Mixing
Prove that if a sequence is unbounded, then the sequence is not Cauchy
So far:
If a sequence is unbounded, therefore it is a monotonic, divergent, sequence. Choosing $epsilon = frac12$, and $Ninmathbb{N}$. Assume n, m $geq N$.
Then $left| a_n - a_{N} right|
< epsilon$ cannot be true, because...
I understand that if the sequence is unbounded, then the difference between $a_n$ and $a_N$ will eventually become greater than $epsilon$, but I'm not sure how to write it out using the language of mathematics.
sequences-and-series proof-explanation cauchy-sequences
edited Nov 21 '18 at 0:18
Tianlalu
3,09621038
asked Nov 20 '18 at 19:23
soyfroggy
133
add a comment |
So far:
If a sequence is unbounded, therefore it is a monotonic, divergent, sequence. Choosing $epsilon = frac12$, and $Ninmathbb{N}$. Assume n, m $geq N$.
Then $left| a_n - a_{N} right|
< epsilon$ cannot be true, because...
I understand that if the sequence is unbounded, then the difference between $a_n$ and $a_N$ will eventually become greater than $epsilon$, but I'm not sure how to write it out using the language of mathematics.
sequences-and-series proof-explanation cauchy-sequences
edited Nov 21 '18 at 0:18
Tianlalu
3,09621038
asked Nov 20 '18 at 19:23
soyfroggy
133
1
The sequence $1,-2,3,-4,5,-6,ldots$ is unbounded, but not monotonic.
– José Carlos Santos
Nov 20 '18 at 19:28
add a comment |
So far:
If a sequence is unbounded, therefore it is a monotonic, divergent, sequence. Choosing $epsilon = frac12$, and $Ninmathbb{N}$. Assume n, m $geq N$.
Then $left| a_n - a_{N} right|
< epsilon$ cannot be true, because...
I understand that if the sequence is unbounded, then the difference between $a_n$ and $a_N$ will eventually become greater than $epsilon$, but I'm not sure how to write it out using the language of mathematics.
sequences-and-series proof-explanation cauchy-sequences
edited Nov 21 '18 at 0:18
Tianlalu
3,09621038
asked Nov 20 '18 at 19:23
soyfroggy
133
So far:
If a sequence is unbounded, therefore it is a monotonic, divergent, sequence. Choosing $epsilon = frac12$, and $Ninmathbb{N}$. Assume n, m $geq N$.
Then $left| a_n - a_{N} right|
< epsilon$ cannot be true, because...
I understand that if the sequence is unbounded, then the difference between $a_n$ and $a_N$ will eventually become greater than $epsilon$, but I'm not sure how to write it out using the language of mathematics.
sequences-and-series proof-explanation cauchy-sequences
sequences-and-series proof-explanation cauchy-sequences
edited Nov 21 '18 at 0:18
Tianlalu
3,09621038
asked Nov 20 '18 at 19:23
soyfroggy
133
edited Nov 21 '18 at 0:18
Tianlalu
3,09621038
asked Nov 20 '18 at 19:23
soyfroggy
133
edited Nov 21 '18 at 0:18
Tianlalu
3,09621038
edited Nov 21 '18 at 0:18
Tianlalu
3,09621038
edited Nov 21 '18 at 0:18
Tianlalu
3,09621038
3,09621038
asked Nov 20 '18 at 19:23
soyfroggy
133
asked Nov 20 '18 at 19:23
soyfroggy
133
asked Nov 20 '18 at 19:23
soyfroggy
133
133
1
The sequence $1,-2,3,-4,5,-6,ldots$ is unbounded, but not monotonic.
– José Carlos Santos
Nov 20 '18 at 19:28
add a comment |
1
The sequence $1,-2,3,-4,5,-6,ldots$ is unbounded, but not monotonic.
– José Carlos Santos
Nov 20 '18 at 19:28
1
1
The sequence $1,-2,3,-4,5,-6,ldots$ is unbounded, but not monotonic.
– José Carlos Santos
Nov 20 '18 at 19:28
The sequence $1,-2,3,-4,5,-6,ldots$ is unbounded, but not monotonic.
