Prove U (Ai X Bi) [itex]\subseteq[/itex] (U Ai) X (U Bi)

[IntroAnalysis]

Homework Statement

Suppose {Ai l i \inI} and {Bi l i \inI} are indexed families of sets.

Prove that U i \inI(Ai X Bi) \subseteq (Ui \inIAi) X (Ui \inIBi)


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

The Attempt at a Solution

Let (x, y) be arbitrary. Suppose (x, y) \in \bigcupi\inI (Ai X Bi).

Since (x, y) \in\bigcupi\inI(Ai X Bi), there exists an i\inI with x\inAi and y\inBi.

So x \in{xl\existsi\inI(x\inAi)} and
y\in{yl\existsi\inI(y\inBi)}

Therefore, x \in\bigcupi\inI Ai and y\in<br /> \bigcupi\inI Bi.

This is equivalent to (\bigcupi\inI Ai) X (\bigcupi\inI Bi). Hence, \bigcupi\inI (Ai X Bi)\subseteq(Ui\inI Ai) X (\bigcupi\inI Bi).

 

[micromass]

That looks very good!

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

 

[IntroAnalysis]

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