Set notation what does the bar symbol mean?












1












$begingroup$


In set notation could somebody explain the meaning of $mid$ in the equation below please? How does it read?



I read it as $s$ and $j$ are an element of $E$ but what does the $j mid$ mean?



summation



Similarly in this equation what does the comma mean?



flow






elementary-set-theory notation







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Short answer: such that. Respectively, $j$ such that the ordered pair $(j,i)$ is in $E$, $j$ such that the ordered pair $(s,j)$ is in $E$.
    $endgroup$
    – Steven Wagter
    Dec 26 '18 at 12:48








  • 2




    $begingroup$
    See Set-builder notation.
    $endgroup$
    – Mauro ALLEGRANZA
    Dec 26 '18 at 12:53






  • 1




    $begingroup$
    @StevenWagter: You should post your comment as an answer. (And explain why commas are used, and often omitted, in lists.)
    $endgroup$
    – John Bentin
    Dec 26 '18 at 13:03
















1












$begingroup$


In set notation could somebody explain the meaning of $mid$ in the equation below please? How does it read?



I read it as $s$ and $j$ are an element of $E$ but what does the $j mid$ mean?



summation



Similarly in this equation what does the comma mean?



flow






elementary-set-theory notation







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    Short answer: such that. Respectively, $j$ such that the ordered pair $(j,i)$ is in $E$, $j$ such that the ordered pair $(s,j)$ is in $E$.
    $endgroup$
    – Steven Wagter
    Dec 26 '18 at 12:48








  • 2




    $begingroup$
    See Set-builder notation.
    $endgroup$
    – Mauro ALLEGRANZA
    Dec 26 '18 at 12:53






  • 1




    $begingroup$
    @StevenWagter: You should post your comment as an answer. (And explain why commas are used, and often omitted, in lists.)
    $endgroup$
    – John Bentin
    Dec 26 '18 at 13:03














1












1








1





$begingroup$


In set notation could somebody explain the meaning of $mid$ in the equation below please? How does it read?



I read it as $s$ and $j$ are an element of $E$ but what does the $j mid$ mean?



summation



Similarly in this equation what does the comma mean?



flow






elementary-set-theory notation







share|cite|improve this question











$endgroup$




In set notation could somebody explain the meaning of $mid$ in the equation below please? How does it read?



I read it as $s$ and $j$ are an element of $E$ but what does the $j mid$ mean?



summation



Similarly in this equation what does the comma mean?



flow






elementary-set-theory notation




elementary-set-theory notation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 26 '18 at 12:53









Mauro ALLEGRANZA

66.5k449115




66.5k449115










asked Dec 26 '18 at 12:45









cherry aldicherry aldi

203




203








  • 3




    $begingroup$
    Short answer: such that. Respectively, $j$ such that the ordered pair $(j,i)$ is in $E$, $j$ such that the ordered pair $(s,j)$ is in $E$.
    $endgroup$
    – Steven Wagter
    Dec 26 '18 at 12:48








  • 2




    $begingroup$
    See Set-builder notation.
    $endgroup$
    – Mauro ALLEGRANZA
    Dec 26 '18 at 12:53






  • 1




    $begingroup$
    @StevenWagter: You should post your comment as an answer. (And explain why commas are used, and often omitted, in lists.)
    $endgroup$
    – John Bentin
    Dec 26 '18 at 13:03














  • 3




    $begingroup$
    Short answer: such that. Respectively, $j$ such that the ordered pair $(j,i)$ is in $E$, $j$ such that the ordered pair $(s,j)$ is in $E$.
    $endgroup$
    – Steven Wagter
    Dec 26 '18 at 12:48








  • 2




    $begingroup$
    See Set-builder notation.
    $endgroup$
    – Mauro ALLEGRANZA
    Dec 26 '18 at 12:53






  • 1




    $begingroup$
    @StevenWagter: You should post your comment as an answer. (And explain why commas are used, and often omitted, in lists.)
    $endgroup$
    – John Bentin
    Dec 26 '18 at 13:03








