# Set notation what does the bar symbol mean?

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?

• 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 '18 at 12:48
• – Mauro ALLEGRANZA Dec 26 '18 at 12:53
• @StevenWagter: You should post your comment as an answer. (And explain why commas are used, and often omitted, in lists.) – 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.

• Where did the $I$ come from? Don't you mean $E$? – Shaun Dec 28 '18 at 5:32
• @Shaun added some clarification, does it make more sense now? – Steven Wagter Dec 28 '18 at 8:42
• Yes. Thank you, @StevenWagter :) – Shaun Dec 28 '18 at 8:44