Proving Limsup of AUB = Limsup(A) U Limsup(B)

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SUMMARY

The discussion centers on proving that the limit superior of the union of two sequences of sets, A and B, is equal to the union of their individual limit superiors: Limsup(A ∪ B) = Limsup(A) ∪ Limsup(B). The approach involves using the definition of limit superior, where an element x belongs to Limsup(A) if it appears in infinitely many An. The proof demonstrates that if x is in Limsup(A ∪ B), it must also be in either Limsup(A) or Limsup(B), confirming the equality through double inclusion.

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  • Understanding of limit superior in the context of sequences of sets.
  • Familiarity with set theory, particularly unions and intersections.
  • Knowledge of mathematical proofs, specifically double inclusion proofs.
  • Basic concepts of sequences and their convergence properties.
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  • Study the formal definition of limit superior for sequences of sets.
  • Explore examples of limit superior calculations for different sequences.
  • Learn about the properties of unions and intersections in set theory.
  • Investigate advanced topics in real analysis related to convergence and limits.
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Mathematics students, particularly those studying real analysis or set theory, as well as educators looking to enhance their understanding of limit superior concepts and proofs.

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Homework Statement



Given A = {An} and B = {Bn}. Prove that Limsup ({AUB}) = Limsup(A) U Limsup(B).

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The Attempt at a Solution



The result of both sides is a set, so I attempted to use double inclusion to prove it. Used the definition of limsup, but didn't manage to prove that for x in limsup(AUB), it is either in limsup(A) or limsup(B) and vice versa.
 
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Do you know a formula for the lim sup of a sequence of sets, expressed in terms of unions and intersections?
 
You could use the following:
x is in limsup A if and only if x is in An for infinitely many n.

Then x is in \limsup{A\cup B}, if it is in infinitely many A_n\cup B_n. But then it must of course be in infinitely many An or Bn; thus x is in \limsup A_n\cup \limsup B_n.
The other implication is the same thing...
 

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