Mixing
Let ($x_n$) be a monotone sequence and contain a convergent subsequence. Prove that ($x_n$) is convergent.
$begingroup$
Let ($x_n$) be a monotone sequence and contain a convergent subsequence. Prove that ($x_n$) is convergent.
I know that by the Bolzano-Weierstrass Theorem, every bounded sequence has a convergent subsequence. But I need some hints as to how to prove the question above. Thanks for your help!
real-analysis
asked Sep 25 '14 at 6:00
JasonJason
5816
$endgroup$
add a comment |
$begingroup$
Let ($x_n$) be a monotone sequence and contain a convergent subsequence. Prove that ($x_n$) is convergent.
I know that by the Bolzano-Weierstrass Theorem, every bounded sequence has a convergent subsequence. But I need some hints as to how to prove the question above. Thanks for your help!
real-analysis
asked Sep 25 '14 at 6:00
JasonJason
5816
$endgroup$
$begingroup$
Have you already proved, for example, that a non-decreasing sequence which is bounded above converges?
$endgroup$
– André Nicolas
Sep 25 '14 at 6:22
$begingroup$
@JobinIdiculla 0,2,0,4,0,6.... is not monotone
$endgroup$
– Petite Etincelle
Sep 25 '14 at 6:23
$begingroup$
@LiuGang: Thanks. That was a blunder indeed.
$endgroup$
– Train Heartnet
Sep 25 '14 at 6:26
add a comment |
$begingroup$
Let ($x_n$) be a monotone sequence and contain a convergent subsequence. Prove that ($x_n$) is convergent.
I know that by the Bolzano-Weierstrass Theorem, every bounded sequence has a convergent subsequence. But I need some hints as to how to prove the question above. Thanks for your help!
real-analysis
asked Sep 25 '14 at 6:00
JasonJason
5816
$endgroup$
Let ($x_n$) be a monotone sequence and contain a convergent subsequence. Prove that ($x_n$) is convergent.
I know that by the Bolzano-Weierstrass Theorem, every bounded sequence has a convergent subsequence. But I need some hints as to how to prove the question above. Thanks for your help!
real-analysis
real-analysis
asked Sep 25 '14 at 6:00
JasonJason
5816
asked Sep 25 '14 at 6:00
JasonJason
5816
asked Sep 25 '14 at 6:00
JasonJason
5816
asked Sep 25 '14 at 6:00
JasonJason
5816
asked Sep 25 '14 at 6:00
JasonJason
5816
5816
$begingroup$
Have you already proved, for example, that a non-decreasing sequence which is bounded above converges?
$endgroup$
– André Nicolas
Sep 25 '14 at 6:22
$begingroup$
@JobinIdiculla 0,2,0,4,0,6.... is not monotone
$endgroup$
– Petite Etincelle
Sep 25 '14 at 6:23
$begingroup$
@LiuGang: Thanks. That was a blunder indeed.
$endgroup$
– Train Heartnet
Sep 25 '14 at 6:26
add a comment |
$begingroup$
Have you already proved, for example, that a non-decreasing sequence which is bounded above converges?
$endgroup$
– André Nicolas
Sep 25 '14 at 6:22
$begingroup$
@JobinIdiculla 0,2,0,4,0,6.... is not monotone
$endgroup$
– Petite Etincelle
Sep 25 '14 at 6:23
$begingroup$
@LiuGang: Thanks. That was a blunder indeed.
$endgroup$
– Train Heartnet
Sep 25 '14 at 6:26
$begingroup$
Have you already proved, for example, that a non-decreasing sequence which is bounded above converges?
$endgroup$
– André Nicolas
Sep 25 '14 at 6:22
$begingroup$
Have you already proved, for example, that a non-decreasing sequence which is bounded above converges?
$endgroup$
– André Nicolas
Sep 25 '14 at 6:22
$begingroup$
@JobinIdiculla 0,2,0,4,0,6.... is not monotone
$endgroup$
– Petite Etincelle
Sep 25 '14 at 6:23
$begingroup$
@JobinIdiculla 0,2,0,4,0,6.... is not monotone
$endgroup$
– Petite Etincelle
Sep 25 '14 at 6:23
$begingroup$
@LiuGang: Thanks. That was a blunder indeed.
$endgroup$
– Train Heartnet
Sep 25 '14 at 6:26
$begingroup$
@LiuGang: Thanks. That was a blunder indeed.
