If $(x_n)$ is bounded and diverges, then there exist two subsequences of $(x_n)$ that converge to different limits. [duplicate]

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  • Let $(x_n)$ be a bounded but not convergent sequence. Prove that $(x_n)$ has two subsequences converging to different limits.

    3 answers



Is the following argument correct?




If $(x_n)$ is bounded and diverges, then there exist two subsequences
of $(x_n)$ that converge to different limits.




Proof. Assume $(x_n)$ is bounded and diverges, then by Bolzano-Weierstrass theorem it follows that $(x_n)$ contains a subsequence $(x_n_k)$ such that $(x_n_k)tobeta$ for some $betainmathbfR$. Now assume that every subsequence converges to the same limit, but then $(x_n)tobeta$, a contradiction, thus we must have another subsequence $(x_n_r)$ such that $(x_n_r)toalpha$ and $alphaneqbeta$.



$blacksquare$




real-analysis sequences-and-series proof-verification convergence




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marked as duplicate by rtybase, Adrian Keister, Taroccoesbrocco, Leucippus, Xander Henderson Aug 2 at 1:00


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  • I presume this question only really considers sequences in $mathbb R$ $($or $Bbb R^n)$. The proposition is not true in general.
    – Fimpellizieri
    Aug 1 at 21:47














up vote
0
down vote

favorite













This question already has an answer here:



  • Let $(x_n)$ be a bounded but not convergent sequence. Prove that $(x_n)$ has two subsequences converging to different limits.

    3 answers



Is the following argument correct?




If $(x_n)$ is bounded and diverges, then there exist two subsequences
of $(x_n)$ that converge to different limits.




Proof. Assume $(x_n)$ is bounded and diverges, then by Bolzano-Weierstrass theorem it follows that $(x_n)$ contains a subsequence $(x_n_k)$ such that $(x_n_k)tobeta$ for some $betainmathbfR$. Now assume that every subsequence converges to the same limit, but then $(x_n)tobeta$, a contradiction, thus we must have another subsequence $(x_n_r)$ such that $(x_n_r)toalpha$ and $alphaneqbeta$.



$blacksquare$




real-analysis sequences-and-series proof-verification convergence




share|cite|improve this question













marked as duplicate by rtybase, Adrian Keister, Taroccoesbrocco, Leucippus, Xander Henderson Aug 2 at 1:00


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • I presume this question only really considers sequences in $mathbb R$ $($or $Bbb R^n)$. The proposition is not true in general.
    – Fimpellizieri
    Aug 1 at 21:47












up vote
0
down vote

favorite









up vote
0
down vote

favorite












This question already has an answer here:



  • Let $(x_n)$ be a bounded but not convergent sequence. Prove that $(x_n)$ has two subsequences converging to different limits.

    3 answers



Is the following argument correct?




If $(x_n)$ is bounded and diverges, then there exist two subsequences
of $(x_n)$ that converge to different limits.




Proof. Assume $(x_n)$ is bounded and diverges, then by Bolzano-Weierstrass theorem it follows that $(x_n)$ contains a subsequence $(x_n_k)$ such that $(x_n_k)tobeta$ for some $betainmathbfR$. Now assume that every subsequence converges to the same limit, but then $(x_n)tobeta$, a contradiction, thus we must have another subsequence $(x_n_r)$ such that $(x_n_r)toalpha$ and $alphaneqbeta$.



$blacksquare$




real-analysis sequences-and-series proof-verification convergence




share|cite|improve this question














This question already has an answer here:



  • Let $(x_n)$ be a bounded but not convergent sequence. Prove that $(x_n)$ has two subsequences converging to different limits.

    3 answers



Is the following argument correct?




If $(x_n)$ is bounded and diverges, then there exist two subsequences
of $(x_n)$ that converge to different limits.




Proof. Assume $(x_n)$ is bounded and diverges, then by Bolzano-Weierstrass theorem it follows that $(x_n)$ contains a subsequence $(x_n_k)$ such that $(x_n_k)tobeta$ for some $betainmathbfR$. Now assume that every subsequence converges to the same limit, but then $(x_n)tobeta$, a contradiction, thus we must have another subsequence $(x_n_r)$ such that $(x_n_r)toalpha$ and $alphaneqbeta$.



