Problem in understanding solution to problem 4a, chapter 22 of Spivak Calculus
In chapter 22 of Spivak calculus, problem 4a is
Prove that if a subsequence of a Cauchy sequence converges to $l$, then the sequence itself converges to $l$.
This is the solution in Combined answer book
Since $a_{n_k}$ converges to $l$. There exits $Jinmathbb{N}$ such that
$$
k>JRightarrow |a_{n_k}-l|<frac{epsilon}{2}
$$
Since $a_n$ is Cauchy. There exists $N_1inmathbb{N}$ such that
$$
n,m>N_1Rightarrow|a_n-a_m|<frac{epsilon}{2}
$$
Let $N=max(N_1,n_J)$. If $n>N$, then
$$
|a_n-a_{n_{j+1}}|<frac{epsilon}{2}
$$
and
$$
|a_{n_{j+1}}-l|<frac{epsilon}{2}
$$
Consequently $|a_n-l|<epsilon$.
I do not understand, how when $n>N$, then
$$
|a_n-a_{n_{j+1}}|<frac{epsilon}{2}
$$
and
$$
|a_{n_{j+1}}-l|<frac{epsilon}{2}
$$
Could someone please explain it to me?
calculus real-analysis sequences-and-series
add a comment |
In chapter 22 of Spivak calculus, problem 4a is
Prove that if a subsequence of a Cauchy sequence converges to $l$, then the sequence itself converges to $l$.
This is the solution in Combined answer book
Since $a_{n_k}$ converges to $l$. There exits $Jinmathbb{N}$ such that
$$
k>JRightarrow |a_{n_k}-l|<frac{epsilon}{2}
$$
Since $a_n$ is Cauchy. There exists $N_1inmathbb{N}$ such that
$$
n,m>N_1Rightarrow|a_n-a_m|<frac{epsilon}{2}
$$
Let $N=max(N_1,n_J)$. If $n>N$, then
$$
|a_n-a_{n_{j+1}}|<frac{epsilon}{2}
$$
and
$$
|a_{n_{j+1}}-l|<frac{epsilon}{2}
$$
Consequently $|a_n-l|<epsilon$.
I do not understand, how when $n>N$, then
$$
|a_n-a_{n_{j+1}}|<frac{epsilon}{2}
$$
and
$$
|a_{n_{j+1}}-l|<frac{epsilon}{2}
$$
Could someone please explain it to me?
calculus real-analysis sequences-and-series
add a comment |
In chapter 22 of Spivak calculus, problem 4a is
Prove that if a subsequence of a Cauchy sequence converges to $l$, then the sequence itself converges to $l$.
This is the solution in Combined answer book
Since $a_{n_k}$ converges to $l$. There exits $Jinmathbb{N}$ such that
$$
k>JRightarrow |a_{n_k}-l|<frac{epsilon}{2}
$$
Since $a_n$ is Cauchy. There exists $N_1inmathbb{N}$ such that
$$
n,m>N_1Rightarrow|a_n-a_m|<frac{epsilon}{2}
$$
Let $N=max(N_1,n_J)$. If $n>N$, then
$$
|a_n-a_{n_{j+1}}|<frac{epsilon}{2}
$$
and
$$
|a_{n_{j+1}}-l|<frac{epsilon}{2}
$$
Consequently $|a_n-l|<epsilon$.
I do not understand, how when $n>N$, then
$$
|a_n-a_{n_{j+1}}|<frac{epsilon}{2}
$$
and
$$
|a_{n_{j+1}}-l|<frac{epsilon}{2}
$$
Could someone please explain it to me?
calculus real-analysis sequences-and-series
In chapter 22 of Spivak calculus, problem 4a is
Prove that if a subsequence of a Cauchy sequence converges to $l$, then the sequence itself converges to $l$.
This is the solution in Combined answer book
Since $a_{n_k}$ converges to $l$. There exits $Jinmathbb{N}$ such that
$$
k>JRightarrow |a_{n_k}-l|<frac{epsilon}{2}
$$
Since $a_n$ is Cauchy. There exists $N_1inmathbb{N}$ such that
$$
n,m>N_1Rightarrow|a_n-a_m|<frac{epsilon}{2}
$$
Let $N=max(N_1,n_J)$. If $n>N$, then
$$
|a_n-a_{n_{j+1}}|<frac{epsilon}{2}
$$
and
$$
|a_{n_{j+1}}-l|<frac{epsilon}{2}
$$
Consequently $|a_n-l|<epsilon$.
I do not understand, how when $n>N$, then
$$
|a_n-a_{n_{j+1}}|<frac{epsilon}{2}
$$
and
$$
|a_{n_{j+1}}-l|<frac{epsilon}{2}
$$
Could someone please explain it to me?
calculus real-analysis sequences-and-series
calculus real-analysis sequences-and-series
asked Apr 7 '17 at 10:09
user214302
add a comment |
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2 Answers
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The proof seems to be incorrect.