$endgroup$
– Train Heartnet
Sep 25 '14 at 6:26
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Suppose ${x_n}$ is increasing and has a subsequence ${x_{n_k}}$ which converges to $L$. We will prove that ${x_n}$ itself converges to $L$.
For any $epsilon > 0$, we want to find an integer $N_epsilon$ such that $|x_n - L| leq epsilon$ for any $n geq N_epsilon$.
Since ${x_{n_k}}$ is increasing and converges to $L$, we can find $k_epsilon$ such that for any $k geq k_epsilon$, $ -epsilon < x_{n_k} - L <0$.
Take $N_epsilon = n_{k_epsilon}$, then for any $n geq N_epsilon,$ $ x_{n_{k_epsilon}}leq x_n leq L$, so $-epsilon leq x_{n_{k_epsilon}} - Lleq x_n - L leq 0$.
Similarly, we can prove when ${x_n}$ is decreasing
edited Feb 15 '18 at 4:15
Jack Moody
16811
answered Sep 25 '14 at 6:21
Petite EtincellePetite Etincelle
12.5k12149
$endgroup$
$begingroup$
Why is $x_n leq L$ for all $n geq N_epsilon$?
$endgroup$
– jodag
Feb 9 '18 at 7:53
$begingroup$
Because $x_{n}$ is monotonically increasing to the limit $L$.
$endgroup$
– Jack Moody
Feb 15 '18 at 3:37
$begingroup$
But that's what you're trying to show. Why is it assumed?
$endgroup$
– Rafael Vergnaud
Jan 28 at 0:01
$begingroup$
It's also not apparent to why you can post $x_n leq L.$
$endgroup$
– Rafael Vergnaud
Jan 28 at 0:02
add a comment |
$begingroup$
Let $(y_k)$ be a convergent subsequence with a limit $A$. Since $(x_n)$ is monotone, $(y_k)$ is monotone as well. Then every $y$ is smaller than $A$. Then We know that for every $eps > 0$ there exists $K: k > K Rightarrow y_k > A - eps$. Then for every x after $y_k$ in original sequence it's also true since sequence is monotone.Also, for every x there exists an y which stands further in $(x_n)$. Then every $x$ is $< A$. Then the sequence converges to A. (It was for increasing sequence, for decreasing anlogically).
answered Sep 25 '14 at 6:29
Petr NaryshkinPetr Naryshkin
74659
$endgroup$
add a comment |
$begingroup$
Suppose $(x_n)$ is monotone and contains a convergent subsequence $(x_{n_i}).$
Given that $(x_{n_i})$ is convergent, it is bounded above by some upper bound $b in mathbb{N},$ that is, $x_{n_i} leq b,$ $forall i in mathbb{N}.$
Suppose $(x_n)$ is divergent. Given its monotonicity, it follows that $(x_n)$ is unbounded, that is, $forall M in mathbb{N},$ $exists N in mathbb{N}$ such that $forall n geq N$ it follows that $x_n > M$ (at some point "$N$," the sequence passes the boundary $M$ for any boundary $M in mathbb{N}).$ If $(x_n)$ is bounded, it is necessarily the case that $(x_n),$ given that it is monotone increasing, IS convergent (you can prove this).
So, $exists N in mathbb{N}$ such that $forall n geq N$ it follows that $x_n > b.$
Given that $(x_{n_i})$ is bounded above by $b,$ this means that $forall i in mathbb{N},$ $n_i < N.$
Hence, as by definition of a subsequence it is the case that $n_1 < n_2 < cdots n_i,$ the subsequence $(x_{n_i})$ contains fewer than $N$ elements (at most $N - 1$ elements).
However, a subsequence is defined as a function whose domain is the natural numbers. Given that $(x_{n_i})$ contains fewer than $N$ elements, its domain is a finite, proper subset of the natural numbers, that is, its domain is not the natural numbers. Contradiction.
Therefore, $(x_n)$ is convergent.
answered Jan 28 at 0:16
Rafael VergnaudRafael Vergnaud
357217
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Suppose ${x_n}$ is increasing and has a subsequence ${x_{n_k}}$ which converges to $L$. We will prove that ${x_n}$ itself converges to $L$.
For any $epsilon > 0$, we want to find an integer $N_epsilon$ such that $|x_n - L| leq epsilon$ for any $n geq N_epsilon$.
Since ${x_{n_k}}$ is increasing and converges to $L$, we can find $k_epsilon$ such that for any $k geq k_epsilon$, $ -epsilon < x_{n_k} - L <0$.
