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Prove that $(a_n) preccurlyeq_1 (b_n) iff (a_n) preccurlyeq_2 (b_n)$ or $(a_n) approx (b_n)$ for Cauchy...
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Does my attempt look fine or contain logical flaws/gaps?
To prove $langle a_n rangle preccurlyeq_1 langle b_n rangle Longrightarrow langle a_n rangle preccurlyeq_2 langle b_n rangle$ or $langle a_n rangle approx langle b_n rangle$, I use contrapositive. I would like to ask a direct and more concise proof.
Any suggestion is greatly appreciated. Thank you for your help!
Let $mathcal{C}$ be the set of Cauchy sequences of rationals and $d_n=a_n-b_n$. We define relations $preccurlyeq_1$, $preccurlyeq_2$, and $approx$ as follows:
$$begin{align}langle a_n rangle preccurlyeq_1 langle b_n rangle &iff forall epsilon >0, exists N, forall n>N: d_n < epsilon\ langle a_n rangle preccurlyeq_2 langle b_n rangle &iff exists N, forall n>N: d_n le 0\ langle a_n rangle approx langle b_n rangle &iff forall epsilon >0, exists N, forall n>N: |d_n| < epsilon end{align}$$
Theorem: $langle a_n rangle preccurlyeq_1 langle b_n rangle iff langle a_n rangle preccurlyeq_2 langle b_n rangle$ or $langle a_n rangle approx langle b_n rangle$
My attempt:
$langle a_n rangle preccurlyeq_1 langle b_n rangle Longleftarrow langle a_n rangle preccurlyeq_2 langle b_n rangle$ or $langle a_n rangle approx langle b_n rangle$
$langle a_n rangle preccurlyeq_2 langle b_n rangle implies exists N, forall n>N: d_n le 0 implies$ $forall epsilon >0, exists N, forall n>N: d_n < epsilon implies$ $langle a_n rangle preccurlyeq_1 langle b_n rangle$.
$langle a_n rangle approx langle b_n rangle implies forall epsilon >0, exists N, forall n>N: |d_n| < epsilon implies forall epsilon >0, exists N, forall n>N:$ $-epsilon < d_n < epsilon implies forall epsilon >0, exists N, forall n>N: d_n < epsilon implies langle a_n rangle preccurlyeq_1 langle b_n rangle$.
$langle a_n rangle preccurlyeq_1 langle b_n rangle Longrightarrow langle a_n rangle preccurlyeq_2 langle b_n rangle$ or $langle a_n rangle approx langle b_n rangle$
Assume that $langle a_n rangle not preccurlyeq_2 langle b_n rangle$ and $langle a_n rangle not approx langle b_n rangle$. It follows that
$(1) space forall N, exists n>N: d_n > 0$.
$(2) space exists epsilon>0, forall N, exists n>N: |d_n| ge epsilon$.
We then define mappings $f:Bbb N to Bbb N$ by $$f(N)= min {n in Bbb N mid n> N wedge d_n > 0}$$ and $h:Bbb N to Bbb N$ by $$h(N)= min left {m in Bbb N ,middlevert, forall n > m: |a_n - a_{f(N)}| < dfrac{d_{f(N)}}{2} wedge |b_n - b_{f(N)}| < dfrac{d_{f(N)}}{2} right }$$
Such $f(N)$ exists by $(1)$. Such $h(N)$ exists by the fact that $langle a_n rangle$ and $langle b_n rangle$ are Cauchy sequences and that $d_{f(N)}>0$. It follows that $forall n > h(N): d_n > 0$. Moreover, $exists n > h(N): |d_n| ge epsilon$ by $(2)$. As a result, $exists n > h(N):$ $d_n ge epsilon$.
To sum up, $exists epsilon >0, forall N, exists n>N:d_n ge epsilon$. As a result, $langle a_n rangle not preccurlyeq_1 langle b_n rangle$. Hence $langle a_n rangle not preccurlyeq_2 langle b_n rangle$ and $langle a_n rangle not approx langle b_n rangle$ $implies$ $langle a_n rangle not preccurlyeq_1 langle b_n rangle$. Thus $langle a_n rangle preccurlyeq_1 langle b_n rangle Longrightarrow langle a_n rangle preccurlyeq_2 langle b_n rangle$ or $langle a_n rangle approx langle b_n rangle$.
real-analysis proof-verification alternative-proof cauchy-sequences
asked Jan 31 at 10:57
Le Anh DungLe Anh Dung
1,4511621
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add a comment |
$begingroup$
Does my attempt look fine or contain logical flaws/gaps?
