$(x_n)_n$ is a bounded sequence in $(a,b)$

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If $(x_n)_n$ is a bounded sequence in $(a,b)$ and $f (x) colon (a,b) to mathbbR$ is a uniformly continuous function then show that $(f (x_n))_n$ has a convergent subsequence.




I know that the above holds true if $(x_n)_n$ is Cauchy but if only it is bounded then is it true?




real-analysis sequences-and-series




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    So take a convergent subsequence of $x_n$.
    – uniquesolution
    Jul 27 at 9:23














up vote
0
down vote

favorite













If $(x_n)_n$ is a bounded sequence in $(a,b)$ and $f (x) colon (a,b) to mathbbR$ is a uniformly continuous function then show that $(f (x_n))_n$ has a convergent subsequence.




I know that the above holds true if $(x_n)_n$ is Cauchy but if only it is bounded then is it true?




real-analysis sequences-and-series




share|cite|improve this question

















  • 3




    So take a convergent subsequence of $x_n$.
    – uniquesolution
    Jul 27 at 9:23












up vote
0
down vote

favorite









up vote
0
down vote

favorite












If $(x_n)_n$ is a bounded sequence in $(a,b)$ and $f (x) colon (a,b) to mathbbR$ is a uniformly continuous function then show that $(f (x_n))_n$ has a convergent subsequence.




I know that the above holds true if $(x_n)_n$ is Cauchy but if only it is bounded then is it true?




real-analysis sequences-and-series




share|cite|improve this question














If $(x_n)_n$ is a bounded sequence in $(a,b)$ and $f (x) colon (a,b) to mathbbR$ is a uniformly continuous function then show that $(f (x_n))_n$ has a convergent subsequence.




I know that the above holds true if $(x_n)_n$ is Cauchy but if only it is bounded then is it true?





real-analysis sequences-and-series





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 27 at 9:27









Taroccoesbrocco

3,36941331




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asked Jul 27 at 9:21









Shrimon Mukherjee

175




175







  • 3




    So take a convergent subsequence of $x_n$.
    – uniquesolution
    Jul 27 at 9:23












  • 3




    So take a convergent subsequence of $x_n$.
    – uniquesolution
    Jul 27 at 9:23







3




3




So take a convergent subsequence of $x_n$.
– uniquesolution
Jul 27 at 9:23




So take a convergent subsequence of $x_n$.
– uniquesolution
Jul 27 at 9:23










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$(x_n)$ is bounded, hence $(x_n)$ contains a convergent subsequence $(x_n_k)$.



$(x_n_k)$ is Cauchy, hence $(f(x_n_k))$ contains a convergent subsequence, which is a subsequence of $(f(x_n))$.






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    up vote
    3
    down vote













    $(x_n)$ is bounded, hence $(x_n)$ contains a convergent subsequence $(x_n_k)$.



    $(x_n_k)$ is Cauchy, hence $(f(x_n_k))$ contains a convergent subsequence, which is a subsequence of $(f(x_n))$.






    share|cite|improve this answer

























      up vote
      3
      down vote













      $(x_n)$ is bounded, hence $(x_n)$ contains a convergent subsequence $(x_n_k)$.



      $(x_n_k)$ is Cauchy, hence $(f(x_n_k))$ contains a convergent subsequence, which is a subsequence of $(f(x_n))$.






      share|cite|improve this answer























        up vote
        3
        down vote










        up vote
        3
        down vote









        $(x_n)$ is bounded, hence $(x_n)$ contains a convergent subsequence $(x_n_k)$.



        $(x_n_k)$ is Cauchy, hence $(f(x_n_k))$ contains a convergent subsequence, which is a subsequence of $(f(x_n))$.






        share|cite|improve this answer













        $(x_n)$ is bounded, hence $(x_n)$ contains a convergent subsequence $(x_n_k)$.



        $(x_n_k)$ is Cauchy, hence $(f(x_n_k))$ contains a convergent subsequence, which is a subsequence of $(f(x_n))$.







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Jul 27 at 9:28









        Fred

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