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

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SUMMARY

The discussion centers on proving the set inclusion U i ∈I (Ai X Bi) ⊆ (U i ∈I Ai) X (U i ∈I Bi) for indexed families of sets {Ai | i ∈ I} and {Bi | i ∈ I}. The proof establishes that if (x, y) belongs to the union of the Cartesian products Ai X Bi, then x must belong to the union of Ai and y must belong to the union of Bi. This leads to the conclusion that the original inclusion holds true. The discussion also suggests exploring counterexamples for the reverse inclusion.

PREREQUISITES
  • Understanding of indexed families of sets
  • Familiarity with Cartesian products of sets
  • Knowledge of set notation and operations
  • Basic proof techniques in set theory
NEXT STEPS
  • Study the properties of Cartesian products in set theory
  • Learn about indexed unions and their implications
  • Explore counterexamples in set theory to understand inclusion relations
  • Review proof strategies in "How to Prove It, 2nd Edition" by Daniel J. Velleman
USEFUL FOR

Mathematics students, particularly those studying set theory and proof techniques, as well as educators seeking to clarify concepts related to indexed families and Cartesian products.

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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).
 
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That looks very good!

It might be instructive to find a counterexample to the reverse inclusion...
 
Thanks for commenting, I appreciate it. I have a very picky professor.
 

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