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IntroAnalysis
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Homework Statement
Suppose {Ai l i [itex]\in[/itex]I} and {Bi l i [itex]\in[/itex]I} are indexed families of sets.
Prove that U i [itex]\in[/itex]I(Ai X Bi) [itex]\subseteq[/itex] (Ui [itex]\in[/itex]IAi) X (Ui [itex]\in[/itex]IBi)
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) [itex]\in[/itex] [itex]\bigcup[/itex]i[itex]\in[/itex]I (Ai X Bi).
Since (x, y) [itex]\in[/itex][itex]\bigcup[/itex]i[itex]\in[/itex]I(Ai X Bi), there exists an i[itex]\in[/itex]I with x[itex]\in[/itex]Ai and y[itex]\in[/itex]Bi.
So x [itex]\in[/itex]{xl[itex]\exists[/itex]i[itex]\in[/itex]I(x[itex]\in[/itex]Ai)} and
y[itex]\in[/itex]{yl[itex]\exists[/itex]i[itex]\in[/itex]I(y[itex]\in[/itex]Bi)}
Therefore, x [itex]\in[/itex][itex]\bigcup[/itex]i[itex]\in[/itex]I Ai and y[itex]\in[/itex][itex]
\bigcup[/itex]i[itex]\in[/itex]I Bi.
This is equivalent to ([itex]\bigcup[/itex]i[itex]\in[/itex]I Ai) X ([itex]\bigcup[/itex]i[itex]\in[/itex]I Bi). Hence, [itex]\bigcup[/itex]i[itex]\in[/itex]I (Ai X Bi)[itex]\subseteq[/itex](Ui[itex]\in[/itex]I Ai) X ([itex]\bigcup[/itex]i[itex]\in[/itex]I Bi).