A sequence that has no Cauchy subsequence
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Give an example of a sequence in $mathbb{R}$ which has no subsequence which is a Cauchy sequence.
I can find out a sequence that is not a Cauchy sequence such as ${ln(n)}$ once $|ln(n)-ln(n+1)|=0$ but $|ln(n)-ln(2n)|=|ln(frac{1}{2})|>epsilon$
$forall epsilon<ln(frac{1}{2})$
I can still find a subsequence of the type ${ln(2n)}_{2ninmathbb{N}}$ such that $|ln(2n)-ln(2n+1)|=0$
Question:
What should I do to get a sequence that has no Cauchy subsequence?
Thanks in advance!
real-analysis sequences-and-series general-topology metric-spaces cauchy-sequences
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up vote
2
down vote
favorite
Give an example of a sequence in $mathbb{R}$ which has no subsequence which is a Cauchy sequence.
I can find out a sequence that is not a Cauchy sequence such as ${ln(n)}$ once $|ln(n)-ln(n+1)|=0$ but $|ln(n)-ln(2n)|=|ln(frac{1}{2})|>epsilon$
$forall epsilon<ln(frac{1}{2})$
I can still find a subsequence of the type ${ln(2n)}_{2ninmathbb{N}}$ such that $|ln(2n)-ln(2n+1)|=0$
Question:
What should I do to get a sequence that has no Cauchy subsequence?
Thanks in advance!
real-analysis sequences-and-series general-topology metric-spaces cauchy-sequences
In the first place, you should only be looking at sequences which are not convergent, right?
– MPW
5 hours ago
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Give an example of a sequence in $mathbb{R}$ which has no subsequence which is a Cauchy sequence.
I can find out a sequence that is not a Cauchy sequence such as ${ln(n)}$ once $|ln(n)-ln(n+1)|=0$ but $|ln(n)-ln(2n)|=|ln(frac{1}{2})|>epsilon$
$forall epsilon<ln(frac{1}{2})$
I can still find a subsequence of the type ${ln(2n)}_{2ninmathbb{N}}$ such that $|ln(2n)-ln(2n+1)|=0$
Question:
What should I do to get a sequence that has no Cauchy subsequence?
Thanks in advance!
real-analysis sequences-and-series general-topology metric-spaces cauchy-sequences
Give an example of a sequence in $mathbb{R}$ which has no subsequence which is a Cauchy sequence.
I can find out a sequence that is not a Cauchy sequence such as ${ln(n)}$ once $|ln(n)-ln(n+1)|=0$ but $|ln(n)-ln(2n)|=|ln(frac{1}{2})|>epsilon$
$forall epsilon<ln(frac{1}{2})$
I can still find a subsequence of the type ${ln(2n)}_{2ninmathbb{N}}$ such that $|ln(2n)-ln(2n+1)|=0$
Question:
What should I do to get a sequence that has no Cauchy subsequence?
Thanks in advance!
real-analysis sequences-and-series general-topology metric-spaces cauchy-sequences
real-analysis sequences-and-series general-topology metric-spaces cauchy-sequences
edited 4 hours ago
Bernard
115k637107
115k637107
asked 6 hours ago
Pedro Gomes
1,5622619
1,5622619
In the first place, you should only be looking at sequences which are not convergent, right?
– MPW
5 hours ago
add a comment |
In the first place, you should only be looking at sequences which are not convergent, right?
– MPW
5 hours ago
In the first place, you should only be looking at sequences which are not convergent, right?
– MPW
5 hours ago
In the first place, you should only be looking at sequences which are not convergent, right?
– MPW
5 hours ago
add a comment |
3 Answers
3
active
oldest
votes
up vote
8
down vote
Take $a_n=n$. Then for any subsequence $n_k$, $|n_k-n_{k-1}|geq 1$. So, it has no Cauchy sub-sequence.
New contributor
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up vote
3
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Take a sequence $(a_n)$ such that $a_nto infty$. Then each subsequence $(a_{n_k})$ is such that $a_{n_k}to infty$ and thus can't be Cauchy since necessarily Cauchy sequences are bounded.
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up vote
2
down vote
Take the sequence $1,2,3,4,ldots$ It has no Cauchy subsequence since the distance between any two distinct terms is at least $1$.
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
Take $a_n=n$. Then for any subsequence $n_k$, $|n_k-n_{k-1}|geq 1$. So, it has no Cauchy sub-sequence.
New contributor
add a comment |
up vote
8
down vote
Take $a_n=n$. Then for any subsequence $n_k$, $|n_k-n_{k-1}|geq 1$. So, it has no Cauchy sub-sequence.
New contributor
add a comment |
up vote
8
down vote
up vote
8
down vote
Take $a_n=n$. Then for any subsequence $n_k$, $|n_k-n_{k-1}|geq 1$. So, it has no Cauchy sub-sequence.
New contributor
Take $a_n=n$. Then for any subsequence $n_k$, $|n_k-n_{k-1}|geq 1$. So, it has no Cauchy sub-sequence.
New contributor
New contributor
answered 6 hours ago
John_Wick
3318
3318
New contributor
New contributor
add a comment |
add a comment |
up vote
3
down vote
Take a sequence $(a_n)$ such that $a_nto infty$. Then each subsequence $(a_{n_k})$ is such that $a_{n_k}to infty$ and thus can't be Cauchy since necessarily Cauchy sequences are bounded.
add a comment |
up vote
3
down vote
Take a sequence $(a_n)$ such that $a_nto infty$. Then each subsequence $(a_{n_k})$ is such that $a_{n_k}to infty$ and thus can't be Cauchy since necessarily Cauchy sequences are bounded.
add a comment |
up vote
3
down vote
up vote
3
down vote
Take a sequence $(a_n)$ such that $a_nto infty$. Then each subsequence $(a_{n_k})$ is such that $a_{n_k}to infty$ and thus can't be Cauchy since necessarily Cauchy sequences are bounded.
Take a sequence $(a_n)$ such that $a_nto infty$. Then each subsequence $(a_{n_k})$ is such that $a_{n_k}to infty$ and thus can't be Cauchy since necessarily Cauchy sequences are bounded.
answered 5 hours ago
Foobaz John
19.4k41249
19.4k41249
add a comment |
add a comment |
up vote
2
down vote
Take the sequence $1,2,3,4,ldots$ It has no Cauchy subsequence since the distance between any two distinct terms is at least $1$.
add a comment |
up vote
2
down vote
Take the sequence $1,2,3,4,ldots$ It has no Cauchy subsequence since the distance between any two distinct terms is at least $1$.
add a comment |
up vote
2
down vote
up vote
2
down vote
Take the sequence $1,2,3,4,ldots$ It has no Cauchy subsequence since the distance between any two distinct terms is at least $1$.
Take the sequence $1,2,3,4,ldots$ It has no Cauchy subsequence since the distance between any two distinct terms is at least $1$.
answered 6 hours ago
José Carlos Santos
140k19111204
140k19111204
add a comment |
add a comment |
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In the first place, you should only be looking at sequences which are not convergent, right?
– MPW
5 hours ago