3




3




$begingroup$
Short answer: such that. Respectively, $j$ such that the ordered pair $(j,i)$ is in $E$, $j$ such that the ordered pair $(s,j)$ is in $E$.
$endgroup$
– Steven Wagter
Dec 26 '18 at 12:48






$begingroup$
Short answer: such that. Respectively, $j$ such that the ordered pair $(j,i)$ is in $E$, $j$ such that the ordered pair $(s,j)$ is in $E$.
$endgroup$
– Steven Wagter
Dec 26 '18 at 12:48






2




2




$begingroup$
See Set-builder notation.
$endgroup$
– Mauro ALLEGRANZA
Dec 26 '18 at 12:53




$begingroup$
See Set-builder notation.
$endgroup$
– Mauro ALLEGRANZA
Dec 26 '18 at 12:53




1




1




$begingroup$
@StevenWagter: You should post your comment as an answer. (And explain why commas are used, and often omitted, in lists.)
$endgroup$
– John Bentin
Dec 26 '18 at 13:03




$begingroup$
@StevenWagter: You should post your comment as an answer. (And explain why commas are used, and often omitted, in lists.)
$endgroup$
– John Bentin
Dec 26 '18 at 13:03










1 Answer
1






active

oldest

votes


















2












$begingroup$

Suppose the $m_{j,i}$'s are elements of the set $M$, and that each pair $(j,i) subset I subset Bbb{N}^2$ is an index of an element in $M$, where $I$ is an indexing set. Then



$$Inputs(i) = sum_{j|(j,i) in E} m_{j,i}$$ means "Given some fixed input value $i$, sum those elements in $M$ whose index $(j,i)$ satisfies $(j,i) in E$.



Here $E$ is some subset of $Bbb{N}^2$. So we loop over all the $j$'s ($i$ is fixed) and add the terms corresponding to those $j$'s which satisfy $(j,i) in E$.



In the second picture, the comma can be read as a vertical bar. They have the same meaning in this case.



EDIT - in response to Shaun:



$I$ is some set used for indexing the elements in $M$, which must exist since otherwise the subscript of ordered pairs doesn't make sense. For example, take a $2times2$-matrix. Then if you want to take the matrix element of the first row, second column, one writes $a_{1,2}$. But what is really going on, is that you have the index set $I = { (1,1),(1,2),(2,1),(2,2)}$, where there is a one-to-one correspondence between elements in $I$ and the set of matrix elements.



For example, returning to our situation, taking the same indexing set $I$ as above and letting $E = {(1,1),(2,1),(2,2)}$, then $Inputs(2) = m_{2,2}$.



Another example of indexing sets: $$1/2+1/4+1/8 +... = sum_{n in Bbb{N} } frac{1}{2^n}$$



Here $Bbb{N}$ is the indexing set.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Where did the $I$ come from? Don't you mean $E$?
    $endgroup$
    – Shaun
    Dec 28 '18 at 5:32








  • 1




    $begingroup$
    @Shaun added some clarification, does it make more sense now?
    $endgroup$
    – Steven Wagter
    Dec 28 '18 at 8:42










  • $begingroup$
    Yes. Thank you, @StevenWagter :)
    $endgroup$
    – Shaun
    Dec 28 '18 at 8:44











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1 Answer
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oldest

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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









2












$begingroup$

Suppose the $m_{j,i}$'s are elements of the set $M$, and that each pair $(j,i) subset I subset Bbb{N}^2$ is an index of an element in $M$, where $I$ is an indexing set. Then



$$Inputs(i) = sum_{j|(j,i) in E} m_{j,i}$$ means "Given some fixed input value $i$, sum those elements in $M$ whose index $(j,i)$ satisfies $(j,i) in E$.



Here $E$ is some subset of $Bbb{N}^2$. So we loop over all the $j$'s ($i$ is fixed) and add the terms corresponding to those $j$'s which satisfy $(j,i) in E$.



In the second picture, the comma can be read as a vertical bar. They have the same meaning in this case.