– José Carlos Santos
Nov 20 '18 at 19:28
add a comment |
2 Answers
2
active
oldest
votes
It's simpler to, as Federico has done, assume it's Cauchy and prove that it's bounded, but if you want to assume that it's unbounded and prove that's not Cauchy, you just have to use the definition of "unbounded". Unbounded means that for all $M$, there exists $n$ such that $|a_n|>M$. For this to occur, at least one of the following must be true: "$forall M exists n: a_n>M$" or "$forall M exists n: a_n<-M$". WLOG, we can assume the former. So simply choose $M=a_N+epsilon$ and then you can use the triangle inequality to show that $|a_n-a_N|>epsilon$.
So this shows that for all $a_N$ and $epsilon$, there is some $a_n$ that is more than $epsilon$ away from $a_N$. Technically, that's not quite enough to show this isn't Cauchy, as we need $n>N$. However, unboundedness requires an infinite number of terms, so it's not possible that all of the terms larger than $M$ have index less than $N$. To be rigorous, we could take $M$ to be the maximum of $a_n+epsilon$ for $n leq N$.
answered Nov 20 '18 at 19:48
Acccumulation
6,8062617
@Federico I said in my first line that if we just want a proof of the claim, your proof is simpler. But I was addressing how the OP could continue their current line of thinking to a proof. And $N$ is the index of an arbitrary term.
– Acccumulation
Nov 21 '18 at 16:05
Sorry, I misread the first line. Really sorry :)
– Federico
Nov 21 '18 at 16:06
add a comment |
By contradiction, let $(a_n)_{ninmathbb N}$ be Cauchy. There is $n_0$ such that $|a_n-a_{n_0}|leq1$ for $ngeq n_0$. But then
$$
|a_n| leq max{|a_1|,dots,|a_{n_0-1}|,|a_{n_0}|+1}, quad forall ninmathbb N,
$$
so the sequence is bounded.
answered Nov 20 '18 at 19:28
Federico
4,709514
@Acccumulation I don't understant. I provided an upper bound for $|a_n|$, so it's both an upper bound and a lower bound for $a_n$.
– Federico
Nov 21 '18 at 16:17
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
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votes
It's simpler to, as Federico has done, assume it's Cauchy and prove that it's bounded, but if you want to assume that it's unbounded and prove that's not Cauchy, you just have to use the definition of "unbounded". Unbounded means that for all $M$, there exists $n$ such that $|a_n|>M$. For this to occur, at least one of the following must be true: "$forall M exists n: a_n>M$" or "$forall M exists n: a_n<-M$". WLOG, we can assume the former. So simply choose $M=a_N+epsilon$ and then you can use the triangle inequality to show that $|a_n-a_N|>epsilon$.
So this shows that for all $a_N$ and $epsilon$, there is some $a_n$ that is more than $epsilon$ away from $a_N$. Technically, that's not quite enough to show this isn't Cauchy, as we need $n>N$. However, unboundedness requires an infinite number of terms, so it's not possible that all of the terms larger than $M$ have index less than $N$. To be rigorous, we could take $M$ to be the maximum of $a_n+epsilon$ for $n leq N$.
answered Nov 20 '18 at 19:48
Acccumulation
6,8062617
@Federico I said in my first line that if we just want a proof of the claim, your proof is simpler. But I was addressing how the OP could continue their current line of thinking to a proof. And $N$ is the index of an arbitrary term.
– Acccumulation
Nov 21 '18 at 16:05
Sorry, I misread the first line. Really sorry :)
– Federico
Nov 21 '18 at 16:06
add a comment |
It's simpler to, as Federico has done, assume it's Cauchy and prove that it's bounded, but if you want to assume that it's unbounded and prove that's not Cauchy, you just have to use the definition of "unbounded". Unbounded means that for all $M$, there exists $n$ such that $|a_n|>M$. For this to occur, at least one of the following must be true: "$forall M exists n: a_n>M$" or "$forall M exists n: a_n<-M$". WLOG, we can assume the former. So simply choose $M=a_N+epsilon$ and then you can use the triangle inequality to show that $|a_n-a_N|>epsilon$.