In order to deduce
begin{align*}
|a_n-a_{n_{J+1}}|<frac{epsilon}{2}tag{1}
end{align*}
we have to assure that $n>N_1$ and $n_{J+1}>N_1$.
Since $n>N=max(N_1,n_J)geq N_1$ we have $n>N_1$. But we don't know if $n_{J+1}>N_1$.
We know $n_{J+1}>n_J$ which is sufficient to show
begin{align*}
|a_{n_{J+1}}-l|<frac{epsilon}{2}
end{align*}
but not sufficient to show (1).
answered Apr 7 '17 at 20:07
Markus Scheuer
60k455143
add a comment |
Proof goes as follows:
Let $a_{n}$ be a Cauchy sequence.
Suppose that $a_{j_{n}}$ is a convergent subsequence of $a_{n}$.
First, note that the sequence $j_{n} geq n$ for all $n in mathbb{N}$ (You can prove this as an exercise). (*)
We want to proof that $forall epsilon > 0$ there exist a $N in mathbb{N}$ such that for all $n geq N$, $|a_{n}-L|<epsilon$ (for some $L$ where $L$ is the limit we want).
Let $epsilon > 0$.
Since $a_{n}$ is a Cauchy sequence, then exist an $N_{1} in mathbb{N}$ such that for all $m,n geq N_{1}$, $|a_{n}-a_{m}|<dfrac{epsilon}{2}$.
In the other hand, we have that $a_{j_{n}}$ is convergent (say, to $M$). Then, exist $N_{2}$ such that for all $n geq N_{2}$, $|a_{j_{n}}-M|<dfrac{epsilon}{2}$
If we choose $N_{0}=max(N_{1},N_{2})$, then let $n in mathbb{N}$ such that $n geq N_{0}$.
By (*) we have $j_{n} geq n geq N_{0}$. In particular, $j_{n}, n geq N_{1}$ then, since $a_{n}$ is a Cauchy sequence by hypotesis, $|a_{j_{n}}-a_{n}|< dfrac{epsilon}{2}$
But also $n geq N_{2}$, then $|a_{j_{n}}-M|< dfrac{epsilon}{2}$.
By triangle inequality, we have $$|a_{n}-M|=|(a_{n}-a_{j_{n}})+(a_{j_{n}}-M)|leq |a_{n}-a_{j_{n}}|+|a_{j_{n}}-M|=|a_{j_{n}}-a_{n}|+|a_{j_{n}}-M|<dfrac{epsilon}{2}+dfrac{epsilon}{2}=epsilon$$
In other words: $$|a_{n}-M|<epsilon$$
And $M$ is the $L$ we wanted at the begin of the proof.
answered Dec 1 at 2:48
rowcol
734
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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active
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active
oldest
votes
The proof seems to be incorrect.
In order to deduce
begin{align*}
|a_n-a_{n_{J+1}}|<frac{epsilon}{2}tag{1}
end{align*}
we have to assure that $n>N_1$ and $n_{J+1}>N_1$.
Since $n>N=max(N_1,n_J)geq N_1$ we have $n>N_1$. But we don't know if $n_{J+1}>N_1$.
We know $n_{J+1}>n_J$ which is sufficient to show
begin{align*}
|a_{n_{J+1}}-l|<frac{epsilon}{2}
end{align*}
but not sufficient to show (1).
answered Apr 7 '17 at 20:07
Markus Scheuer
60k455143
add a comment |
The proof seems to be incorrect.
In order to deduce
begin{align*}
|a_n-a_{n_{J+1}}|<frac{epsilon}{2}tag{1}
end{align*}
we have to assure that $n>N_1$ and $n_{J+1}>N_1$.
Since $n>N=max(N_1,n_J)geq N_1$ we have $n>N_1$. But we don't know if $n_{J+1}>N_1$.
We know $n_{J+1}>n_J$ which is sufficient to show
begin{align*}
|a_{n_{J+1}}-l|<frac{epsilon}{2}
end{align*}
but not sufficient to show (1).
answered Apr 7 '17 at 20:07
Markus Scheuer
60k455143
add a comment |
The proof seems to be incorrect.
In order to deduce
begin{align*}
|a_n-a_{n_{J+1}}|<frac{epsilon}{2}tag{1}
end{align*}
we have to assure that $n>N_1$ and $n_{J+1}>N_1$.
Since $n>N=max(N_1,n_J)geq N_1$ we have $n>N_1$. But we don't know if $n_{J+1}>N_1$.
