Defining a sequence [on hold]











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Just a quick question.



Can a sequence, say $s_n$ be defined as a union of two or more sequences, say $s_1,s_2,s_3$?






real-analysis sequences-and-series







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put on hold as unclear what you're asking by Martin R, vrugtehagel, Mark, Vasya, Jyrki Lahtonen 22 hours ago


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.















  • You can define whatever you like.
    – Martin R
    yesterday










  • but the concept of sequence defined without ordering
    – ehsan eskandari
    yesterday















up vote
0
down vote

favorite












Just a quick question.



Can a sequence, say $s_n$ be defined as a union of two or more sequences, say $s_1,s_2,s_3$?






real-analysis sequences-and-series







share|cite|improve this question













put on hold as unclear what you're asking by Martin R, vrugtehagel, Mark, Vasya, Jyrki Lahtonen 22 hours ago


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.















  • You can define whatever you like.
    – Martin R
    yesterday










  • but the concept of sequence defined without ordering
    – ehsan eskandari
    yesterday













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Just a quick question.



Can a sequence, say $s_n$ be defined as a union of two or more sequences, say $s_1,s_2,s_3$?






real-analysis sequences-and-series







share|cite|improve this question













Just a quick question.



Can a sequence, say $s_n$ be defined as a union of two or more sequences, say $s_1,s_2,s_3$?






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 yesterday









Allorja

298




298




put on hold as unclear what you're asking by Martin R, vrugtehagel, Mark, Vasya, Jyrki Lahtonen 22 hours ago


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.






put on hold as unclear what you're asking by Martin R, vrugtehagel, Mark, Vasya, Jyrki Lahtonen 22 hours ago


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.














  • You can define whatever you like.
    – Martin R
    yesterday










  • but the concept of sequence defined without ordering
    – ehsan eskandari
    yesterday


















  • You can define whatever you like.
    – Martin R
    yesterday










  • but the concept of sequence defined without ordering
    – ehsan eskandari
    yesterday
















You can define whatever you like.
– Martin R
yesterday




You can define whatever you like.
– Martin R
yesterday












but the concept of sequence defined without ordering
– ehsan eskandari
yesterday




but the concept of sequence defined without ordering
– ehsan eskandari
yesterday










2 Answers
2






active

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0
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A sequence is a function



$$s_n: Asubseteqmathbb{N} to mathbb{R}$$



then we need to specify what we mean with union of $s_1$, $s_2$, $s_3$.



Therefore, if we are referring to the definition used for the union of sets, we can't apply that to sequences since those are not sets but functions.






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    It should be noted that a sequence is a range of a function like $f:N rightarrow X$(where $X$ is any space)and if we put $f(n)=s_n$ then {$s_n$} is a sequence.Hence each sequence is a countable set...now if{$a_n$},{$b_n$},{$c_n$} are sequences then the union of them is acountable set and hence it is asequence






    share|cite|improve this answer





















    • This is not true. A sequence is not the range of a mapping of the natural numbers into some set; it is the mapping itself. It is therefore also incorrect to say that a sequence is a countable set. Even though the union of the images of two sequences is countably infinite and may therefore be put in one-to-one correspondence with the natural numbers, there is no canonical bijection from the natural numbers to this union of sets.
      – Qeeko
      yesterday










    • yeah......but we exactly we show a sequence with{$s_n$} where $S:N rightarrow X$ and $S(n)=s_n$
      – ehsan eskandari
      yesterday












    • I cannot make sense of your reply.
      – Qeeko
      yesterday










    • I get an example right now
      – ehsan eskandari
      yesterday










    • We know that {$2n$},{$4n+10$},...are sequences.in fact the function $f:N rightarrow R$ s.t $f(n)=2n$.......so each countable set is a sequence
      – ehsan eskandari
      yesterday


















    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    0
    down vote













    A sequence is a function



    $$s_n: Asubseteqmathbb{N} to mathbb{R}$$



    then we need to specify what we mean with union of $s_1$, $s_2$, $s_3$.