So this shows that for all $a_N$ and $epsilon$, there is some $a_n$ that is more than $epsilon$ away from $a_N$. Technically, that's not quite enough to show this isn't Cauchy, as we need $n>N$. However, unboundedness requires an infinite number of terms, so it's not possible that all of the terms larger than $M$ have index less than $N$. To be rigorous, we could take $M$ to be the maximum of $a_n+epsilon$ for $n leq N$.
answered Nov 20 '18 at 19:48
Acccumulation
6,8062617
@Federico I said in my first line that if we just want a proof of the claim, your proof is simpler. But I was addressing how the OP could continue their current line of thinking to a proof. And $N$ is the index of an arbitrary term.
– Acccumulation
Nov 21 '18 at 16:05
Sorry, I misread the first line. Really sorry :)
– Federico
Nov 21 '18 at 16:06
add a comment |
It's simpler to, as Federico has done, assume it's Cauchy and prove that it's bounded, but if you want to assume that it's unbounded and prove that's not Cauchy, you just have to use the definition of "unbounded". Unbounded means that for all $M$, there exists $n$ such that $|a_n|>M$. For this to occur, at least one of the following must be true: "$forall M exists n: a_n>M$" or "$forall M exists n: a_n<-M$". WLOG, we can assume the former. So simply choose $M=a_N+epsilon$ and then you can use the triangle inequality to show that $|a_n-a_N|>epsilon$.
So this shows that for all $a_N$ and $epsilon$, there is some $a_n$ that is more than $epsilon$ away from $a_N$. Technically, that's not quite enough to show this isn't Cauchy, as we need $n>N$. However, unboundedness requires an infinite number of terms, so it's not possible that all of the terms larger than $M$ have index less than $N$. To be rigorous, we could take $M$ to be the maximum of $a_n+epsilon$ for $n leq N$.
answered Nov 20 '18 at 19:48
Acccumulation
6,8062617
It's simpler to, as Federico has done, assume it's Cauchy and prove that it's bounded, but if you want to assume that it's unbounded and prove that's not Cauchy, you just have to use the definition of "unbounded". Unbounded means that for all $M$, there exists $n$ such that $|a_n|>M$. For this to occur, at least one of the following must be true: "$forall M exists n: a_n>M$" or "$forall M exists n: a_n<-M$". WLOG, we can assume the former. So simply choose $M=a_N+epsilon$ and then you can use the triangle inequality to show that $|a_n-a_N|>epsilon$.
So this shows that for all $a_N$ and $epsilon$, there is some $a_n$ that is more than $epsilon$ away from $a_N$. Technically, that's not quite enough to show this isn't Cauchy, as we need $n>N$. However, unboundedness requires an infinite number of terms, so it's not possible that all of the terms larger than $M$ have index less than $N$. To be rigorous, we could take $M$ to be the maximum of $a_n+epsilon$ for $n leq N$.
answered Nov 20 '18 at 19:48
Acccumulation
6,8062617
edited Nov 21 '18 at 16:21
answered Nov 20 '18 at 19:48
Acccumulation
6,8062617
answered Nov 20 '18 at 19:48
Acccumulation
6,8062617
answered Nov 20 '18 at 19:48
Acccumulation
6,8062617
6,8062617
@Federico I said in my first line that if we just want a proof of the claim, your proof is simpler. But I was addressing how the OP could continue their current line of thinking to a proof. And $N$ is the index of an arbitrary term.
– Acccumulation
Nov 21 '18 at 16:05
Sorry, I misread the first line. Really sorry :)
– Federico
Nov 21 '18 at 16:06
add a comment |
@Federico I said in my first line that if we just want a proof of the claim, your proof is simpler. But I was addressing how the OP could continue their current line of thinking to a proof. And $N$ is the index of an arbitrary term.