We know $n_{J+1}>n_J$ which is sufficient to show
begin{align*}
|a_{n_{J+1}}-l|<frac{epsilon}{2}
end{align*}
but not sufficient to show (1).
answered Apr 7 '17 at 20:07
Markus Scheuer
60k455143
The proof seems to be incorrect.
In order to deduce
begin{align*}
|a_n-a_{n_{J+1}}|<frac{epsilon}{2}tag{1}
end{align*}
we have to assure that $n>N_1$ and $n_{J+1}>N_1$.
Since $n>N=max(N_1,n_J)geq N_1$ we have $n>N_1$. But we don't know if $n_{J+1}>N_1$.
We know $n_{J+1}>n_J$ which is sufficient to show
begin{align*}
|a_{n_{J+1}}-l|<frac{epsilon}{2}
end{align*}
but not sufficient to show (1).
answered Apr 7 '17 at 20:07
Markus Scheuer
60k455143
answered Apr 7 '17 at 20:07
Markus Scheuer
60k455143
answered Apr 7 '17 at 20:07
Markus Scheuer
60k455143
answered Apr 7 '17 at 20:07
Markus Scheuer
60k455143
60k455143
add a comment |
add a comment |
Proof goes as follows:
Let $a_{n}$ be a Cauchy sequence.
Suppose that $a_{j_{n}}$ is a convergent subsequence of $a_{n}$.
First, note that the sequence $j_{n} geq n$ for all $n in mathbb{N}$ (You can prove this as an exercise). (*)
We want to proof that $forall epsilon > 0$ there exist a $N in mathbb{N}$ such that for all $n geq N$, $|a_{n}-L|<epsilon$ (for some $L$ where $L$ is the limit we want).
Let $epsilon > 0$.
Since $a_{n}$ is a Cauchy sequence, then exist an $N_{1} in mathbb{N}$ such that for all $m,n geq N_{1}$, $|a_{n}-a_{m}|<dfrac{epsilon}{2}$.
In the other hand, we have that $a_{j_{n}}$ is convergent (say, to $M$). Then, exist $N_{2}$ such that for all $n geq N_{2}$, $|a_{j_{n}}-M|<dfrac{epsilon}{2}$
If we choose $N_{0}=max(N_{1},N_{2})$, then let $n in mathbb{N}$ such that $n geq N_{0}$.
By (*) we have $j_{n} geq n geq N_{0}$. In particular, $j_{n}, n geq N_{1}$ then, since $a_{n}$ is a Cauchy sequence by hypotesis, $|a_{j_{n}}-a_{n}|< dfrac{epsilon}{2}$
But also $n geq N_{2}$, then $|a_{j_{n}}-M|< dfrac{epsilon}{2}$.
By triangle inequality, we have $$|a_{n}-M|=|(a_{n}-a_{j_{n}})+(a_{j_{n}}-M)|leq |a_{n}-a_{j_{n}}|+|a_{j_{n}}-M|=|a_{j_{n}}-a_{n}|+|a_{j_{n}}-M|<dfrac{epsilon}{2}+dfrac{epsilon}{2}=epsilon$$
In other words: $$|a_{n}-M|<epsilon$$
And $M$ is the $L$ we wanted at the begin of the proof.
answered Dec 1 at 2:48
rowcol
734
add a comment |
Proof goes as follows:
Let $a_{n}$ be a Cauchy sequence.
Suppose that $a_{j_{n}}$ is a convergent subsequence of $a_{n}$.
First, note that the sequence $j_{n} geq n$ for all $n in mathbb{N}$ (You can prove this as an exercise). (*)
We want to proof that $forall epsilon > 0$ there exist a $N in mathbb{N}$ such that for all $n geq N$, $|a_{n}-L|<epsilon$ (for some $L$ where $L$ is the limit we want).
Let $epsilon > 0$.
Since $a_{n}$ is a Cauchy sequence, then exist an $N_{1} in mathbb{N}$ such that for all $m,n geq N_{1}$, $|a_{n}-a_{m}|<dfrac{epsilon}{2}$.
In the other hand, we have that $a_{j_{n}}$ is convergent (say, to $M$). Then, exist $N_{2}$ such that for all $n geq N_{2}$, $|a_{j_{n}}-M|<dfrac{epsilon}{2}$
If we choose $N_{0}=max(N_{1},N_{2})$, then let $n in mathbb{N}$ such that $n geq N_{0}$.
By (*) we have $j_{n} geq n geq N_{0}$. In particular, $j_{n}, n geq N_{1}$ then, since $a_{n}$ is a Cauchy sequence by hypotesis, $|a_{j_{n}}-a_{n}|< dfrac{epsilon}{2}$
But also $n geq N_{2}$, then $|a_{j_{n}}-M|< dfrac{epsilon}{2}$.