    Therefore, if we are referring to the definition used for the union of sets, we can't apply that to sequences since those are not sets but functions.






    share|cite|improve this answer

























      up vote
      0
      down vote













      A sequence is a function



      $$s_n: Asubseteqmathbb{N} to mathbb{R}$$



      then we need to specify what we mean with union of $s_1$, $s_2$, $s_3$.



      Therefore, if we are referring to the definition used for the union of sets, we can't apply that to sequences since those are not sets but functions.






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        A sequence is a function



        $$s_n: Asubseteqmathbb{N} to mathbb{R}$$



        then we need to specify what we mean with union of $s_1$, $s_2$, $s_3$.



        Therefore, if we are referring to the definition used for the union of sets, we can't apply that to sequences since those are not sets but functions.






        share|cite|improve this answer












        A sequence is a function



        $$s_n: Asubseteqmathbb{N} to mathbb{R}$$



        then we need to specify what we mean with union of $s_1$, $s_2$, $s_3$.



        Therefore, if we are referring to the definition used for the union of sets, we can't apply that to sequences since those are not sets but functions.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered yesterday









        gimusi

        85.5k74294




        85.5k74294






















            up vote
            0
            down vote













            It should be noted that a sequence is a range of a function like $f:N rightarrow X$(where $X$ is any space)and if we put $f(n)=s_n$ then {$s_n$} is a sequence.Hence each sequence is a countable set...now if{$a_n$},{$b_n$},{$c_n$} are sequences then the union of them is acountable set and hence it is asequence






            share|cite|improve this answer





















            • This is not true. A sequence is not the range of a mapping of the natural numbers into some set; it is the mapping itself. It is therefore also incorrect to say that a sequence is a countable set. Even though the union of the images of two sequences is countably infinite and may therefore be put in one-to-one correspondence with the natural numbers, there is no canonical bijection from the natural numbers to this union of sets.
              – Qeeko
              yesterday










            • yeah......but we exactly we show a sequence with{$s_n$} where $S:N rightarrow X$ and $S(n)=s_n$
              – ehsan eskandari
              yesterday












            • I cannot make sense of your reply.
              – Qeeko
              yesterday










            • I get an example right now
              – ehsan eskandari
              yesterday










            • We know that {$2n$},{$4n+10$},...are sequences.in fact the function $f:N rightarrow R$ s.t $f(n)=2n$.......so each countable set is a sequence
              – ehsan eskandari
              yesterday















            up vote
            0
            down vote













            It should be noted that a sequence is a range of a function like $f:N rightarrow X$(where $X$ is any space)and if we put $f(n)=s_n$ then {$s_n$} is a sequence.Hence each sequence is a countable set...now if{$a_n$},{$b_n$},{$c_n$} are sequences then the union of them is acountable set and hence it is asequence






            share|cite|improve this answer





















            • This is not true. A sequence is not the range of a mapping of the natural numbers into some set; it is the mapping itself. It is therefore also incorrect to say that a sequence is a countable set. Even though the union of the images of two sequences is countably infinite and may therefore be put in one-to-one correspondence with the natural numbers, there is no canonical bijection from the natural numbers to this union of sets.
              – Qeeko
              yesterday










            • yeah......but we exactly we show a sequence with{$s_n$} where $S:N rightarrow X$ and $S(n)=s_n$
              – ehsan eskandari
              yesterday












            • I cannot make sense of your reply.
              – Qeeko
              yesterday










            • I get an example right now
              – ehsan eskandari
              yesterday










            • We know that {$2n$},{$4n+10$},...are sequences.in fact the function $f:N rightarrow R$ s.t $f(n)=2n$.......so each countable set is a sequence
              – ehsan eskandari
              yesterday













            up vote
            0
            down vote










            up vote
            0
            down vote









            It should be noted that a sequence is a range of a function like $f:N rightarrow X$(where $X$ is any space)and if we put $f(n)=s_n$ then {$s_n$} is a sequence.Hence each sequence is a countable set...now if{$a_n$},{$b_n$},{$c_n$} are sequences then the union of them is acountable set and hence it is asequence






            share|cite|improve this answer












            It should be noted that a sequence is a range of a function like $f:N rightarrow X$(where $X$ is any space)and if we put $f(n)=s_n$ then {$s_n$} is a sequence.Hence each sequence is a countable set...now if{$a_n$},{$b_n$},{$c_n$} are sequences then the union of them is acountable set and hence it is asequence







