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I've been learning a lot from this book, but have found myself entirely stuck with this seemingly trivial question. I think it's due to a lack of understanding of how to treat universal quantifiers with indexed families of sets.

The question reads:

Suppose $\{A_{i} |i \in I \}$ is an indexed family of sets and $I \neq \emptyset$. Prove that $ \cap_{i\in I}A_{i} \in \cap_{i\in I}\mathscr{P}\left(A_{i}\right)$.

I understand that the logical statement to prove is the same as:

$\{x| \forall i \in I \left(x \in A_i\right)\} \in \{X| \forall i \in I \left(X \in\mathscr{P}\left(A_i\right)\right)\} $

I wrote an example of this using actual sets and also understand that $x$ is a set, and $X$ is a set of sets, but am stuck in working out how to even begin proving this.

Given the universal quantifier, I'm guessing I can start by saying "Let $i$ be arbitrary", but I wouldn't know how to structure the following statement given that we're talking about an indexed family of sets. I've also flicked back in the book, but he doesn't seem to have covered methods for treating such a case.

Any help with my lack of understanding here would help enormously.

abvariant
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  • You want to show that $A = \bigcap_{i \in I}A_i$ is an element of a set which is an intersection over all $i \in I$. So you're right that you start with some arbitrary $i$. What does it mean for $A$ to be an element of this intersection? It means that $A$ is an element of $\scr{P}$ $(A_i)$ for every $i$. So you have to show that for an arbitrary $i$, $A \in \scr{P}$ $(A_i)$, i.e. that $A$ is a subset of $A_i$. – JackR Jan 09 '18 at 13:44
  • In this case I can't help but think all I would be doing in this case is just translating the statement directly into logical notation: If $A = \bigcap_{i \in I}A_i$, so that $A \in\bigcap_{i\in I}\mathscr P(A_i)$, by the original statement to prove. Then $\forall i\in I, A \subseteq A_{i}$ by the definition of a power set. Is this sufficient? – abvariant Jan 11 '18 at 13:21

1 Answers1

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You have $$\forall i\in I,\qquad\bigcap_{i\in I}A_i\subseteq A_i$$ hence $$\forall i\in I,\qquad\bigcap_{i\in I}A_i\in\mathscr P(A_i)$$ Since this holds for all $i\in I$, you get $$\bigcap_{i\in I}A_i\in\bigcap_{i\in I}\mathscr P(A_i)$$

Recall, in general, that $$x\in\bigcap_{i\in I} A_i\iff\forall i\in I(x\in A_i)$$ $$x\subseteq A_i\iff x\in\mathscr P(A_i)$$

Fabio Lucchini
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  • I have a few questions surrounding the answer you've provided. Firstly, your first statement: $\forall i\in I,\bigcap_{i\in I}A_i\subseteq A_i $, how would this be translated into English? What does the second occurance of $A_{i}$ refer to here? Presumably it is "Any given $A_{i}$"? – abvariant Jan 11 '18 at 12:53
  • First statement, says that the intersection of a family of sets ${A_i:i\in I}$ is a set contained in each member $A_i$ of the family. The second occurrence of $A_i$ is thus one of the set of the family ${A_i:i\in I}$. – Fabio Lucchini Jan 11 '18 at 12:59
  • Fabio, I think you've helped to illuminate exactly why I struggled with this. What confused me here is that $A_{i}$ on the LHS of the "$\subseteq$" is for all i in I, but the RHS only refers to ONE set, despite the "i" subscript. I had no idea it was possible to write this in such a way. With this, and the definitions you've provided that I'm familiar with, this has answered the question. Thank you! – abvariant Jan 11 '18 at 13:48
  • One more thing, in the context of your second line, does the $A_{i}$ in $\mathscr P(A_i)$ also refer to the power set of one set of the family ${A_{i} : i \in I}$? – abvariant Jan 11 '18 at 14:06
  • Yes, it is the same set $A_i $ of the first statement. – Fabio Lucchini Jan 11 '18 at 14:35