By triangle inequality, we have $$|a_{n}-M|=|(a_{n}-a_{j_{n}})+(a_{j_{n}}-M)|leq |a_{n}-a_{j_{n}}|+|a_{j_{n}}-M|=|a_{j_{n}}-a_{n}|+|a_{j_{n}}-M|<dfrac{epsilon}{2}+dfrac{epsilon}{2}=epsilon$$
In other words: $$|a_{n}-M|<epsilon$$
And $M$ is the $L$ we wanted at the begin of the proof.
answered Dec 1 at 2:48
rowcol
734
add a comment |
Proof goes as follows:
Let $a_{n}$ be a Cauchy sequence.
Suppose that $a_{j_{n}}$ is a convergent subsequence of $a_{n}$.
First, note that the sequence $j_{n} geq n$ for all $n in mathbb{N}$ (You can prove this as an exercise). (*)
We want to proof that $forall epsilon > 0$ there exist a $N in mathbb{N}$ such that for all $n geq N$, $|a_{n}-L|<epsilon$ (for some $L$ where $L$ is the limit we want).
Let $epsilon > 0$.
Since $a_{n}$ is a Cauchy sequence, then exist an $N_{1} in mathbb{N}$ such that for all $m,n geq N_{1}$, $|a_{n}-a_{m}|<dfrac{epsilon}{2}$.
In the other hand, we have that $a_{j_{n}}$ is convergent (say, to $M$). Then, exist $N_{2}$ such that for all $n geq N_{2}$, $|a_{j_{n}}-M|<dfrac{epsilon}{2}$
If we choose $N_{0}=max(N_{1},N_{2})$, then let $n in mathbb{N}$ such that $n geq N_{0}$.
By (*) we have $j_{n} geq n geq N_{0}$. In particular, $j_{n}, n geq N_{1}$ then, since $a_{n}$ is a Cauchy sequence by hypotesis, $|a_{j_{n}}-a_{n}|< dfrac{epsilon}{2}$
But also $n geq N_{2}$, then $|a_{j_{n}}-M|< dfrac{epsilon}{2}$.
By triangle inequality, we have $$|a_{n}-M|=|(a_{n}-a_{j_{n}})+(a_{j_{n}}-M)|leq |a_{n}-a_{j_{n}}|+|a_{j_{n}}-M|=|a_{j_{n}}-a_{n}|+|a_{j_{n}}-M|<dfrac{epsilon}{2}+dfrac{epsilon}{2}=epsilon$$
In other words: $$|a_{n}-M|<epsilon$$
And $M$ is the $L$ we wanted at the begin of the proof.
answered Dec 1 at 2:48
rowcol
734
Proof goes as follows:
Let $a_{n}$ be a Cauchy sequence.
Suppose that $a_{j_{n}}$ is a convergent subsequence of $a_{n}$.
First, note that the sequence $j_{n} geq n$ for all $n in mathbb{N}$ (You can prove this as an exercise). (*)
We want to proof that $forall epsilon > 0$ there exist a $N in mathbb{N}$ such that for all $n geq N$, $|a_{n}-L|<epsilon$ (for some $L$ where $L$ is the limit we want).
Let $epsilon > 0$.
Since $a_{n}$ is a Cauchy sequence, then exist an $N_{1} in mathbb{N}$ such that for all $m,n geq N_{1}$, $|a_{n}-a_{m}|<dfrac{epsilon}{2}$.
In the other hand, we have that $a_{j_{n}}$ is convergent (say, to $M$). Then, exist $N_{2}$ such that for all $n geq N_{2}$, $|a_{j_{n}}-M|<dfrac{epsilon}{2}$
If we choose $N_{0}=max(N_{1},N_{2})$, then let $n in mathbb{N}$ such that $n geq N_{0}$.
By (*) we have $j_{n} geq n geq N_{0}$. In particular, $j_{n}, n geq N_{1}$ then, since $a_{n}$ is a Cauchy sequence by hypotesis, $|a_{j_{n}}-a_{n}|< dfrac{epsilon}{2}$
But also $n geq N_{2}$, then $|a_{j_{n}}-M|< dfrac{epsilon}{2}$.
By triangle inequality, we have $$|a_{n}-M|=|(a_{n}-a_{j_{n}})+(a_{j_{n}}-M)|leq |a_{n}-a_{j_{n}}|+|a_{j_{n}}-M|=|a_{j_{n}}-a_{n}|+|a_{j_{n}}-M|<dfrac{epsilon}{2}+dfrac{epsilon}{2}=epsilon$$
In other words: $$|a_{n}-M|<epsilon$$
And $M$ is the $L$ we wanted at the begin of the proof.
answered Dec 1 at 2:48
rowcol
734
answered Dec 1 at 2:48
rowcol
734
answered Dec 1 at 2:48
rowcol
734
answered Dec 1 at 2:48
rowcol
734
734
add a comment |
add a comment |
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