If $sup Anotin A$ then $A$ contains a countably infinite subset












1














I am required to show that if a bounded non empty set $Asubseteq mathbf{R}$ is such that $sup Anotin A$, then $A$ contains a countably infinites subset.



Now my idea is that for each $kinmathbf{N}$ there would exist $a_kin A$ such that $sup A-frac{1}{k}<a_k$ with the required set being $H = {a_1,a_2,dots}$ but how can i modify this construction to ensure that all elements of $H$ are distinct?






real-analysis soft-question







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  • Why do you want the elements to be distinct?
    – drhab
    Nov 24 at 13:05










  • Simply throw out all of the ones that coincide. If there were finitely many such, then since your sequence converges, it must be eventually constant (if there are infinitely many of two different values, then it has a (constant) subsequence converging to each), but it converges to $mathop{mathrm{sup}}A$, so this would mean that $mathop{mathrm{Sup}}A in A$, so we must have infinitely many distinct $a_n$.
    – user3482749
    Nov 24 at 13:07
















1














I am required to show that if a bounded non empty set $Asubseteq mathbf{R}$ is such that $sup Anotin A$, then $A$ contains a countably infinites subset.



Now my idea is that for each $kinmathbf{N}$ there would exist $a_kin A$ such that $sup A-frac{1}{k}<a_k$ with the required set being $H = {a_1,a_2,dots}$ but how can i modify this construction to ensure that all elements of $H$ are distinct?






real-analysis soft-question







share|cite|improve this question






















  • Why do you want the elements to be distinct?
    – drhab
    Nov 24 at 13:05










  • Simply throw out all of the ones that coincide. If there were finitely many such, then since your sequence converges, it must be eventually constant (if there are infinitely many of two different values, then it has a (constant) subsequence converging to each), but it converges to $mathop{mathrm{sup}}A$, so this would mean that $mathop{mathrm{Sup}}A in A$, so we must have infinitely many distinct $a_n$.
    – user3482749
    Nov 24 at 13:07














1












1








1







I am required to show that if a bounded non empty set $Asubseteq mathbf{R}$ is such that $sup Anotin A$, then $A$ contains a countably infinites subset.



Now my idea is that for each $kinmathbf{N}$ there would exist $a_kin A$ such that $sup A-frac{1}{k}<a_k$ with the required set being $H = {a_1,a_2,dots}$ but how can i modify this construction to ensure that all elements of $H$ are distinct?






real-analysis soft-question







share|cite|improve this question













I am required to show that if a bounded non empty set $Asubseteq mathbf{R}$ is such that $sup Anotin A$, then $A$ contains a countably infinites subset.



Now my idea is that for each $kinmathbf{N}$ there would exist $a_kin A$ such that $sup A-frac{1}{k}<a_k$ with the required set being $H = {a_1,a_2,dots}$ but how can i modify this construction to ensure that all elements of $H$ are distinct?






real-analysis soft-question




real-analysis soft-question






share|cite|improve this question













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asked Nov 24 at 13:03









Atif Farooq

3,1422825




3,1422825












  • Why do you want the elements to be distinct?
    – drhab
    Nov 24 at 13:05










  • Simply throw out all of the ones that coincide. If there were finitely many such, then since your sequence converges, it must be eventually constant (if there are infinitely many of two different values, then it has a (constant) subsequence converging to each), but it converges to $mathop{mathrm{sup}}A$, so this would mean that $mathop{mathrm{Sup}}A in A$, so we must have infinitely many distinct $a_n$.
    – user3482749
    Nov 24 at 13:07


















  • Why do you want the elements to be distinct?
    – drhab
    Nov 24 at 13:05










  • Simply throw out all of the ones that coincide. If there were finitely many such, then since your sequence converges, it must be eventually constant (if there are infinitely many of two different values, then it has a (constant) subsequence converging to each), but it converges to $mathop{mathrm{sup}}A$, so this would mean that $mathop{mathrm{Sup}}A in A$, so we must have infinitely many distinct $a_n$.
    – user3482749
    Nov 24 at 13:07
















Why do you want the elements to be distinct?
– drhab
Nov 24 at 13:05




Why do you want the elements to be distinct?
– drhab
Nov 24 at 13:05












Simply throw out all of the ones that coincide. If there were finitely many such, then since your sequence converges, it must be eventually constant (if there are infinitely many of two different values, then it has a (constant) subsequence converging to each), but it converges to $mathop{mathrm{sup}}A$, so this would mean that $mathop{mathrm{Sup}}A in A$, so we must have infinitely many distinct $a_n$.
– user3482749
Nov 24 at 13:07




Simply throw out all of the ones that coincide. If there were finitely many such, then since your sequence converges, it must be eventually constant (if there are infinitely many of two different values, then it has a (constant) subsequence converging to each), but it converges to $mathop{mathrm{sup}}A$, so this would mean that $mathop{mathrm{Sup}}A in A$, so we must have infinitely many distinct $a_n$.
– user3482749
Nov 24 at 13:07










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Try this: if a set $AsubsetBbb R$ is finite and nonempty then $sup Ain A$.






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    1 Answer
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    1 Answer
    1






    active

    oldest

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    active

    oldest

    votes






    active

    oldest

    votes









    2














    Try this: if a set $AsubsetBbb R$ is finite and nonempty then $sup Ain A$.






    share|cite|improve this answer


























      2














      Try this: if a set $AsubsetBbb R$ is finite and nonempty then $sup Ain A$.






      share|cite|improve this answer
























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        2








        2






        Try this: if a set $AsubsetBbb R$ is finite and nonempty then $sup Ain A$.






        share|cite|improve this answer












        Try this: if a set $AsubsetBbb R$ is finite and nonempty then $sup Ain A$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 24 at 13:06









        Masacroso

        12.8k41746




        12.8k41746






























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