# Help indexed family sets proof!

1. Sep 10, 2011

### IntroAnalysis

1. The problem statement, all variables and given/known data
Ai and Bi are indexed families of sets. Prove that Ui (Ai $\bigcap$ Bi) $\subseteq$ (UiAi) $\bigcap$ (UiBi).

2. Relevant equations

3. The attempt at a solution
Suppose arbitrary x. Let x $\in$
{x l $\forall$i$\in$I(x$\in$Ai$\bigcap$Bi)
This means x $\in${x l $\forall$i$\in$I(x$\in$Ai)$\wedge$$\forall$i$\in$I(x$\in$Bi).
1. The problem statement, all variables and given/known data
This means x $\in$$\neg$$\exists$i$\in$I(x$\notin$Ai)$\wedge$$\neg\exists$i$\in$I(x$\in$$\notin$Bi)}
Which is equivalent to: x$\in${x l $\exists$i$\in$I(x$\in$Ai)$\wedge$$\exists$i$\in$I(x$\in$Bi)}
Therefore, x$\in${x l ($\bigcup$i$\in$IAi)$\bigcap$(Ui$\in$IBi)}
Therefore, is equiv. to Ui$\in$I(Ai$\bigcap$Bi), then x$\in$(Ui$\in$IAi)$\bigcap$(Ii$\in$IBi).
Therefore Ui$\in$I(Ai$\bigcap$Bi)$\subseteq$(Ui$\in$IAi)$\bigcap$(Ii$\in$IBi).

2. Sep 11, 2011

### Stephen Tashi

Are you trying to being by taking an arbitrary x in $\bigcup_{i=1}^\infty (A_i \cap B_i)$ ? If so, why are you applying the "$\forall$" quantifier to the index?

I suggest beginning this way:

Let x be an arbitrary element of $\bigcup_{i=1}^\infty (A_i \cap B_i)$

For such an x, we know that there exists an index j such that $x \in A_j \cap B_j$. This follows from the definition and properties of a union of sets.

This implies $x \in A_j$ and $x \in B_j$ by definition of an intersection of two sets.