty} preccurlyeq langle b_n rangle_{n=0}^{infty}$ and $langle b_n rangle_{n=0}^{infty} preccurlyeq langle a_n rangle_{n=0}^{infty}$.[...]
$endgroup$
– Le Anh Dung
Jan 30 at 11:03
$begingroup$
[...] 2. To avoid the use of Axiom of Choice, we can define $f:Bbb N to Bbb N$ by $f(N)=min {ninBbb N mid n>N wedge a_n - b_n ge epsilon }$ for all $NinBbb N$.
$endgroup$
– Le Anh Dung
Jan 30 at 11:03
$begingroup$
Right: it is not an order relation. While thinking about your attempt, I only cared about the specific claim that you were trying to prove.
$endgroup$
– José Carlos Santos
Jan 30 at 11:04
$begingroup$
Please have a check on my Claim 2. too!
$endgroup$
– Le Anh Dung
Jan 30 at 11:05
$begingroup$
It looks correct to me.
$endgroup$
– José Carlos Santos
Jan 30 at 11:28
$begingroup$
Thank you so much for your verification! Please confirm if my understanding is correct. 1. Although @Did edited my title into ... a total order, I think $preccurlyeq$ is NOT an order relation. This is because $preccurlyeq$ is NOT antisymmetric, i.e there exist $langle a_n rangle_{n=0}^{infty} neq langle b_n rangle_{n=0}^{infty}$ such that $langle a_n rangle_{n=0}^{infty} preccurlyeq langle b_n rangle_{n=0}^{infty}$ and $langle b_n rangle_{n=0}^{infty} preccurlyeq langle a_n rangle_{n=0}^{infty}$.[...]
$endgroup$
– Le Anh Dung
Jan 30 at 11:03
$begingroup$
Thank you so much for your verification! Please confirm if my understanding is correct. 1. Although @Did edited my title into ... a total order, I think $preccurlyeq$ is NOT an order relation. This is because $preccurlyeq$ is NOT antisymmetric, i.e there exist $langle a_n rangle_{n=0}^{infty} neq langle b_n rangle_{n=0}^{infty}$ such that $langle a_n rangle_{n=0}^{infty} preccurlyeq langle b_n rangle_{n=0}^{infty}$ and $langle b_n rangle_{n=0}^{infty} preccurlyeq langle a_n rangle_{n=0}^{infty}$.[...]
$endgroup$
– Le Anh Dung
Jan 30 at 11:03
$begingroup$
[...] 2. To avoid the use of Axiom of Choice, we can define $f:Bbb N to Bbb N$ by $f(N)=min {ninBbb N mid n>N wedge a_n - b_n ge epsilon }$ for all $NinBbb N$.
$endgroup$
– Le Anh Dung
Jan 30 at 11:03
$begingroup$
[...] 2. To avoid the use of Axiom of Choice, we can define $f:Bbb N to Bbb N$ by $f(N)=min {ninBbb N mid n>N wedge a_n - b_n ge epsilon }$ for all $NinBbb N$.
$endgroup$
– Le Anh Dung
Jan 30 at 11:03
$begingroup$
Right: it is not an order relation. While thinking about your attempt, I only cared about the specific claim that you were trying to prove.
$endgroup$
– José Carlos Santos
Jan 30 at 11:04
$begingroup$
Right: it is not an order relation. While thinking about your attempt, I only cared about the specific claim that you were trying to prove.
$endgroup$
– José Carlos Santos
Jan 30 at 11:04
$begingroup$
Please have a check on my Claim 2. too!
$endgroup$
– Le Anh Dung
Jan 30 at 11:05
$begingroup$
Please have a check on my Claim 2. too!
$endgroup$
– Le Anh Dung
Jan 30 at 11:05
$begingroup$
It looks correct to me.
$endgroup$
– José Carlos Santos
Jan 30 at 11:28
$begingroup$
It looks correct to me.
$endgroup$
– José Carlos Santos
Jan 30 at 11:28
|
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