– Acccumulation
Nov 21 '18 at 16:05
Sorry, I misread the first line. Really sorry :)
– Federico
Nov 21 '18 at 16:06
@Federico I said in my first line that if we just want a proof of the claim, your proof is simpler. But I was addressing how the OP could continue their current line of thinking to a proof. And $N$ is the index of an arbitrary term.
– Acccumulation
Nov 21 '18 at 16:05
@Federico I said in my first line that if we just want a proof of the claim, your proof is simpler. But I was addressing how the OP could continue their current line of thinking to a proof. And $N$ is the index of an arbitrary term.
– Acccumulation
Nov 21 '18 at 16:05
Sorry, I misread the first line. Really sorry :)
– Federico
Nov 21 '18 at 16:06
Sorry, I misread the first line. Really sorry :)
– Federico
Nov 21 '18 at 16:06
add a comment |
By contradiction, let $(a_n)_{ninmathbb N}$ be Cauchy. There is $n_0$ such that $|a_n-a_{n_0}|leq1$ for $ngeq n_0$. But then
$$
|a_n| leq max{|a_1|,dots,|a_{n_0-1}|,|a_{n_0}|+1}, quad forall ninmathbb N,
$$
so the sequence is bounded.
answered Nov 20 '18 at 19:28
Federico
4,709514
@Acccumulation I don't understant. I provided an upper bound for $|a_n|$, so it's both an upper bound and a lower bound for $a_n$.
– Federico
Nov 21 '18 at 16:17
add a comment |
By contradiction, let $(a_n)_{ninmathbb N}$ be Cauchy. There is $n_0$ such that $|a_n-a_{n_0}|leq1$ for $ngeq n_0$. But then
$$
|a_n| leq max{|a_1|,dots,|a_{n_0-1}|,|a_{n_0}|+1}, quad forall ninmathbb N,
$$
so the sequence is bounded.
answered Nov 20 '18 at 19:28
Federico
4,709514
@Acccumulation I don't understant. I provided an upper bound for $|a_n|$, so it's both an upper bound and a lower bound for $a_n$.
– Federico
Nov 21 '18 at 16:17
add a comment |
By contradiction, let $(a_n)_{ninmathbb N}$ be Cauchy. There is $n_0$ such that $|a_n-a_{n_0}|leq1$ for $ngeq n_0$. But then
$$
|a_n| leq max{|a_1|,dots,|a_{n_0-1}|,|a_{n_0}|+1}, quad forall ninmathbb N,
$$
so the sequence is bounded.
answered Nov 20 '18 at 19:28
Federico
4,709514
By contradiction, let $(a_n)_{ninmathbb N}$ be Cauchy. There is $n_0$ such that $|a_n-a_{n_0}|leq1$ for $ngeq n_0$. But then
$$
|a_n| leq max{|a_1|,dots,|a_{n_0-1}|,|a_{n_0}|+1}, quad forall ninmathbb N,
$$
so the sequence is bounded.
answered Nov 20 '18 at 19:28
Federico
4,709514
answered Nov 20 '18 at 19:28
Federico
4,709514
answered Nov 20 '18 at 19:28
Federico
4,709514
answered Nov 20 '18 at 19:28
Federico
4,709514
4,709514
@Acccumulation I don't understant. I provided an upper bound for $|a_n|$, so it's both an upper bound and a lower bound for $a_n$.
– Federico
Nov 21 '18 at 16:17
add a comment |
@Acccumulation I don't understant. I provided an upper bound for $|a_n|$, so it's both an upper bound and a lower bound for $a_n$.
– Federico
Nov 21 '18 at 16:17
@Acccumulation I don't understant. I provided an upper bound for $|a_n|$, so it's both an upper bound and a lower bound for $a_n$.
– Federico
Nov 21 '18 at 16:17
@Acccumulation I don't understant. I provided an upper bound for $|a_n|$, so it's both an upper bound and a lower bound for $a_n$.
– Federico
Nov 21 '18 at 16:17
add a comment |
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1
The sequence $1,-2,3,-4,5,-6,ldots$ is unbounded, but not monotonic.
– José Carlos Santos
Nov 20 '18 at 19:28