Take $N_epsilon = n_{k_epsilon}$, then for any $n geq N_epsilon,$ $ x_{n_{k_epsilon}}leq x_n leq L$, so $-epsilon leq x_{n_{k_epsilon}} - Lleq x_n - L leq 0$.
Similarly, we can prove when ${x_n}$ is decreasing
edited Feb 15 '18 at 4:15
Jack Moody
16811
answered Sep 25 '14 at 6:21
Petite EtincellePetite Etincelle
12.5k12149
$endgroup$
$begingroup$
Why is $x_n leq L$ for all $n geq N_epsilon$?
$endgroup$
– jodag
Feb 9 '18 at 7:53
$begingroup$
Because $x_{n}$ is monotonically increasing to the limit $L$.
$endgroup$
– Jack Moody
Feb 15 '18 at 3:37
$begingroup$
But that's what you're trying to show. Why is it assumed?
$endgroup$
– Rafael Vergnaud
Jan 28 at 0:01
$begingroup$
It's also not apparent to why you can post $x_n leq L.$
$endgroup$
– Rafael Vergnaud
Jan 28 at 0:02
add a comment |
$begingroup$
Suppose ${x_n}$ is increasing and has a subsequence ${x_{n_k}}$ which converges to $L$. We will prove that ${x_n}$ itself converges to $L$.
For any $epsilon > 0$, we want to find an integer $N_epsilon$ such that $|x_n - L| leq epsilon$ for any $n geq N_epsilon$.
Since ${x_{n_k}}$ is increasing and converges to $L$, we can find $k_epsilon$ such that for any $k geq k_epsilon$, $ -epsilon < x_{n_k} - L <0$.
Take $N_epsilon = n_{k_epsilon}$, then for any $n geq N_epsilon,$ $ x_{n_{k_epsilon}}leq x_n leq L$, so $-epsilon leq x_{n_{k_epsilon}} - Lleq x_n - L leq 0$.
Similarly, we can prove when ${x_n}$ is decreasing
edited Feb 15 '18 at 4:15
Jack Moody
16811
answered Sep 25 '14 at 6:21
Petite EtincellePetite Etincelle
12.5k12149
$endgroup$
$begingroup$
Why is $x_n leq L$ for all $n geq N_epsilon$?
$endgroup$
– jodag
Feb 9 '18 at 7:53
$begingroup$
Because $x_{n}$ is monotonically increasing to the limit $L$.
$endgroup$
– Jack Moody
Feb 15 '18 at 3:37
$begingroup$
But that's what you're trying to show. Why is it assumed?
$endgroup$
– Rafael Vergnaud
Jan 28 at 0:01
$begingroup$
It's also not apparent to why you can post $x_n leq L.$
$endgroup$
– Rafael Vergnaud
Jan 28 at 0:02
add a comment |
$begingroup$
Suppose ${x_n}$ is increasing and has a subsequence ${x_{n_k}}$ which converges to $L$. We will prove that ${x_n}$ itself converges to $L$.
For any $epsilon > 0$, we want to find an integer $N_epsilon$ such that $|x_n - L| leq epsilon$ for any $n geq N_epsilon$.
Since ${x_{n_k}}$ is increasing and converges to $L$, we can find $k_epsilon$ such that for any $k geq k_epsilon$, $ -epsilon < x_{n_k} - L <0$.
Take $N_epsilon = n_{k_epsilon}$, then for any $n geq N_epsilon,$ $ x_{n_{k_epsilon}}leq x_n leq L$, so $-epsilon leq x_{n_{k_epsilon}} - Lleq x_n - L leq 0$.
Similarly, we can prove when ${x_n}$ is decreasing
edited Feb 15 '18 at 4:15
Jack Moody
16811
answered Sep 25 '14 at 6:21
Petite EtincellePetite Etincelle
12.5k12149
$endgroup$
Suppose ${x_n}$ is increasing and has a subsequence ${x_{n_k}}$ which converges to $L$. We will prove that ${x_n}$ itself converges to $L$.
For any $epsilon > 0$, we want to find an integer $N_epsilon$ such that $|x_n - L| leq epsilon$ for any $n geq N_epsilon$.
Since ${x_{n_k}}$ is increasing and converges to $L$, we can find $k_epsilon$ such that for any $k geq k_epsilon$, $ -epsilon < x_{n_k} - L <0$.
Take $N_epsilon = n_{k_epsilon}$, then for any $n geq N_epsilon,$ $ x_{n_{k_epsilon}}leq x_n leq L$, so $-epsilon leq x_{n_{k_epsilon}} - Lleq x_n - L leq 0$.
