Mixing
Let $(a_n)$ and $(b_n)$ be Cauchy sequences of rationals. Then $(a_nb_n)$ is Cauchy sequence
$begingroup$
Let $(a_n)$ and $(b_n)$ be Cauchy sequences of rationals. Then $(a_nb_n)$ is a Cauchy sequence.
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
Lemma: If $(a_n)$ is a Cauchy sequence of rationals, then for all $n in Bbb N$, $|a_n| < A$ for some $A in Bbb Q$.
By lemma, there exists $A$ such that $|a_n| < A$ and $|b_n| < A$ for all $n in Bbb N$.
For a given $epsilon >0$, take an integer $N$ such that $|b_n-b_m|<dfrac{epsilon}{2A}$ and $|a_n-a_m|<dfrac{epsilon}{2A}$ for all $n>N$.
$begin{align}
|a_nb_n-a_mb_m|
&=|a_n(b_n-b_m) + b_m(a_n-a_m)|\
&le |a_n(b_n-b_m)| + |b_m(a_n-a_m)|\
&= |a_n||b_n-b_m| + |b_m||a_n-a_m|\
&< Adfrac{epsilon}{2A}+ Adfrac{epsilon}{2A}\
&=epsilon
end{align}$
Hence $(a_nb_n)$ is a Cauchy sequence.
real-analysis proof-verification cauchy-sequences
edited Jan 30 at 16:02
Adrian Keister
5,26971933
asked Jan 30 at 15:59
Le Anh DungLe Anh Dung
1,4511621
$endgroup$
add a comment |
$begingroup$
Let $(a_n)$ and $(b_n)$ be Cauchy sequences of rationals. Then $(a_nb_n)$ is a Cauchy sequence.
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
Lemma: If $(a_n)$ is a Cauchy sequence of rationals, then for all $n in Bbb N$, $|a_n| < A$ for some $A in Bbb Q$.
By lemma, there exists $A$ such that $|a_n| < A$ and $|b_n| < A$ for all $n in Bbb N$.
For a given $epsilon >0$, take an integer $N$ such that $|b_n-b_m|<dfrac{epsilon}{2A}$ and $|a_n-a_m|<dfrac{epsilon}{2A}$ for all $n>N$.
$begin{align}
|a_nb_n-a_mb_m|
&=|a_n(b_n-b_m) + b_m(a_n-a_m)|\
&le |a_n(b_n-b_m)| + |b_m(a_n-a_m)|\
&= |a_n||b_n-b_m| + |b_m||a_n-a_m|\
&< Adfrac{epsilon}{2A}+ Adfrac{epsilon}{2A}\
&=epsilon
end{align}$
Hence $(a_nb_n)$ is a Cauchy sequence.
real-analysis proof-verification cauchy-sequences
edited Jan 30 at 16:02
Adrian Keister
5,26971933
asked Jan 30 at 15:59
Le Anh DungLe Anh Dung
1,4511621
$endgroup$
5
$begingroup$
It looks fine to me.
$endgroup$
– José Carlos Santos
Jan 30 at 16:00
$begingroup$
By the way: The proof works exactly the same for complex sequences. It is also irrelevant that the number $A$ is rational.
$endgroup$
– Mars Plastic
Jan 30 at 16:06
add a comment |
$begingroup$
Let $(a_n)$ and $(b_n)$ be Cauchy sequences of rationals. Then $(a_nb_n)$ is a Cauchy sequence.
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
Lemma: If $(a_n)$ is a Cauchy sequence of rationals, then for all $n in Bbb N$, $|a_n| < A$ for some $A in Bbb Q$.
By lemma, there exists $A$ such that $|a_n| < A$ and $|b_n| < A$ for all $n in Bbb N$.
For a given $epsilon >0$, take an integer $N$ such that $|b_n-b_m|<dfrac{epsilon}{2A}$ and $|a_n-a_m|<dfrac{epsilon}{2A}$ for all $n>N$.
$begin{align}
|a_nb_n-a_mb_m|
&=|a_n(b_n-b_m) + b_m(a_n-a_m)|\
&le |a_n(b_n-b_m)| + |b_m(a_n-a_m)|\
&= |a_n||b_n-b_m| + |b_m||a_n-a_m|\
&< Adfrac{epsilon}{2A}+ Adfrac{epsilon}{2A}\
&=epsilon
end{align}$
Hence $(a_nb_n)$ is a Cauchy sequence.
real-analysis proof-verification cauchy-sequences
edited Jan 30 at 16:02
Adrian Keister
5,26971933
asked Jan 30 at 15:59
Le Anh DungLe Anh Dung
1,4511621
$endgroup$
Let $(a_n)$ and $(b_n)$ be Cauchy sequences of rationals. Then $(a_nb_n)$ is a Cauchy sequence.
Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!
My attempt:
Lemma: If $(a_n)$ is a Cauchy sequence of rationals, then for all $n in Bbb N$, $|a_n| < A$ for some $A in Bbb Q$.
By lemma, there exists $A$ such that $|a_n| < A$ and $|b_n| < A$ for all $n in Bbb N$.