To prove $langle a_n rangle preccurlyeq_1 langle b_n rangle Longrightarrow langle a_n rangle preccurlyeq_2 langle b_n rangle$ or $langle a_n rangle approx langle b_n rangle$, I use contrapositive. I would like to ask a direct and more concise proof.
Any suggestion is greatly appreciated. Thank you for your help!
Let $mathcal{C}$ be the set of Cauchy sequences of rationals and $d_n=a_n-b_n$. We define relations $preccurlyeq_1$, $preccurlyeq_2$, and $approx$ as follows:
$$begin{align}langle a_n rangle preccurlyeq_1 langle b_n rangle &iff forall epsilon >0, exists N, forall n>N: d_n < epsilon\ langle a_n rangle preccurlyeq_2 langle b_n rangle &iff exists N, forall n>N: d_n le 0\ langle a_n rangle approx langle b_n rangle &iff forall epsilon >0, exists N, forall n>N: |d_n| < epsilon end{align}$$
Theorem: $langle a_n rangle preccurlyeq_1 langle b_n rangle iff langle a_n rangle preccurlyeq_2 langle b_n rangle$ or $langle a_n rangle approx langle b_n rangle$
My attempt:
$langle a_n rangle preccurlyeq_1 langle b_n rangle Longleftarrow langle a_n rangle preccurlyeq_2 langle b_n rangle$ or $langle a_n rangle approx langle b_n rangle$
$langle a_n rangle preccurlyeq_2 langle b_n rangle implies exists N, forall n>N: d_n le 0 implies$ $forall epsilon >0, exists N, forall n>N: d_n < epsilon implies$ $langle a_n rangle preccurlyeq_1 langle b_n rangle$.
$langle a_n rangle approx langle b_n rangle implies forall epsilon >0, exists N, forall n>N: |d_n| < epsilon implies forall epsilon >0, exists N, forall n>N:$ $-epsilon < d_n < epsilon implies forall epsilon >0, exists N, forall n>N: d_n < epsilon implies langle a_n rangle preccurlyeq_1 langle b_n rangle$.
$langle a_n rangle preccurlyeq_1 langle b_n rangle Longrightarrow langle a_n rangle preccurlyeq_2 langle b_n rangle$ or $langle a_n rangle approx langle b_n rangle$
Assume that $langle a_n rangle not preccurlyeq_2 langle b_n rangle$ and $langle a_n rangle not approx langle b_n rangle$. It follows that
$(1) space forall N, exists n>N: d_n > 0$.
$(2) space exists epsilon>0, forall N, exists n>N: |d_n| ge epsilon$.
We then define mappings $f:Bbb N to Bbb N$ by $$f(N)= min {n in Bbb N mid n> N wedge d_n > 0}$$ and $h:Bbb N to Bbb N$ by $$h(N)= min left {m in Bbb N ,middlevert, forall n > m: |a_n - a_{f(N)}| < dfrac{d_{f(N)}}{2} wedge |b_n - b_{f(N)}| < dfrac{d_{f(N)}}{2} right }$$
Such $f(N)$ exists by $(1)$. Such $h(N)$ exists by the fact that $langle a_n rangle$ and $langle b_n rangle$ are Cauchy sequences and that $d_{f(N)}>0$. It follows that $forall n > h(N): d_n > 0$. Moreover, $exists n > h(N): |d_n| ge epsilon$ by $(2)$. As a result, $exists n > h(N):$ $d_n ge epsilon$.
To sum up, $exists epsilon >0, forall N, exists n>N:d_n ge epsilon$. As a result, $langle a_n rangle not preccurlyeq_1 langle b_n rangle$. Hence $langle a_n rangle not preccurlyeq_2 langle b_n rangle$ and $langle a_n rangle not approx langle b_n rangle$ $implies$ $langle a_n rangle not preccurlyeq_1 langle b_n rangle$. Thus $langle a_n rangle preccurlyeq_1 langle b_n rangle Longrightarrow langle a_n rangle preccurlyeq_2 langle b_n rangle$ or $langle a_n rangle approx langle b_n rangle$.
real-analysis proof-verification alternative-proof cauchy-sequences
asked Jan 31 at 10:57
Le Anh DungLe Anh Dung
1,4511621
$endgroup$
add a comment |
$begingroup$
Does my attempt look fine or contain logical flaws/gaps?
