Prove U (Ai X Bi) $\subseteq$ (U Ai) X (U Bi)

1. Oct 5, 2011

IntroAnalysis

1. The problem statement, all variables and given/known data
Suppose {Ai l i $\in$I} and {Bi l i $\in$I} are indexed families of sets.

Prove that U i $\in$I(Ai X Bi) $\subseteq$ (Ui $\in$IAi) X (Ui $\in$IBi)

2. Relevant\subseteq equations
From How to Prove It, 2nd Edition, Sec. 4.1 #11a)

3. The attempt at a solution

Let (x, y) be arbitrary. Suppose (x, y) $\in$ $\bigcup$i$\in$I (Ai X Bi).

Since (x, y) $\in$$\bigcup$i$\in$I(Ai X Bi), there exists an i$\in$I with x$\in$Ai and y$\in$Bi.

So x $\in${xl$\exists$i$\in$I(x$\in$Ai)} and
y$\in${yl$\exists$i$\in$I(y$\in$Bi)}

Therefore, x $\in$$\bigcup$i$\in$I Ai and y$\in$$\bigcup$i$\in$I Bi.

This is equivalent to ($\bigcup$i$\in$I Ai) X ($\bigcup$i$\in$I Bi). Hence, $\bigcup$i$\in$I (Ai X Bi)$\subseteq$(Ui$\in$I Ai) X ($\bigcup$i$\in$I Bi).
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 5, 2011

micromass

Staff Emeritus
That looks very good!!!!

It might be instructive to find a counterexample to the reverse inclusion...

3. Oct 5, 2011

IntroAnalysis

Thanks for commenting, I appreciate it. I have a very picky professor.