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?


Similarly in this equation what does the comma mean?


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

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 $2\times2$-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.

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