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered yesterday









            ehsan eskandari

            885




            885












            • This is not true. A sequence is not the range of a mapping of the natural numbers into some set; it is the mapping itself. It is therefore also incorrect to say that a sequence is a countable set. Even though the union of the images of two sequences is countably infinite and may therefore be put in one-to-one correspondence with the natural numbers, there is no canonical bijection from the natural numbers to this union of sets.
              – Qeeko
              yesterday










            • yeah......but we exactly we show a sequence with{$s_n$} where $S:N rightarrow X$ and $S(n)=s_n$
              – ehsan eskandari
              yesterday












            • I cannot make sense of your reply.
              – Qeeko
              yesterday










            • I get an example right now
              – ehsan eskandari
              yesterday










            • We know that {$2n$},{$4n+10$},...are sequences.in fact the function $f:N rightarrow R$ s.t $f(n)=2n$.......so each countable set is a sequence
              – ehsan eskandari
              yesterday


















            • This is not true. A sequence is not the range of a mapping of the natural numbers into some set; it is the mapping itself. It is therefore also incorrect to say that a sequence is a countable set. Even though the union of the images of two sequences is countably infinite and may therefore be put in one-to-one correspondence with the natural numbers, there is no canonical bijection from the natural numbers to this union of sets.
              – Qeeko
              yesterday










            • yeah......but we exactly we show a sequence with{$s_n$} where $S:N rightarrow X$ and $S(n)=s_n$
              – ehsan eskandari
              yesterday












            • I cannot make sense of your reply.
              – Qeeko
              yesterday










            • I get an example right now
              – ehsan eskandari
              yesterday










            • We know that {$2n$},{$4n+10$},...are sequences.in fact the function $f:N rightarrow R$ s.t $f(n)=2n$.......so each countable set is a sequence
              – ehsan eskandari
              yesterday
















            This is not true. A sequence is not the range of a mapping of the natural numbers into some set; it is the mapping itself. It is therefore also incorrect to say that a sequence is a countable set. Even though the union of the images of two sequences is countably infinite and may therefore be put in one-to-one correspondence with the natural numbers, there is no canonical bijection from the natural numbers to this union of sets.
            – Qeeko
            yesterday




            This is not true. A sequence is not the range of a mapping of the natural numbers into some set; it is the mapping itself. It is therefore also incorrect to say that a sequence is a countable set. Even though the union of the images of two sequences is countably infinite and may therefore be put in one-to-one correspondence with the natural numbers, there is no canonical bijection from the natural numbers to this union of sets.
            – Qeeko
            yesterday












            yeah......but we exactly we show a sequence with{$s_n$} where $S:N rightarrow X$ and $S(n)=s_n$
            – ehsan eskandari
            yesterday






            yeah......but we exactly we show a sequence with{$s_n$} where $S:N rightarrow X$ and $S(n)=s_n$
            – ehsan eskandari
            yesterday














            I cannot make sense of your reply.
            – Qeeko
            yesterday




            I cannot make sense of your reply.
            – Qeeko
            yesterday












            I get an example right now
            – ehsan eskandari
            yesterday




            I get an example right now
            – ehsan eskandari
            yesterday












            We know that {$2n$},{$4n+10$},...are sequences.in fact the function $f:N rightarrow R$ s.t $f(n)=2n$.......so each countable set is a sequence
            – ehsan eskandari
            yesterday




            We know that {$2n$},{$4n+10$},...are sequences.in fact the function $f:N rightarrow R$ s.t $f(n)=2n$.......so each countable set is a sequence
            – ehsan eskandari
            yesterday



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