simple question about properly divergent sequence












2












$begingroup$


Is it true that if a sequence ${x_{n}}$ has no properly divergent subsequence, then the sequence is bounded?



I think it is true intuitively considering subsequential limit but i’m not sure about that.



Give me some counter-example or comment. Thank you.






real-analysis sequences-and-series







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    If the sequence is not bounded then either there is a subsequence strictly increasing to $infty$ or there is a subsequence strictly decreasing to $-infty$
    $endgroup$
    – Kavi Rama Murthy
    Jan 3 at 6:33










  • $begingroup$
    Properly divergent sequence? What is the meaning of this, please?
    $endgroup$
    – dmtri
    Jan 3 at 6:49










  • $begingroup$
    @KaviRamaMurthy Actually, I tried to show your statement in another way using the assumption as above leads a contradiction.
    $endgroup$
    – Primavera
    Jan 4 at 7:25










  • $begingroup$
    @dmtri diverges at plus/minus infinity not oscillate
    $endgroup$
    – Primavera
    Jan 4 at 7:26
















2












$begingroup$


Is it true that if a sequence ${x_{n}}$ has no properly divergent subsequence, then the sequence is bounded?



I think it is true intuitively considering subsequential limit but i’m not sure about that.



Give me some counter-example or comment. Thank you.






real-analysis sequences-and-series







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    If the sequence is not bounded then either there is a subsequence strictly increasing to $infty$ or there is a subsequence strictly decreasing to $-infty$
    $endgroup$
    – Kavi Rama Murthy
    Jan 3 at 6:33










  • $begingroup$
    Properly divergent sequence? What is the meaning of this, please?
    $endgroup$
    – dmtri
    Jan 3 at 6:49










  • $begingroup$
    @KaviRamaMurthy Actually, I tried to show your statement in another way using the assumption as above leads a contradiction.
    $endgroup$
    – Primavera
    Jan 4 at 7:25










  • $begingroup$
    @dmtri diverges at plus/minus infinity not oscillate
    $endgroup$
    – Primavera
    Jan 4 at 7:26














2












2








2





$begingroup$


Is it true that if a sequence ${x_{n}}$ has no properly divergent subsequence, then the sequence is bounded?



I think it is true intuitively considering subsequential limit but i’m not sure about that.



Give me some counter-example or comment. Thank you.






real-analysis sequences-and-series







share|cite|improve this question









$endgroup$




Is it true that if a sequence ${x_{n}}$ has no properly divergent subsequence, then the sequence is bounded?



I think it is true intuitively considering subsequential limit but i’m not sure about that.



Give me some counter-example or comment. Thank you.






real-analysis sequences-and-series




real-analysis sequences-and-series






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 3 at 6:31









PrimaveraPrimavera

31619




31619








  • 1




    $begingroup$
    If the sequence is not bounded then either there is a subsequence strictly increasing to $infty$ or there is a subsequence strictly decreasing to $-infty$
    $endgroup$
    – Kavi Rama Murthy
    Jan 3 at 6:33










  • $begingroup$
    Properly divergent sequence? What is the meaning of this, please?
    $endgroup$
    – dmtri
    Jan 3 at 6:49










  • $begingroup$
    @KaviRamaMurthy Actually, I tried to show your statement in another way using the assumption as above leads a contradiction.
    $endgroup$
    – Primavera
    Jan 4 at 7:25










  • $begingroup$
    @dmtri diverges at plus/minus infinity not oscillate
    $endgroup$
    – Primavera
    Jan 4 at 7:26














  • 1




    $begingroup$
    If the sequence is not bounded then either there is a subsequence strictly increasing to $infty$ or there is a subsequence strictly decreasing to $-infty$
    $endgroup$
    – Kavi Rama Murthy
    Jan 3 at 6:33










  • $begingroup$
    Properly divergent sequence? What is the meaning of this, please?
    $endgroup$
    – dmtri
    Jan 3 at 6:49










  • $begingroup$
    @KaviRamaMurthy Actually, I tried to show your statement in another way using the assumption as above leads a contradiction.
    $endgroup$
    – Primavera
    Jan 4 at 7:25










  • $begingroup$
    @dmtri diverges at plus/minus infinity not oscillate
    $endgroup$
    – Primavera
    Jan 4 at 7:26








1




1




$begingroup$
If the sequence is not bounded then either there is a subsequence strictly increasing to $infty$ or there is a subsequence strictly decreasing to $-infty$
$endgroup$
– Kavi Rama Murthy
Jan 3 at 6:33




$begingroup$
If the sequence is not bounded then either there is a subsequence strictly increasing to $infty$ or there is a subsequence strictly decreasing to $-infty$
$endgroup$
– Kavi Rama Murthy
Jan 3 at 6:33












$begingroup$
Properly divergent sequence? What is the meaning of this, please?
$endgroup$
– dmtri
Jan 3 at 6:49




$begingroup$
Properly divergent sequence? What is the meaning of this, please?
$endgroup$
– dmtri
Jan 3 at 6:49












$begingroup$
@KaviRamaMurthy Actually, I tried to show your statement in another way using the assumption as above leads a contradiction.
$endgroup$
– Primavera
Jan 4 at 7:25




$begingroup$
@KaviRamaMurthy Actually, I tried to show your statement in another way using the assumption as above leads a contradiction.
$endgroup$
– Primavera
Jan 4 at 7:25












$begingroup$
@dmtri diverges at plus/minus infinity not oscillate
$endgroup$
– Primavera
Jan 4 at 7:26




$begingroup$
@dmtri diverges at plus/minus infinity not oscillate
$endgroup$
– Primavera
Jan 4 at 7:26










1 Answer
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1












$begingroup$

Equivalently, if a sequence is unbounded, it has a sub-sequence strictly increasing to $infty$ or strictly decreasing to $-infty.$



Let $(x_j)_{jin Bbb N}$ be an unbounded sequence in $Bbb R$.



