How Can Intersection of Indexed Family Sets Belong to Their Power Sets?

AI Thread Summary
The discussion centers on proving that the intersection of an indexed family of sets, denoted as {Ai | i ∈ I}, belongs to the power set of those sets when I is not empty. Participants clarify the notation and the conditions of the problem, emphasizing the need to show that the intersection is a subset of the power set. There is some confusion regarding the notation and whether I is empty or not, which is critical for the proof. The conversation includes attempts to establish the relationship between the intersection and the power set through logical reasoning. Ultimately, the focus remains on the mathematical proof and the correct interpretation of the indexed family of sets.
nike5
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Homework Statement


Suppose {Ai| i \in I} is an indexed family of sets and I does
equal an empty set. Prove that \bigcap i \in I Ai
\in \bigcap i\in I P(Ai ) and P(Ai) is the
power set of Ai

Homework Equations


none


The Attempt at a Solution


Suppose x \in {Ai| i \in I}. Let i be an arbitrary element of
I where x \in Ai . Then let y be an arbitrary element of x. Since x
is an element of Ai and y \in x it follows that ...

maybe i want to show that \bigcap i \in I Ai \subseteq \bigcap i \in I Ai and then
I could say that \bigcap i \in I Ai \in \bigcap i\in I P(Ai )
 
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Let \left\{ A_{i} \right\}_{i \in I} be your indexed set of family.

Do you mean this \bigcap_{i=1} A_i = \left\{ x : \forall i \in I: x \in A_i \right\}?
 
Yes sry about the horrible looking symbols
 
nike5 said:

Homework Statement


Suppose {Ai| i \in I} is an indexed family of sets and I does
equal an empty set.
Did you mean "does not equal and empty set"?
 
I \neq \oslash is what I meant
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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