Similarly, we can prove when ${x_n}$ is decreasing
edited Feb 15 '18 at 4:15
Jack Moody
16811
answered Sep 25 '14 at 6:21
Petite EtincellePetite Etincelle
12.5k12149
edited Feb 15 '18 at 4:15
Jack Moody
16811
edited Feb 15 '18 at 4:15
Jack Moody
16811
edited Feb 15 '18 at 4:15
Jack Moody
16811
16811
answered Sep 25 '14 at 6:21
Petite EtincellePetite Etincelle
12.5k12149
answered Sep 25 '14 at 6:21
Petite EtincellePetite Etincelle
12.5k12149
answered Sep 25 '14 at 6:21
Petite EtincellePetite Etincelle
12.5k12149
12.5k12149
$begingroup$
Why is $x_n leq L$ for all $n geq N_epsilon$?
$endgroup$
– jodag
Feb 9 '18 at 7:53
$begingroup$
Because $x_{n}$ is monotonically increasing to the limit $L$.
$endgroup$
– Jack Moody
Feb 15 '18 at 3:37
$begingroup$
But that's what you're trying to show. Why is it assumed?
$endgroup$
– Rafael Vergnaud
Jan 28 at 0:01
$begingroup$
It's also not apparent to why you can post $x_n leq L.$
$endgroup$
– Rafael Vergnaud
Jan 28 at 0:02
add a comment |
$begingroup$
Why is $x_n leq L$ for all $n geq N_epsilon$?
$endgroup$
– jodag
Feb 9 '18 at 7:53
$begingroup$
Because $x_{n}$ is monotonically increasing to the limit $L$.
$endgroup$
– Jack Moody
Feb 15 '18 at 3:37
$begingroup$
But that's what you're trying to show. Why is it assumed?
$endgroup$
– Rafael Vergnaud
Jan 28 at 0:01
$begingroup$
It's also not apparent to why you can post $x_n leq L.$
$endgroup$
– Rafael Vergnaud
Jan 28 at 0:02
$begingroup$
Why is $x_n leq L$ for all $n geq N_epsilon$?
$endgroup$
– jodag
Feb 9 '18 at 7:53
$begingroup$
Why is $x_n leq L$ for all $n geq N_epsilon$?
$endgroup$
– jodag
Feb 9 '18 at 7:53
$begingroup$
Because $x_{n}$ is monotonically increasing to the limit $L$.
$endgroup$
– Jack Moody
Feb 15 '18 at 3:37
$begingroup$
Because $x_{n}$ is monotonically increasing to the limit $L$.
$endgroup$
– Jack Moody
Feb 15 '18 at 3:37
$begingroup$
But that's what you're trying to show. Why is it assumed?
$endgroup$
– Rafael Vergnaud
Jan 28 at 0:01
$begingroup$
But that's what you're trying to show. Why is it assumed?
$endgroup$
– Rafael Vergnaud
Jan 28 at 0:01
$begingroup$
It's also not apparent to why you can post $x_n leq L.$
$endgroup$
– Rafael Vergnaud
Jan 28 at 0:02
$begingroup$
It's also not apparent to why you can post $x_n leq L.$
$endgroup$
– Rafael Vergnaud
Jan 28 at 0:02
add a comment |
$begingroup$
Let $(y_k)$ be a convergent subsequence with a limit $A$. Since $(x_n)$ is monotone, $(y_k)$ is monotone as well. Then every $y$ is smaller than $A$. Then We know that for every $eps > 0$ there exists $K: k > K Rightarrow y_k > A - eps$. Then for every x after $y_k$ in original sequence it's also true since sequence is monotone.Also, for every x there exists an y which stands further in $(x_n)$. Then every $x$ is $< A$. Then the sequence converges to A. (It was for increasing sequence, for decreasing anlogically).
answered Sep 25 '14 at 6:29
Petr NaryshkinPetr Naryshkin
74659
$endgroup$
add a comment |
$begingroup$
Let $(y_k)$ be a convergent subsequence with a limit $A$. Since $(x_n)$ is monotone, $(y_k)$ is monotone as well. Then every $y$ is smaller than $A$. Then We know that for every $eps > 0$ there exists $K: k > K Rightarrow y_k > A - eps$. Then for every x after $y_k$ in original sequence it's also true since sequence is monotone.Also, for every x there exists an y which stands further in $(x_n)$. Then every $x$ is $< A$. Then the sequence converges to A. (It was for increasing sequence, for decreasing anlogically).