$blacksquare$





This question already has an answer here:



  • Let $(x_n)$ be a bounded but not convergent sequence. Prove that $(x_n)$ has two subsequences converging to different limits.

    3 answers





real-analysis sequences-and-series proof-verification convergence





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share|cite|improve this question




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edited Aug 1 at 21:34









José Carlos Santos

112k1696172




112k1696172









asked Aug 1 at 21:25









Atif Farooq

2,7352824




2,7352824




marked as duplicate by rtybase, Adrian Keister, Taroccoesbrocco, Leucippus, Xander Henderson Aug 2 at 1:00


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






marked as duplicate by rtybase, Adrian Keister, Taroccoesbrocco, Leucippus, Xander Henderson Aug 2 at 1:00


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.













  • I presume this question only really considers sequences in $mathbb R$ $($or $Bbb R^n)$. The proposition is not true in general.
    – Fimpellizieri
    Aug 1 at 21:47
















  • I presume this question only really considers sequences in $mathbb R$ $($or $Bbb R^n)$. The proposition is not true in general.
    – Fimpellizieri
    Aug 1 at 21:47















I presume this question only really considers sequences in $mathbb R$ $($or $Bbb R^n)$. The proposition is not true in general.
– Fimpellizieri
Aug 1 at 21:47




I presume this question only really considers sequences in $mathbb R$ $($or $Bbb R^n)$. The proposition is not true in general.
– Fimpellizieri
Aug 1 at 21:47










1 Answer
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No, it is not correct. You are trying to prove that some convergent subsequence converges to a number different from $beta$. Denying this means that every subsequence either diverges or converges to $beta$ too. But you only considered the second possibility.



You can prove the theorem as follows: since the sequence diverges, $limsup_nx_nneqliminf_nx_n$. But $(x_n)_ninmathbb N$ has a subsequence whose limit is $limsup_nx_n$ and a subsequence whose limit is $liminf_nx_n$.






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    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    5
    down vote



    accepted










    No, it is not correct. You are trying to prove that some convergent subsequence converges to a number different from $beta$. Denying this means that every subsequence either diverges or converges to $beta$ too. But you only considered the second possibility.



    You can prove the theorem as follows: since the sequence diverges, $limsup_nx_nneqliminf_nx_n$. But $(x_n)_ninmathbb N$ has a subsequence whose limit is $limsup_nx_n$ and a subsequence whose limit is $liminf_nx_n$.






    share|cite|improve this answer

























      up vote
      5
      down vote



      accepted










      No, it is not correct. You are trying to prove that some convergent subsequence converges to a number different from $beta$. Denying this means that every subsequence either diverges or converges to $beta$ too. But you only considered the second possibility.



      You can prove the theorem as follows: since the sequence diverges, $limsup_nx_nneqliminf_nx_n$. But $(x_n)_ninmathbb N$ has a subsequence whose limit is $limsup_nx_n$ and a subsequence whose limit is $liminf_nx_n$.






      share|cite|improve this answer























        up vote
        5
        down vote



        accepted







        up vote
        5
        down vote



        accepted






        No, it is not correct. You are trying to prove that some convergent subsequence converges to a number different from $beta$. Denying this means that every subsequence either diverges or converges to $beta$ too. But you only considered the second possibility.



        You can prove the theorem as follows: since the sequence diverges, $limsup_nx_nneqliminf_nx_n$. But $(x_n)_ninmathbb N$ has a subsequence whose limit is $limsup_nx_n$ and a subsequence whose limit is $liminf_nx_n$.






        share|cite|improve this answer













        No, it is not correct. You are trying to prove that some convergent subsequence converges to a number different from $beta$. Denying this means that every subsequence either diverges or converges to $beta$ too. But you only considered the second possibility.



        You can prove the theorem as follows: since the sequence diverges, $limsup_nx_nneqliminf_nx_n$. But $(x_n)_ninmathbb N$ has a subsequence whose limit is $limsup_nx_n$ and a subsequence whose limit is $liminf_nx_n$.







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Aug 1 at 21:30









        José Carlos Santos

        112k1696172




        112k1696172












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