For a given $epsilon >0$, take an integer $N$ such that $|b_n-b_m|<dfrac{epsilon}{2A}$ and $|a_n-a_m|<dfrac{epsilon}{2A}$ for all $n>N$.
$begin{align}
|a_nb_n-a_mb_m|
&=|a_n(b_n-b_m) + b_m(a_n-a_m)|\
&le |a_n(b_n-b_m)| + |b_m(a_n-a_m)|\
&= |a_n||b_n-b_m| + |b_m||a_n-a_m|\
&< Adfrac{epsilon}{2A}+ Adfrac{epsilon}{2A}\
&=epsilon
end{align}$
Hence $(a_nb_n)$ is a Cauchy sequence.
real-analysis proof-verification cauchy-sequences
real-analysis proof-verification cauchy-sequences
edited Jan 30 at 16:02
Adrian Keister
5,26971933
asked Jan 30 at 15:59
Le Anh DungLe Anh Dung
1,4511621
edited Jan 30 at 16:02
Adrian Keister
5,26971933
asked Jan 30 at 15:59
Le Anh DungLe Anh Dung
1,4511621
edited Jan 30 at 16:02
Adrian Keister
5,26971933
edited Jan 30 at 16:02
Adrian Keister
5,26971933
edited Jan 30 at 16:02
Adrian Keister
5,26971933
5,26971933
asked Jan 30 at 15:59
Le Anh DungLe Anh Dung
1,4511621
asked Jan 30 at 15:59
Le Anh DungLe Anh Dung
1,4511621
asked Jan 30 at 15:59
Le Anh DungLe Anh Dung
1,4511621
1,4511621
5
$begingroup$
It looks fine to me.
$endgroup$
– José Carlos Santos
Jan 30 at 16:00
$begingroup$
By the way: The proof works exactly the same for complex sequences. It is also irrelevant that the number $A$ is rational.
$endgroup$
– Mars Plastic
Jan 30 at 16:06
add a comment |
5
$begingroup$
It looks fine to me.
$endgroup$
– José Carlos Santos
Jan 30 at 16:00
$begingroup$
By the way: The proof works exactly the same for complex sequences. It is also irrelevant that the number $A$ is rational.
$endgroup$
– Mars Plastic
Jan 30 at 16:06
5
5
$begingroup$
It looks fine to me.
$endgroup$
– José Carlos Santos
Jan 30 at 16:00
$begingroup$
It looks fine to me.
$endgroup$
– José Carlos Santos
Jan 30 at 16:00
$begingroup$
By the way: The proof works exactly the same for complex sequences. It is also irrelevant that the number $A$ is rational.
$endgroup$
– Mars Plastic
Jan 30 at 16:06
$begingroup$
By the way: The proof works exactly the same for complex sequences. It is also irrelevant that the number $A$ is rational.
$endgroup$
– Mars Plastic
Jan 30 at 16:06
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This looks good. Well done.
Just because there is always room for improvement, I have a tiny nit-pick. When you take $N$, you could mention explicitly that it is guaranteed to exist because $a_n$ and $b_n$ are both Cauchy.
answered Jan 30 at 16:05
ArthurArthur
122k7122210
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
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$begingroup$
This looks good. Well done.
Just because there is always room for improvement, I have a tiny nit-pick. When you take $N$, you could mention explicitly that it is guaranteed to exist because $a_n$ and $b_n$ are both Cauchy.
answered Jan 30 at 16:05
ArthurArthur
122k7122210
$endgroup$
add a comment |
$begingroup$
This looks good. Well done.
Just because there is always room for improvement, I have a tiny nit-pick. When you take $N$, you could mention explicitly that it is guaranteed to exist because $a_n$ and $b_n$ are both Cauchy.
answered Jan 30 at 16:05
ArthurArthur
122k7122210
$endgroup$
add a comment |
$begingroup$
This looks good. Well done.
Just because there is always room for improvement, I have a tiny nit-pick. When you take $N$, you could mention explicitly that it is guaranteed to exist because $a_n$ and $b_n$ are both Cauchy.
answered Jan 30 at 16:05
ArthurArthur
122k7122210
$endgroup$
This looks good. Well done.
Just because there is always room for improvement, I have a tiny nit-pick. When you take $N$, you could mention explicitly that it is guaranteed to exist because $a_n$ and $b_n$ are both Cauchy.
answered Jan 30 at 16:05
ArthurArthur
122k7122210
answered Jan 30 at 16:05
ArthurArthur
122k7122210
answered Jan 30 at 16:05
ArthurArthur
122k7122210
answered Jan 30 at 16:05
ArthurArthur
122k7122210
122k7122210
add a comment |
add a comment |
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5
$begingroup$
It looks fine to me.
$endgroup$
– José Carlos Santos
Jan 30 at 16:00
$begingroup$
By the way: The proof works exactly the same for complex sequences. It is also irrelevant that the number $A$ is rational.
$endgroup$
– Mars Plastic
Jan 30 at 16:06