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Having a hard time with this one. As I understand it ∩F equals $$\{ x: ∀A ∈ F, x ∈ A)\}$$ which is also equivalent to $\underset{i\in I}{∩} A_i$. And this would mean if F were {{1,2,4},{4,7,8}} then the resulting set of all x ∈ $\underset{i\in I}{∩} A_i$ would be {4}. Is that wrong?

If the above is right, then what is the resulting set of x ∈ (B \ ($\underset{i\in I}{∩} A_i$)) where set B = {1,2,4,7,8,9} and F is the same as above {{1,2,4},{4,7,8}}? With my above logic I would compute $\underset{i\in I}{∩} A_i$ as {4} and subtract this from set B to produce the resulting set of {1,2,7,8,9}. BUT... I ran into a statement that says otherwise so I think my understandings above are botched.

Any help in clearing this one up is greatly appreciated!

user21820
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maybedave
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    On this site we use MathJaX to format our maths. Here you can find a basic tutorial. – gebruiker Jun 10 '16 at 14:50
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    Welcome to Math.SE! More context on exactly what the sets $A, I,$ and $F$ are would be helpful in answering your problem. I tried to improve the formatting somewhat, but I was not able to improve the formatting completely because I did not fully understand what was being said. Please feel free to rollback my edit and fix the formatting in a manner more appropriate, if applicable. – Chill2Macht Jun 10 '16 at 14:57
  • I'm really trying to find out if how I am computing the intersection of a family set wrong and then if not how to compute the result for disjointed set of B. Each A is a subset of F and I listed F as being {{1,2,4},{4,7,8}}. Is the result of the intersection of this family set {4}? Also, I looked at MathJax but the documentation talks about calling it from web pages. I couldn't find anything clear on how to install and use it in a text editor where I'm creating this stuff (MS Word). – maybedave Jun 10 '16 at 15:13
  • Click on the link in gebruiker's comment. You don't need to install anything to use LaTeX via MathJax on Math SE. – user21820 Jun 10 '16 at 15:14
  • correction, each A is NOT a subset of F. Each A exists within a subscript (index) of F. Not sure of an easy way how to say that. – maybedave Jun 10 '16 at 15:14
  • "But...I ran into a statement that says otherwise..." what statement? Perhaps it was a typo or slightly different notation. – JMoravitz Jun 10 '16 at 15:29

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Given a collection of sets with labels, for example $A_1, A_2, A_3, \dots$ the intersection of the family of sets $\bigcap\limits_{i\in I} A_i$ is defined as $\{x~:~(\forall i\in I)~~ x\in A_i\}$.

In particular, the indexing set here, $I$, need not be the entirety of the indexes used for $A_1,A_2,\dots$. $I$ could merely be a subset of the indices used.

Furthermore, the indices and $I$ need not necessarily be countable and the indices need not necessarily be numbers.

I could have the collection of sets $B_{\text{dog}},B_{\text{cat}},B_{\text{rabbit}},B_{\text{human}},B_{\text{anime characters}}\dots$ where each $B$ contains the set of popular names for each respective category.

With $I=\{\text{dog},\text{anime characters}\}$ you could have something like $\bigcap\limits_{i\in I} B_i = B_{\text{dog}}\cap B_{\text{anime characters}}=\{\text{Jo-jo}, \text{Kiba}, \text{Toboe},\dots\}$

As you noticed, this could also have been described using other notation.

$\bigcap\limits_{i\in I} A_i = \bigcap \left\{A_i~:~i\in I\right\}$

For your specific example, $A_1=\{1,2,4\}$ and $A_2=\{4,7,8\}$, and $I=\{1,2\}$, you would indeed have $\bigcap\limits_{i\in I} A_i = A_1\cap A_2 = \{4\}$.

Further, setsubtracting this from your $B$ gives exactly as you say.

JMoravitz
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