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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  • In the first place, you should only be looking at sequences which are not convergent, right?
    – MPW
    5 hours ago















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







share|cite|improve this question
























  • In the first place, you should only be looking at sequences which are not convergent, right?
    – MPW
    5 hours ago













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







share|cite|improve this question
















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






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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


















  • 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










3 Answers
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Take $a_n=n$. Then for any subsequence $n_k$, $|n_k-n_{k-1}|geq 1$. So, it has no Cauchy sub-sequence.






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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
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      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
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        3 Answers
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        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.






        share|cite|improve this answer








        New contributor




        John_Wick is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
        Check out our Code of Conduct.






















          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.






          share|cite|improve this answer








          New contributor




          John_Wick is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.




















            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.






            share|cite|improve this answer








            New contributor




            John_Wick is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.









            Take $a_n=n$. Then for any subsequence $n_k$, $|n_k-n_{k-1}|geq 1$. So, it has no Cauchy sub-sequence.







            share|cite|improve this answer








            New contributor




            John_Wick is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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            share|cite|improve this answer



            share|cite|improve this answer






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            answered 6 hours ago









            John_Wick

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            3318




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                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.






                share|cite|improve this answer

























                  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.






                  share|cite|improve this answer























                    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.






                    share|cite|improve this answer












                    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.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 5 hours ago









                    Foobaz John

                    19.4k41249




                    19.4k41249






















                        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 $1$.






                        share|cite|improve this answer

























                          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 $1$.






                          share|cite|improve this answer























                            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 $1$.






                            share|cite|improve this answer












                            Take the sequence $1,2,3,4,ldots$ It has no Cauchy subsequence since the distance between any two distinct terms is at least $1$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 6 hours ago









                            José Carlos Santos

                            140k19111204




                            140k19111204






























                                 

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