answered Sep 25 '14 at 6:29
Petr NaryshkinPetr Naryshkin
74659
$endgroup$
add a comment |
$begingroup$
Let $(y_k)$ be a convergent subsequence with a limit $A$. Since $(x_n)$ is monotone, $(y_k)$ is monotone as well. Then every $y$ is smaller than $A$. Then We know that for every $eps > 0$ there exists $K: k > K Rightarrow y_k > A - eps$. Then for every x after $y_k$ in original sequence it's also true since sequence is monotone.Also, for every x there exists an y which stands further in $(x_n)$. Then every $x$ is $< A$. Then the sequence converges to A. (It was for increasing sequence, for decreasing anlogically).
answered Sep 25 '14 at 6:29
Petr NaryshkinPetr Naryshkin
74659
$endgroup$
Let $(y_k)$ be a convergent subsequence with a limit $A$. Since $(x_n)$ is monotone, $(y_k)$ is monotone as well. Then every $y$ is smaller than $A$. Then We know that for every $eps > 0$ there exists $K: k > K Rightarrow y_k > A - eps$. Then for every x after $y_k$ in original sequence it's also true since sequence is monotone.Also, for every x there exists an y which stands further in $(x_n)$. Then every $x$ is $< A$. Then the sequence converges to A. (It was for increasing sequence, for decreasing anlogically).
answered Sep 25 '14 at 6:29
Petr NaryshkinPetr Naryshkin
74659
answered Sep 25 '14 at 6:29
Petr NaryshkinPetr Naryshkin
74659
answered Sep 25 '14 at 6:29
Petr NaryshkinPetr Naryshkin
74659
answered Sep 25 '14 at 6:29
Petr NaryshkinPetr Naryshkin
74659
74659
add a comment |
add a comment |
$begingroup$
Suppose $(x_n)$ is monotone and contains a convergent subsequence $(x_{n_i}).$
Given that $(x_{n_i})$ is convergent, it is bounded above by some upper bound $b in mathbb{N},$ that is, $x_{n_i} leq b,$ $forall i in mathbb{N}.$
Suppose $(x_n)$ is divergent. Given its monotonicity, it follows that $(x_n)$ is unbounded, that is, $forall M in mathbb{N},$ $exists N in mathbb{N}$ such that $forall n geq N$ it follows that $x_n > M$ (at some point "$N$," the sequence passes the boundary $M$ for any boundary $M in mathbb{N}).$ If $(x_n)$ is bounded, it is necessarily the case that $(x_n),$ given that it is monotone increasing, IS convergent (you can prove this).
So, $exists N in mathbb{N}$ such that $forall n geq N$ it follows that $x_n > b.$
Given that $(x_{n_i})$ is bounded above by $b,$ this means that $forall i in mathbb{N},$ $n_i < N.$
Hence, as by definition of a subsequence it is the case that $n_1 < n_2 < cdots n_i,$ the subsequence $(x_{n_i})$ contains fewer than $N$ elements (at most $N - 1$ elements).
However, a subsequence is defined as a function whose domain is the natural numbers. Given that $(x_{n_i})$ contains fewer than $N$ elements, its domain is a finite, proper subset of the natural numbers, that is, its domain is not the natural numbers. Contradiction.
Therefore, $(x_n)$ is convergent.
answered Jan 28 at 0:16
Rafael VergnaudRafael Vergnaud
357217
$endgroup$
add a comment |
$begingroup$
Suppose $(x_n)$ is monotone and contains a convergent subsequence $(x_{n_i}).$
Given that $(x_{n_i})$ is convergent, it is bounded above by some upper bound $b in mathbb{N},$ that is, $x_{n_i} leq b,$ $forall i in mathbb{N}.$
Suppose $(x_n)$ is divergent. Given its monotonicity, it follows that $(x_n)$ is unbounded, that is, $forall M in mathbb{N},$ $exists N in mathbb{N}$ such that $forall n geq N$ it follows that $x_n > M$ (at some point "$N$," the sequence passes the boundary $M$ for any boundary $M in mathbb{N}).$ If $(x_n)$ is bounded, it is necessarily the case that $(x_n),$ given that it is monotone increasing, IS convergent (you can prove this).
So, $exists N in mathbb{N}$ such that $forall n geq N$ it follows that $x_n > b.$
Given that $(x_{n_i})$ is bounded above by $b,$ this means that $forall i in mathbb{N},$ $n_i < N.$
Hence, as by definition of a subsequence it is the case that $n_1 < n_2 < cdots n_i,$ the subsequence $(x_{n_i})$ contains fewer than $N$ elements (at most $N - 1$ elements).
