How should superscript N and subscript n=1 be thought of in set theory?
$begingroup$
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!
elementary-set-theory notation
edited Dec 16 '18 at 22:00
Andrés E. Caicedo
65.7k8160250
asked Dec 16 '18 at 16:50
KenpachiKenpachi
33
$endgroup$
add a comment |
$begingroup$
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!
elementary-set-theory notation
edited Dec 16 '18 at 22:00
Andrés E. Caicedo
65.7k8160250
asked Dec 16 '18 at 16:50
KenpachiKenpachi
33
$endgroup$
$begingroup$
Maybe ${ (x^1, l^1), ldots, (x^N, l^N) }$
$endgroup$
– Mauro ALLEGRANZA
Dec 16 '18 at 16:54
add a comment |
$begingroup$
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!
elementary-set-theory notation
edited Dec 16 '18 at 22:00
Andrés E. Caicedo
65.7k8160250
asked Dec 16 '18 at 16:50
KenpachiKenpachi
33
$endgroup$
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
elementary-set-theory notation
edited Dec 16 '18 at 22:00
Andrés E. Caicedo
65.7k8160250
asked Dec 16 '18 at 16:50
KenpachiKenpachi
33
edited Dec 16 '18 at 22:00
Andrés E. Caicedo
65.7k8160250
asked Dec 16 '18 at 16:50
KenpachiKenpachi
33
edited Dec 16 '18 at 22:00
Andrés E. Caicedo
65.7k8160250
edited Dec 16 '18 at 22:00
Andrés E. Caicedo
65.7k8160250
edited Dec 16 '18 at 22:00
Andrés E. Caicedo
65.7k8160250
65.7k8160250
asked Dec 16 '18 at 16:50
KenpachiKenpachi
33
asked Dec 16 '18 at 16:50
KenpachiKenpachi
33
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
add a comment |
$begingroup$
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
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
That means that $D$ is the set of all $(x^n,l^n)$, where $n$ varies from $1$ to $N$.
answered Dec 16 '18 at 16:53
user3482749user3482749
4,296919
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 16 '18 at 17:12
fleabloodfleablood
72.2k22687
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
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votes
$begingroup$
That means that $D$ is the set of all $(x^n,l^n)$, where $n$ varies from $1$ to $N$.
answered Dec 16 '18 at 16:53
user3482749user3482749
4,296919
$endgroup$
add a comment |
$begingroup$
That means that $D$ is the set of all $(x^n,l^n)$, where $n$ varies from $1$ to $N$.
answered Dec 16 '18 at 16:53
user3482749user3482749
4,296919
$endgroup$
add a comment |
$begingroup$
That means that $D$ is the set of all $(x^n,l^n)$, where $n$ varies from $1$ to $N$.
answered Dec 16 '18 at 16:53
user3482749user3482749
4,296919
$endgroup$
That means that $D$ is the set of all $(x^n,l^n)$, where $n$ varies from $1$ to $N$.
answered Dec 16 '18 at 16:53
user3482749user3482749
4,296919
answered Dec 16 '18 at 16:53
user3482749user3482749
4,296919
answered Dec 16 '18 at 16:53
user3482749user3482749
4,296919
answered Dec 16 '18 at 16:53
user3482749user3482749
4,296919
4,296919
add a comment |
add a comment |
$begingroup$
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.
answered Dec 16 '18 at 17:12
fleabloodfleablood
72.2k22687
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 16 '18 at 17:12
fleabloodfleablood
72.2k22687
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 16 '18 at 17:12
fleabloodfleablood
72.2k22687
$endgroup$
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.
answered Dec 16 '18 at 17:12
fleabloodfleablood
72.2k22687
answered Dec 16 '18 at 17:12
fleabloodfleablood
72.2k22687
answered Dec 16 '18 at 17:12
fleabloodfleablood
72.2k22687
answered Dec 16 '18 at 17:12
fleabloodfleablood
72.2k22687
72.2k22687
add a comment |
add a comment |
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$begingroup$
Maybe ${ (x^1, l^1), ldots, (x^N, l^N) }$
$endgroup$
– Mauro ALLEGRANZA
Dec 16 '18 at 16:54