EDIT - in response to Shaun:



$I$ is some set used for indexing the elements in $M$, which must exist since otherwise the subscript of ordered pairs doesn't make sense. For example, take a $2times2$-matrix. Then if you want to take the matrix element of the first row, second column, one writes $a_{1,2}$. But what is really going on, is that you have the index set $I = { (1,1),(1,2),(2,1),(2,2)}$, where there is a one-to-one correspondence between elements in $I$ and the set of matrix elements.



For example, returning to our situation, taking the same indexing set $I$ as above and letting $E = {(1,1),(2,1),(2,2)}$, then $Inputs(2) = m_{2,2}$.



Another example of indexing sets: $$1/2+1/4+1/8 +... = sum_{n in Bbb{N} } frac{1}{2^n}$$



Here $Bbb{N}$ is the indexing set.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Where did the $I$ come from? Don't you mean $E$?
    $endgroup$
    – Shaun
    Dec 28 '18 at 5:32








  • 1




    $begingroup$
    @Shaun added some clarification, does it make more sense now?
    $endgroup$
    – Steven Wagter
    Dec 28 '18 at 8:42










  • $begingroup$
    Yes. Thank you, @StevenWagter :)
    $endgroup$
    – Shaun
    Dec 28 '18 at 8:44
















2












$begingroup$

Suppose the $m_{j,i}$'s are elements of the set $M$, and that each pair $(j,i) subset I subset Bbb{N}^2$ is an index of an element in $M$, where $I$ is an indexing set. Then



$$Inputs(i) = sum_{j|(j,i) in E} m_{j,i}$$ means "Given some fixed input value $i$, sum those elements in $M$ whose index $(j,i)$ satisfies $(j,i) in E$.



Here $E$ is some subset of $Bbb{N}^2$. So we loop over all the $j$'s ($i$ is fixed) and add the terms corresponding to those $j$'s which satisfy $(j,i) in E$.



In the second picture, the comma can be read as a vertical bar. They have the same meaning in this case.



EDIT - in response to Shaun:



$I$ is some set used for indexing the elements in $M$, which must exist since otherwise the subscript of ordered pairs doesn't make sense. For example, take a $2times2$-matrix. Then if you want to take the matrix element of the first row, second column, one writes $a_{1,2}$. But what is really going on, is that you have the index set $I = { (1,1),(1,2),(2,1),(2,2)}$, where there is a one-to-one correspondence between elements in $I$ and the set of matrix elements.



For example, returning to our situation, taking the same indexing set $I$ as above and letting $E = {(1,1),(2,1),(2,2)}$, then $Inputs(2) = m_{2,2}$.



Another example of indexing sets: $$1/2+1/4+1/8 +... = sum_{n in Bbb{N} } frac{1}{2^n}$$



Here $Bbb{N}$ is the indexing set.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Where did the $I$ come from? Don't you mean $E$?
    $endgroup$
    – Shaun
    Dec 28 '18 at 5:32








  • 1




    $begingroup$
    @Shaun added some clarification, does it make more sense now?
    $endgroup$
    – Steven Wagter
    Dec 28 '18 at 8:42










  • $begingroup$
    Yes. Thank you, @StevenWagter :)
    $endgroup$
    – Shaun
    Dec 28 '18 at 8:44














2












2








2





$begingroup$

Suppose the $m_{j,i}$'s are elements of the set $M$, and that each pair $(j,i) subset I subset Bbb{N}^2$ is an index of an element in $M$, where $I$ is an indexing set. Then



$$Inputs(i) = sum_{j|(j,i) in E} m_{j,i}$$ means "Given some fixed input value $i$, sum those elements in $M$ whose index $(j,i)$ satisfies $(j,i) in E$.



Here $E$ is some subset of $Bbb{N}^2$. So we loop over all the $j$'s ($i$ is fixed) and add the terms corresponding to those $j$'s which satisfy $(j,i) in E$.



In the second picture, the comma can be read as a vertical bar. They have the same meaning in this case.



EDIT - in response to Shaun:



$I$ is some set used for indexing the elements in $M$, which must exist since otherwise the subscript of ordered pairs doesn't make sense. For example, take a $2times2$-matrix. Then if you want to take the matrix element of the first row, second column, one writes $a_{1,2}$. But what is really going on, is that you have the index set $I = { (1,1),(1,2),(2,1),(2,2)}$, where there is a one-to-one correspondence between elements in $I$ and the set of matrix elements.