However, a subsequence is defined as a function whose domain is the natural numbers. Given that $(x_{n_i})$ contains fewer than $N$ elements, its domain is a finite, proper subset of the natural numbers, that is, its domain is not the natural numbers. Contradiction.
Therefore, $(x_n)$ is convergent.
answered Jan 28 at 0:16
Rafael VergnaudRafael Vergnaud
357217
$endgroup$
add a comment |
$begingroup$
Suppose $(x_n)$ is monotone and contains a convergent subsequence $(x_{n_i}).$
Given that $(x_{n_i})$ is convergent, it is bounded above by some upper bound $b in mathbb{N},$ that is, $x_{n_i} leq b,$ $forall i in mathbb{N}.$
Suppose $(x_n)$ is divergent. Given its monotonicity, it follows that $(x_n)$ is unbounded, that is, $forall M in mathbb{N},$ $exists N in mathbb{N}$ such that $forall n geq N$ it follows that $x_n > M$ (at some point "$N$," the sequence passes the boundary $M$ for any boundary $M in mathbb{N}).$ If $(x_n)$ is bounded, it is necessarily the case that $(x_n),$ given that it is monotone increasing, IS convergent (you can prove this).
So, $exists N in mathbb{N}$ such that $forall n geq N$ it follows that $x_n > b.$
Given that $(x_{n_i})$ is bounded above by $b,$ this means that $forall i in mathbb{N},$ $n_i < N.$
Hence, as by definition of a subsequence it is the case that $n_1 < n_2 < cdots n_i,$ the subsequence $(x_{n_i})$ contains fewer than $N$ elements (at most $N - 1$ elements).
However, a subsequence is defined as a function whose domain is the natural numbers. Given that $(x_{n_i})$ contains fewer than $N$ elements, its domain is a finite, proper subset of the natural numbers, that is, its domain is not the natural numbers. Contradiction.
Therefore, $(x_n)$ is convergent.
answered Jan 28 at 0:16
Rafael VergnaudRafael Vergnaud
357217
$endgroup$
Suppose $(x_n)$ is monotone and contains a convergent subsequence $(x_{n_i}).$
Given that $(x_{n_i})$ is convergent, it is bounded above by some upper bound $b in mathbb{N},$ that is, $x_{n_i} leq b,$ $forall i in mathbb{N}.$
Suppose $(x_n)$ is divergent. Given its monotonicity, it follows that $(x_n)$ is unbounded, that is, $forall M in mathbb{N},$ $exists N in mathbb{N}$ such that $forall n geq N$ it follows that $x_n > M$ (at some point "$N$," the sequence passes the boundary $M$ for any boundary $M in mathbb{N}).$ If $(x_n)$ is bounded, it is necessarily the case that $(x_n),$ given that it is monotone increasing, IS convergent (you can prove this).
So, $exists N in mathbb{N}$ such that $forall n geq N$ it follows that $x_n > b.$
Given that $(x_{n_i})$ is bounded above by $b,$ this means that $forall i in mathbb{N},$ $n_i < N.$
Hence, as by definition of a subsequence it is the case that $n_1 < n_2 < cdots n_i,$ the subsequence $(x_{n_i})$ contains fewer than $N$ elements (at most $N - 1$ elements).
However, a subsequence is defined as a function whose domain is the natural numbers. Given that $(x_{n_i})$ contains fewer than $N$ elements, its domain is a finite, proper subset of the natural numbers, that is, its domain is not the natural numbers. Contradiction.
Therefore, $(x_n)$ is convergent.
answered Jan 28 at 0:16
Rafael VergnaudRafael Vergnaud
357217
answered Jan 28 at 0:16
Rafael VergnaudRafael Vergnaud
357217
answered Jan 28 at 0:16
Rafael VergnaudRafael Vergnaud
357217
answered Jan 28 at 0:16
Rafael VergnaudRafael Vergnaud
357217
357217
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$begingroup$
Have you already proved, for example, that a non-decreasing sequence which is bounded above converges?
$endgroup$
– André Nicolas
Sep 25 '14 at 6:22
$begingroup$
@JobinIdiculla 0,2,0,4,0,6.... is not monotone
$endgroup$
– Petite Etincelle
Sep 25 '14 at 6:23
$begingroup$
@LiuGang: Thanks. That was a blunder indeed.
$endgroup$
– Train Heartnet
Sep 25 '14 at 6:26