To prove $langle a_n rangle preccurlyeq_1 langle b_n rangle Longrightarrow langle a_n rangle preccurlyeq_2 langle b_n rangle$ or $langle a_n rangle approx langle b_n rangle$, I use contrapositive. I would like to ask a direct and more concise proof.
Any suggestion is greatly appreciated. Thank you for your help!
Let $mathcal{C}$ be the set of Cauchy sequences of rationals and $d_n=a_n-b_n$. We define relations $preccurlyeq_1$, $preccurlyeq_2$, and $approx$ as follows:
$$begin{align}langle a_n rangle preccurlyeq_1 langle b_n rangle &iff forall epsilon >0, exists N, forall n>N: d_n < epsilon\ langle a_n rangle preccurlyeq_2 langle b_n rangle &iff exists N, forall n>N: d_n le 0\ langle a_n rangle approx langle b_n rangle &iff forall epsilon >0, exists N, forall n>N: |d_n| < epsilon end{align}$$
Theorem: $langle a_n rangle preccurlyeq_1 langle b_n rangle iff langle a_n rangle preccurlyeq_2 langle b_n rangle$ or $langle a_n rangle approx langle b_n rangle$
My attempt:
$langle a_n rangle preccurlyeq_1 langle b_n rangle Longleftarrow langle a_n rangle preccurlyeq_2 langle b_n rangle$ or $langle a_n rangle approx langle b_n rangle$
$langle a_n rangle preccurlyeq_2 langle b_n rangle implies exists N, forall n>N: d_n le 0 implies$ $forall epsilon >0, exists N, forall n>N: d_n < epsilon implies$ $langle a_n rangle preccurlyeq_1 langle b_n rangle$.
$langle a_n rangle approx langle b_n rangle implies forall epsilon >0, exists N, forall n>N: |d_n| < epsilon implies forall epsilon >0, exists N, forall n>N:$ $-epsilon < d_n < epsilon implies forall epsilon >0, exists N, forall n>N: d_n < epsilon implies langle a_n rangle preccurlyeq_1 langle b_n rangle$.
$langle a_n rangle preccurlyeq_1 langle b_n rangle Longrightarrow langle a_n rangle preccurlyeq_2 langle b_n rangle$ or $langle a_n rangle approx langle b_n rangle$
Assume that $langle a_n rangle not preccurlyeq_2 langle b_n rangle$ and $langle a_n rangle not approx langle b_n rangle$. It follows that
$(1) space forall N, exists n>N: d_n > 0$.
$(2) space exists epsilon>0, forall N, exists n>N: |d_n| ge epsilon$.
We then define mappings $f:Bbb N to Bbb N$ by $$f(N)= min {n in Bbb N mid n> N wedge d_n > 0}$$ and $h:Bbb N to Bbb N$ by $$h(N)= min left {m in Bbb N ,middlevert, forall n > m: |a_n - a_{f(N)}| < dfrac{d_{f(N)}}{2} wedge |b_n - b_{f(N)}| < dfrac{d_{f(N)}}{2} right }$$
Such $f(N)$ exists by $(1)$. Such $h(N)$ exists by the fact that $langle a_n rangle$ and $langle b_n rangle$ are Cauchy sequences and that $d_{f(N)}>0$. It follows that $forall n > h(N): d_n > 0$. Moreover, $exists n > h(N): |d_n| ge epsilon$ by $(2)$. As a result, $exists n > h(N):$ $d_n ge epsilon$.
To sum up, $exists epsilon >0, forall N, exists n>N:d_n ge epsilon$. As a result, $langle a_n rangle not preccurlyeq_1 langle b_n rangle$. Hence $langle a_n rangle not preccurlyeq_2 langle b_n rangle$ and $langle a_n rangle not approx langle b_n rangle$ $implies$ $langle a_n rangle not preccurlyeq_1 langle b_n rangle$. Thus $langle a_n rangle preccurlyeq_1 langle b_n rangle Longrightarrow langle a_n rangle preccurlyeq_2 langle b_n rangle$ or $langle a_n rangle approx langle b_n rangle$.
real-analysis proof-verification alternative-proof cauchy-sequences
asked Jan 31 at 10:57
Le Anh DungLe Anh Dung
1,4511621
$endgroup$
Does my attempt look fine or contain logical flaws/gaps?
To prove $langle a_n rangle preccurlyeq_1 langle b_n rangle Longrightarrow langle a_n rangle preccurlyeq_2 langle b_n rangle$ or $langle a_n rangle approx langle b_n rangle$, I use contrapositive. I would like to ask a direct and more concise proof.
