How should superscript N and subscript n=1 be thought of in set theory?












-1












$begingroup$


I'm learning set theory while reading a research paper and they use
$$D = {( x^n, l^n)}^N_{n=1}$$



How should this be read? I'm know it would indicate ${( x<^1, l^1)}$ But with the $N$ being a capital $N$ I'm not entirely sure what that would represent.



Thanks for your help!










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  • $begingroup$
    Maybe ${ (x^1, l^1), ldots, (x^N, l^N) }$
    $endgroup$
    – Mauro ALLEGRANZA
    Dec 16 '18 at 16:54


















-1












$begingroup$


I'm learning set theory while reading a research paper and they use
$$D = {( x^n, l^n)}^N_{n=1}$$



How should this be read? I'm know it would indicate ${( x<^1, l^1)}$ But with the $N$ being a capital $N$ I'm not entirely sure what that would represent.



Thanks for your help!










share|cite|improve this question











$endgroup$












  • $begingroup$
    Maybe ${ (x^1, l^1), ldots, (x^N, l^N) }$
    $endgroup$
    – Mauro ALLEGRANZA
    Dec 16 '18 at 16:54
















-1












-1








-1





$begingroup$


I'm learning set theory while reading a research paper and they use
$$D = {( x^n, l^n)}^N_{n=1}$$



How should this be read? I'm know it would indicate ${( x<^1, l^1)}$ But with the $N$ being a capital $N$ I'm not entirely sure what that would represent.



Thanks for your help!










share|cite|improve this question











$endgroup$




I'm learning set theory while reading a research paper and they use
$$D = {( x^n, l^n)}^N_{n=1}$$



How should this be read? I'm know it would indicate ${( x<^1, l^1)}$ But with the $N$ being a capital $N$ I'm not entirely sure what that would represent.



Thanks for your help!







elementary-set-theory notation






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edited Dec 16 '18 at 22:00









Andrés E. Caicedo

65.7k8160250




65.7k8160250










asked Dec 16 '18 at 16:50









KenpachiKenpachi

33




33












  • $begingroup$
    Maybe ${ (x^1, l^1), ldots, (x^N, l^N) }$
    $endgroup$
    – Mauro ALLEGRANZA
    Dec 16 '18 at 16:54




















  • $begingroup$
    Maybe ${ (x^1, l^1), ldots, (x^N, l^N) }$
    $endgroup$
    – Mauro ALLEGRANZA
    Dec 16 '18 at 16:54


















$begingroup$
Maybe ${ (x^1, l^1), ldots, (x^N, l^N) }$
$endgroup$
– Mauro ALLEGRANZA
Dec 16 '18 at 16:54






$begingroup$
Maybe ${ (x^1, l^1), ldots, (x^N, l^N) }$
$endgroup$
– Mauro ALLEGRANZA
Dec 16 '18 at 16:54












2 Answers
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$begingroup$

That means that $D$ is the set of all $(x^n,l^n)$, where $n$ varies from $1$ to $N$.






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$endgroup$





















    0












    $begingroup$

    That index notation whether in the form $sum_{n=1}^Na_n$ or $prod_{n=1}^Na_n$ or ${a_n}_{n=1}^N a_n$ usually means to evaluate for $a_1,a_2,....$ upto $a_{N-2},a_{N-1},a_N$.



    So ${(x^n,l^n)}_{n=1}^N$ probably (but might not) means ${(x^1, l^1),(x^2,l^2),......,(x^N, l^N)}$



    At least that is my guess. Is $N$ used as a constant value elsewhere? Does the $N$ look like the symbol for the natural numbers, $mathbb N$? It might mean something else but I doubt it.






    share|cite|improve this answer









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      2 Answers
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      2 Answers
      2






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      active

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      1












      $begingroup$

      That means that $D$ is the set of all $(x^n,l^n)$, where $n$ varies from $1$ to $N$.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        That means that $D$ is the set of all $(x^n,l^n)$, where $n$ varies from $1$ to $N$.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          That means that $D$ is the set of all $(x^n,l^n)$, where $n$ varies from $1$ to $N$.






          share|cite|improve this answer









          $endgroup$



          That means that $D$ is the set of all $(x^n,l^n)$, where $n$ varies from $1$ to $N$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 16 '18 at 16:53









          user3482749user3482749

          4,296919




          4,296919























              0












              $begingroup$

              That index notation whether in the form $sum_{n=1}^Na_n$ or $prod_{n=1}^Na_n$ or ${a_n}_{n=1}^N a_n$ usually means to evaluate for $a_1,a_2,....$ upto $a_{N-2},a_{N-1},a_N$.



              So ${(x^n,l^n)}_{n=1}^N$ probably (but might not) means ${(x^1, l^1),(x^2,l^2),......,(x^N, l^N)}$



              At least that is my guess. Is $N$ used as a constant value elsewhere? Does the $N$ look like the symbol for the natural numbers, $mathbb N$? It might mean something else but I doubt it.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                That index notation whether in the form $sum_{n=1}^Na_n$ or $prod_{n=1}^Na_n$ or ${a_n}_{n=1}^N a_n$ usually means to evaluate for $a_1,a_2,....$ upto $a_{N-2},a_{N-1},a_N$.



                So ${(x^n,l^n)}_{n=1}^N$ probably (but might not) means ${(x^1, l^1),(x^2,l^2),......,(x^N, l^N)}$



                At least that is my guess. Is $N$ used as a constant value elsewhere? Does the $N$ look like the symbol for the natural numbers, $mathbb N$? It might mean something else but I doubt it.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  That index notation whether in the form $sum_{n=1}^Na_n$ or $prod_{n=1}^Na_n$ or ${a_n}_{n=1}^N a_n$ usually means to evaluate for $a_1,a_2,....$ upto $a_{N-2},a_{N-1},a_N$.



                  So ${(x^n,l^n)}_{n=1}^N$ probably (but might not) means ${(x^1, l^1),(x^2,l^2),......,(x^N, l^N)}$



                  At least that is my guess. Is $N$ used as a constant value elsewhere? Does the $N$ look like the symbol for the natural numbers, $mathbb N$? It might mean something else but I doubt it.






                  share|cite|improve this answer









                  $endgroup$



                  That index notation whether in the form $sum_{n=1}^Na_n$ or $prod_{n=1}^Na_n$ or ${a_n}_{n=1}^N a_n$ usually means to evaluate for $a_1,a_2,....$ upto $a_{N-2},a_{N-1},a_N$.



                  So ${(x^n,l^n)}_{n=1}^N$ probably (but might not) means ${(x^1, l^1),(x^2,l^2),......,(x^N, l^N)}$



                  At least that is my guess. Is $N$ used as a constant value elsewhere? Does the $N$ look like the symbol for the natural numbers, $mathbb N$? It might mean something else but I doubt it.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 16 '18 at 17:12









                  fleabloodfleablood

                  72.2k22687




                  72.2k22687






























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