Let $n_1$ be the least $nin Bbb N$ such that $|x_n|>1.$ Define $n_{j+1}$ recursively as follows: Let $$y_j=max (j+1, max { |x_i|:1leq ileq n_j}).$$ Let $n_{j+1}$ be the least $nin Bbb N$ such that $|a_n|>y_j.$



For every $jin Bbb N$ we have $jleq n_j<n_{j+1}$ and $j<|x_{n_j}|<|x_{n_{j+1}}|.$



Let $S={jin Bbb N: x_{n_j}>j}$ and $T={jin Bbb N: x_{n_j}<-j}.$ We have $Scup T=Bbb N.$



If $S$ is an infinite set then the sequence $(x_{n_j})_{jin S}$ strictly increases to $infty.$



If $T$ is an infinite set then $(x_{n_j})_{jin T}$ strictly decreases to $-infty.$






share|cite|improve this answer











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    $begingroup$

    Equivalently, if a sequence is unbounded, it has a sub-sequence strictly increasing to $infty$ or strictly decreasing to $-infty.$



    Let $(x_j)_{jin Bbb N}$ be an unbounded sequence in $Bbb R$.



    Let $n_1$ be the least $nin Bbb N$ such that $|x_n|>1.$ Define $n_{j+1}$ recursively as follows: Let $$y_j=max (j+1, max { |x_i|:1leq ileq n_j}).$$ Let $n_{j+1}$ be the least $nin Bbb N$ such that $|a_n|>y_j.$



    For every $jin Bbb N$ we have $jleq n_j<n_{j+1}$ and $j<|x_{n_j}|<|x_{n_{j+1}}|.$



    Let $S={jin Bbb N: x_{n_j}>j}$ and $T={jin Bbb N: x_{n_j}<-j}.$ We have $Scup T=Bbb N.$



    If $S$ is an infinite set then the sequence $(x_{n_j})_{jin S}$ strictly increases to $infty.$



    If $T$ is an infinite set then $(x_{n_j})_{jin T}$ strictly decreases to $-infty.$






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Equivalently, if a sequence is unbounded, it has a sub-sequence strictly increasing to $infty$ or strictly decreasing to $-infty.$



      Let $(x_j)_{jin Bbb N}$ be an unbounded sequence in $Bbb R$.



      Let $n_1$ be the least $nin Bbb N$ such that $|x_n|>1.$ Define $n_{j+1}$ recursively as follows: Let $$y_j=max (j+1, max { |x_i|:1leq ileq n_j}).$$ Let $n_{j+1}$ be the least $nin Bbb N$ such that $|a_n|>y_j.$



      For every $jin Bbb N$ we have $jleq n_j<n_{j+1}$ and $j<|x_{n_j}|<|x_{n_{j+1}}|.$



      Let $S={jin Bbb N: x_{n_j}>j}$ and $T={jin Bbb N: x_{n_j}<-j}.$ We have $Scup T=Bbb N.$



      If $S$ is an infinite set then the sequence $(x_{n_j})_{jin S}$ strictly increases to $infty.$



      If $T$ is an infinite set then $(x_{n_j})_{jin T}$ strictly decreases to $-infty.$






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Equivalently, if a sequence is unbounded, it has a sub-sequence strictly increasing to $infty$ or strictly decreasing to $-infty.$



        Let $(x_j)_{jin Bbb N}$ be an unbounded sequence in $Bbb R$.



        Let $n_1$ be the least $nin Bbb N$ such that $|x_n|>1.$ Define $n_{j+1}$ recursively as follows: Let $$y_j=max (j+1, max { |x_i|:1leq ileq n_j}).$$ Let $n_{j+1}$ be the least $nin Bbb N$ such that $|a_n|>y_j.$



        For every $jin Bbb N$ we have $jleq n_j<n_{j+1}$ and $j<|x_{n_j}|<|x_{n_{j+1}}|.$



        Let $S={jin Bbb N: x_{n_j}>j}$ and $T={jin Bbb N: x_{n_j}<-j}.$ We have $Scup T=Bbb N.$



        If $S$ is an infinite set then the sequence $(x_{n_j})_{jin S}$ strictly increases to $infty.$



        If $T$ is an infinite set then $(x_{n_j})_{jin T}$ strictly decreases to $-infty.$






        share|cite|improve this answer











        $endgroup$



        Equivalently, if a sequence is unbounded, it has a sub-sequence strictly increasing to $infty$ or strictly decreasing to $-infty.$



        Let $(x_j)_{jin Bbb N}$ be an unbounded sequence in $Bbb R$.



        Let $n_1$ be the least $nin Bbb N$ such that $|x_n|>1.$ Define $n_{j+1}$ recursively as follows: Let $$y_j=max (j+1, max { |x_i|:1leq ileq n_j}).$$ Let $n_{j+1}$ be the least $nin Bbb N$ such that $|a_n|>y_j.$



        For every $jin Bbb N$ we have $jleq n_j<n_{j+1}$ and $j<|x_{n_j}|<|x_{n_{j+1}}|.$



        Let $S={jin Bbb N: x_{n_j}>j}$ and $T={jin Bbb N: x_{n_j}<-j}.$ We have $Scup T=Bbb N.$



        If $S$ is an infinite set then the sequence $(x_{n_j})_{jin S}$ strictly increases to $infty.$



        If $T$ is an infinite set then $(x_{n_j})_{jin T}$ strictly decreases to $-infty.$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 3 at 9:15

























        answered Jan 3 at 9:10









        DanielWainfleetDanielWainfleet

        35.7k31648




        35.7k31648






























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