Set notation what does the bar symbol mean?












1














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







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  • 3




    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 $E$.
    – Steven Wagter
    Dec 26 at 12:48








  • 2




    See Set-builder notation.
    – Mauro ALLEGRANZA
    Dec 26 at 12:53






  • 1




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
















1














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









New contributor




cherry aldi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
















  • 3




    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 $E$.
    – Steven Wagter
    Dec 26 at 12:48








  • 2




    See Set-builder notation.
    – Mauro ALLEGRANZA
    Dec 26 at 12:53






  • 1




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














1












1








1







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









New contributor




cherry aldi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











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









New contributor




cherry aldi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




cherry aldi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited Dec 26 at 12:53









Mauro ALLEGRANZA

64.2k448112




64.2k448112






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asked Dec 26 at 12:45









cherry aldi

132




132




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Check out our Code of Conduct.





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cherry aldi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






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Check out our Code of Conduct.








  • 3




    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 $E$.
    – Steven Wagter
    Dec 26 at 12:48








  • 2




    See Set-builder notation.
    – Mauro ALLEGRANZA
    Dec 26 at 12:53






  • 1




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














  • 3




    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 $E$.
    – Steven Wagter
    Dec 26 at 12:48








  • 2




    See Set-builder notation.
    – Mauro ALLEGRANZA
    Dec 26 at 12:53






  • 1




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








3




3




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 $E$.
– Steven Wagter
Dec 26 at 12:48






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 $E$.
– Steven Wagter
Dec 26 at 12:48






2




2




See Set-builder notation.
– Mauro ALLEGRANZA
Dec 26 at 12:53




See Set-builder notation.
– Mauro ALLEGRANZA
Dec 26 at 12:53




1




1




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




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










1 Answer
1






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oldest

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2














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























  • Where did the $I$ come from? Don't you mean $E$?
    – Shaun
    2 days ago








  • 1




    @Shaun added some clarification, does it make more sense now?
    – Steven Wagter
    2 days ago










  • Yes. Thank you, @StevenWagter :)
    – Shaun
    2 days ago











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1 Answer
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1 Answer
1






active

oldest

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active

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active

oldest

votes









2














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























  • Where did the $I$ come from? Don't you mean $E$?
    – Shaun
    2 days ago








  • 1




    @Shaun added some clarification, does it make more sense now?
    – Steven Wagter
    2 days ago










  • Yes. Thank you, @StevenWagter :)
    – Shaun
    2 days ago
















2














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























  • Where did the $I$ come from? Don't you mean $E$?
    – Shaun
    2 days ago








  • 1




    @Shaun added some clarification, does it make more sense now?
    – Steven Wagter
    2 days ago










  • Yes. Thank you, @StevenWagter :)
    – Shaun
    2 days ago














2












2








2






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














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 2 days ago

























answered Dec 26 at 13:34









Steven Wagter

1178




1178












  • Where did the $I$ come from? Don't you mean $E$?
    – Shaun
    2 days ago








  • 1




    @Shaun added some clarification, does it make more sense now?
    – Steven Wagter
    2 days ago










  • Yes. Thank you, @StevenWagter :)
    – Shaun
    2 days ago


















  • Where did the $I$ come from? Don't you mean $E$?
    – Shaun
    2 days ago








  • 1




    @Shaun added some clarification, does it make more sense now?
    – Steven Wagter
    2 days ago










  • Yes. Thank you, @StevenWagter :)
    – Shaun
    2 days ago
















Where did the $I$ come from? Don't you mean $E$?
– Shaun
2 days ago






Where did the $I$ come from? Don't you mean $E$?
– Shaun
2 days ago






1




1




@Shaun added some clarification, does it make more sense now?
– Steven Wagter
2 days ago




@Shaun added some clarification, does it make more sense now?
– Steven Wagter
2 days ago












Yes. Thank you, @StevenWagter :)
– Shaun
2 days ago




Yes. Thank you, @StevenWagter :)
– Shaun
2 days ago










cherry aldi is a new contributor. Be nice, and check out our Code of Conduct.










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cherry aldi is a new contributor. Be nice, and check out our Code of Conduct.













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