Confusion arising from the 'infiniteness' of a sequence.












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Let $ X = { i | i in mathbb{N} } = mathbb{N}$. Then is there an infinite sequence $ {t_i } $ with terms from $X$ such that $forall i t_i > t_{i+1} $ viz. a strictly decreasing sequence. If not, how does a formal proof showing this go about?



My thought process was follows:
If $X_m = { i | i in mathbb{N}, i leq m }$ was a finite set, then definitely a strictly decreasing sequence from $X_m$ with the same size$^*$ as $X_m$ would exist - simply by 'reversing' the order in which the elements in $X_m$ had been enumerated - i.e. element 1 from $X_m$ would be the last term of the sequence $ {t_i } $, 2 the second to last and so on. But the same argument of 'reversing' does not work for when $X$ is (countably) infinite. Constructing a sequence (as in the case of $|X| < infty$) is easier than showing that no such sequence can be constructed (as in the case of $|X| = infty$) - a process I am clueless about.



*size - size of a finite set is the cardinality of that set, size of a finite sequence is the number of terms of that sequence.






sequences-and-series infinity







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    0












    $begingroup$


    Let $ X = { i | i in mathbb{N} } = mathbb{N}$. Then is there an infinite sequence $ {t_i } $ with terms from $X$ such that $forall i t_i > t_{i+1} $ viz. a strictly decreasing sequence. If not, how does a formal proof showing this go about?



    My thought process was follows:
    If $X_m = { i | i in mathbb{N}, i leq m }$ was a finite set, then definitely a strictly decreasing sequence from $X_m$ with the same size$^*$ as $X_m$ would exist - simply by 'reversing' the order in which the elements in $X_m$ had been enumerated - i.e. element 1 from $X_m$ would be the last term of the sequence $ {t_i } $, 2 the second to last and so on. But the same argument of 'reversing' does not work for when $X$ is (countably) infinite. Constructing a sequence (as in the case of $|X| < infty$) is easier than showing that no such sequence can be constructed (as in the case of $|X| = infty$) - a process I am clueless about.



    *size - size of a finite set is the cardinality of that set, size of a finite sequence is the number of terms of that sequence.






    sequences-and-series infinity







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Let $ X = { i | i in mathbb{N} } = mathbb{N}$. Then is there an infinite sequence $ {t_i } $ with terms from $X$ such that $forall i t_i > t_{i+1} $ viz. a strictly decreasing sequence. If not, how does a formal proof showing this go about?



      My thought process was follows:
      If $X_m = { i | i in mathbb{N}, i leq m }$ was a finite set, then definitely a strictly decreasing sequence from $X_m$ with the same size$^*$ as $X_m$ would exist - simply by 'reversing' the order in which the elements in $X_m$ had been enumerated - i.e. element 1 from $X_m$ would be the last term of the sequence $ {t_i } $, 2 the second to last and so on. But the same argument of 'reversing' does not work for when $X$ is (countably) infinite. Constructing a sequence (as in the case of $|X| < infty$) is easier than showing that no such sequence can be constructed (as in the case of $|X| = infty$) - a process I am clueless about.



      *size - size of a finite set is the cardinality of that set, size of a finite sequence is the number of terms of that sequence.






      sequences-and-series infinity







      share|cite|improve this question









      $endgroup$




      Let $ X = { i | i in mathbb{N} } = mathbb{N}$. Then is there an infinite sequence $ {t_i } $ with terms from $X$ such that $forall i t_i > t_{i+1} $ viz. a strictly decreasing sequence. If not, how does a formal proof showing this go about?



      My thought process was follows:
      If $X_m = { i | i in mathbb{N}, i leq m }$ was a finite set, then definitely a strictly decreasing sequence from $X_m$ with the same size$^*$ as $X_m$ would exist - simply by 'reversing' the order in which the elements in $X_m$ had been enumerated - i.e. element 1 from $X_m$ would be the last term of the sequence $ {t_i } $, 2 the second to last and so on. But the same argument of 'reversing' does not work for when $X$ is (countably) infinite. Constructing a sequence (as in the case of $|X| < infty$) is easier than showing that no such sequence can be constructed (as in the case of $|X| = infty$) - a process I am clueless about.



      *size - size of a finite set is the cardinality of that set, size of a finite sequence is the number of terms of that sequence.






      sequences-and-series infinity




      sequences-and-series infinity






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      share|cite|improve this question











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      asked Dec 30 '18 at 8:44









      KaindKaind

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      749414






















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          $begingroup$

          Assume that such a sequence $(t_i)_i$ exists.



          $T:={t_imid iinmathbb N}subseteqmathbb N$ is a non-empty subset of well-ordered $mathbb N$ hence has a smallest element $k$.



          Some $iinmathbb N$ exists such that $t_i=k$.



          Then $t_{i+1}<t_i=k$ contradicting that $k$ is smallest element of $T$.






          share|cite|improve this answer











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            1 Answer
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            active

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            1 Answer
            1






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            active

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            active

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            2












            $begingroup$

            Assume that such a sequence $(t_i)_i$ exists.



            $T:={t_imid iinmathbb N}subseteqmathbb N$ is a non-empty subset of well-ordered $mathbb N$ hence has a smallest element $k$.



            Some $iinmathbb N$ exists such that $t_i=k$.



            Then $t_{i+1}<t_i=k$ contradicting that $k$ is smallest element of $T$.






            share|cite|improve this answer











            $endgroup$


















              2












              $begingroup$

              Assume that such a sequence $(t_i)_i$ exists.



              $T:={t_imid iinmathbb N}subseteqmathbb N$ is a non-empty subset of well-ordered $mathbb N$ hence has a smallest element $k$.



              Some $iinmathbb N$ exists such that $t_i=k$.



              Then $t_{i+1}<t_i=k$ contradicting that $k$ is smallest element of $T$.






              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                Assume that such a sequence $(t_i)_i$ exists.



                $T:={t_imid iinmathbb N}subseteqmathbb N$ is a non-empty subset of well-ordered $mathbb N$ hence has a smallest element $k$.



                Some $iinmathbb N$ exists such that $t_i=k$.



                Then $t_{i+1}<t_i=k$ contradicting that $k$ is smallest element of $T$.






                share|cite|improve this answer











                $endgroup$



                Assume that such a sequence $(t_i)_i$ exists.



                $T:={t_imid iinmathbb N}subseteqmathbb N$ is a non-empty subset of well-ordered $mathbb N$ hence has a smallest element $k$.



                Some $iinmathbb N$ exists such that $t_i=k$.



                Then $t_{i+1}<t_i=k$ contradicting that $k$ is smallest element of $T$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 30 '18 at 9:07

























                answered Dec 30 '18 at 8:58









                drhabdrhab

                103k545136




                103k545136






























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