Any suggestion is greatly appreciated. Thank you for your help!
Let $mathcal{C}$ be the set of Cauchy sequences of rationals and $d_n=a_n-b_n$. We define relations $preccurlyeq_1$, $preccurlyeq_2$, and $approx$ as follows:
$$begin{align}langle a_n rangle preccurlyeq_1 langle b_n rangle &iff forall epsilon >0, exists N, forall n>N: d_n < epsilon\ langle a_n rangle preccurlyeq_2 langle b_n rangle &iff exists N, forall n>N: d_n le 0\ langle a_n rangle approx langle b_n rangle &iff forall epsilon >0, exists N, forall n>N: |d_n| < epsilon end{align}$$
Theorem: $langle a_n rangle preccurlyeq_1 langle b_n rangle iff langle a_n rangle preccurlyeq_2 langle b_n rangle$ or $langle a_n rangle approx langle b_n rangle$
My attempt:
$langle a_n rangle preccurlyeq_1 langle b_n rangle Longleftarrow langle a_n rangle preccurlyeq_2 langle b_n rangle$ or $langle a_n rangle approx langle b_n rangle$
$langle a_n rangle preccurlyeq_2 langle b_n rangle implies exists N, forall n>N: d_n le 0 implies$ $forall epsilon >0, exists N, forall n>N: d_n < epsilon implies$ $langle a_n rangle preccurlyeq_1 langle b_n rangle$.
$langle a_n rangle approx langle b_n rangle implies forall epsilon >0, exists N, forall n>N: |d_n| < epsilon implies forall epsilon >0, exists N, forall n>N:$ $-epsilon < d_n < epsilon implies forall epsilon >0, exists N, forall n>N: d_n < epsilon implies langle a_n rangle preccurlyeq_1 langle b_n rangle$.
$langle a_n rangle preccurlyeq_1 langle b_n rangle Longrightarrow langle a_n rangle preccurlyeq_2 langle b_n rangle$ or $langle a_n rangle approx langle b_n rangle$
Assume that $langle a_n rangle not preccurlyeq_2 langle b_n rangle$ and $langle a_n rangle not approx langle b_n rangle$. It follows that
$(1) space forall N, exists n>N: d_n > 0$.
$(2) space exists epsilon>0, forall N, exists n>N: |d_n| ge epsilon$.
We then define mappings $f:Bbb N to Bbb N$ by $$f(N)= min {n in Bbb N mid n> N wedge d_n > 0}$$ and $h:Bbb N to Bbb N$ by $$h(N)= min left {m in Bbb N ,middlevert, forall n > m: |a_n - a_{f(N)}| < dfrac{d_{f(N)}}{2} wedge |b_n - b_{f(N)}| < dfrac{d_{f(N)}}{2} right }$$
Such $f(N)$ exists by $(1)$. Such $h(N)$ exists by the fact that $langle a_n rangle$ and $langle b_n rangle$ are Cauchy sequences and that $d_{f(N)}>0$. It follows that $forall n > h(N): d_n > 0$. Moreover, $exists n > h(N): |d_n| ge epsilon$ by $(2)$. As a result, $exists n > h(N):$ $d_n ge epsilon$.
To sum up, $exists epsilon >0, forall N, exists n>N:d_n ge epsilon$. As a result, $langle a_n rangle not preccurlyeq_1 langle b_n rangle$. Hence $langle a_n rangle not preccurlyeq_2 langle b_n rangle$ and $langle a_n rangle not approx langle b_n rangle$ $implies$ $langle a_n rangle not preccurlyeq_1 langle b_n rangle$. Thus $langle a_n rangle preccurlyeq_1 langle b_n rangle Longrightarrow langle a_n rangle preccurlyeq_2 langle b_n rangle$ or $langle a_n rangle approx langle b_n rangle$.
real-analysis proof-verification alternative-proof cauchy-sequences
real-analysis proof-verification alternative-proof cauchy-sequences
asked Jan 31 at 10:57
Le Anh DungLe Anh Dung
1,4511621
asked Jan 31 at 10:57
Le Anh DungLe Anh Dung
1,4511621
edited Jan 31 at 11:08
Le Anh Dung
asked Jan 31 at 10:57
Le Anh DungLe Anh Dung
1,4511621
asked Jan 31 at 10:57
Le Anh DungLe Anh Dung
1,4511621
asked Jan 31 at 10:57
Le Anh DungLe Anh Dung
1,4511621
1,4511621
add a comment |
add a comment |
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