For example, returning to our situation, taking the same indexing set $I$ as above and letting $E = {(1,1),(2,1),(2,2)}$, then $Inputs(2) = m_{2,2}$.



Another example of indexing sets: $$1/2+1/4+1/8 +... = sum_{n in Bbb{N} } frac{1}{2^n}$$



Here $Bbb{N}$ is the indexing set.






share|cite|improve this answer











$endgroup$



Suppose the $m_{j,i}$'s are elements of the set $M$, and that each pair $(j,i) subset I subset Bbb{N}^2$ is an index of an element in $M$, where $I$ is an indexing set. Then



$$Inputs(i) = sum_{j|(j,i) in E} m_{j,i}$$ means "Given some fixed input value $i$, sum those elements in $M$ whose index $(j,i)$ satisfies $(j,i) in E$.



Here $E$ is some subset of $Bbb{N}^2$. So we loop over all the $j$'s ($i$ is fixed) and add the terms corresponding to those $j$'s which satisfy $(j,i) in E$.



In the second picture, the comma can be read as a vertical bar. They have the same meaning in this case.



EDIT - in response to Shaun:



$I$ is some set used for indexing the elements in $M$, which must exist since otherwise the subscript of ordered pairs doesn't make sense. For example, take a $2times2$-matrix. Then if you want to take the matrix element of the first row, second column, one writes $a_{1,2}$. But what is really going on, is that you have the index set $I = { (1,1),(1,2),(2,1),(2,2)}$, where there is a one-to-one correspondence between elements in $I$ and the set of matrix elements.



For example, returning to our situation, taking the same indexing set $I$ as above and letting $E = {(1,1),(2,1),(2,2)}$, then $Inputs(2) = m_{2,2}$.



Another example of indexing sets: $$1/2+1/4+1/8 +... = sum_{n in Bbb{N} } frac{1}{2^n}$$



Here $Bbb{N}$ is the indexing set.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 28 '18 at 8:55

























answered Dec 26 '18 at 13:34









Steven WagterSteven Wagter

1569




1569












  • $begingroup$
    Where did the $I$ come from? Don't you mean $E$?
    $endgroup$
    – Shaun
    Dec 28 '18 at 5:32








  • 1




    $begingroup$
    @Shaun added some clarification, does it make more sense now?
    $endgroup$
    – Steven Wagter
    Dec 28 '18 at 8:42










  • $begingroup$
    Yes. Thank you, @StevenWagter :)
    $endgroup$
    – Shaun
    Dec 28 '18 at 8:44


















  • $begingroup$
    Where did the $I$ come from? Don't you mean $E$?
    $endgroup$
    – Shaun
    Dec 28 '18 at 5:32








  • 1




    $begingroup$
    @Shaun added some clarification, does it make more sense now?
    $endgroup$
    – Steven Wagter
    Dec 28 '18 at 8:42










  • $begingroup$
    Yes. Thank you, @StevenWagter :)
    $endgroup$
    – Shaun
    Dec 28 '18 at 8:44
















$begingroup$
Where did the $I$ come from? Don't you mean $E$?
$endgroup$
– Shaun
Dec 28 '18 at 5:32






$begingroup$
Where did the $I$ come from? Don't you mean $E$?
$endgroup$
– Shaun
Dec 28 '18 at 5:32






1




1




$begingroup$
@Shaun added some clarification, does it make more sense now?
$endgroup$
– Steven Wagter
Dec 28 '18 at 8:42




$begingroup$
@Shaun added some clarification, does it make more sense now?
$endgroup$
– Steven Wagter
Dec 28 '18 at 8:42












$begingroup$
Yes. Thank you, @StevenWagter :)
$endgroup$
– Shaun
Dec 28 '18 at 8:44




$begingroup$
Yes. Thank you, @StevenWagter :)
$endgroup$
– Shaun
Dec 28 '18 at 8:44


















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