How to prove $\limsup(\{A_n \cup B_n\}) = \limsup(\{A_n\}) \cup \limsup(\{B_n\})$? Thanks!
Proof: Limit superior intersection
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0@gaer: The union of families *also* does not make sense. Presumably, you meant the family of unions, and I've edited as such. – 2010-11-08
2 Answers
Use the definition, and double inclusion; that is, show that every element of $\limsup(A\cup B)$ must be either an element of $\limsup(A)$ or of $\limsup(B)$; then show that every element of $\limsup(A)$ must be in $\limsup(A\cup B)$ and that every element of $\limsup(B)$ must be in $\limsup(A\cup B)$.
Of course, one must assume that you mean your "$A$" to be a sequence of sets and your "$B$" to likewise be a sequence of sets... Otherwise, what you write does not really make much sense (limit superior and limit inferior of a single set is not usually defined).
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0@gaer: if what you are missing is how to write the proof correctly, then write what you have so people can tell you where you went right and where you went wrong. You might start by giving a *correct and accurate* statement of the problem (see the discussion on why what you wrote makes no sense as written). Then write what you have. I for one will not give you a "complete proof for this problem", because I have no desire to do your homework for you. So edit the question and write down what you've done, and where (and why) you are stuck; people will take it from there. – 2010-11-08
Another nice way is to use characteristic functions:
The map $\chi : \mathcal{P}(\Omega) \to \{0,1\}^\Omega$ assigns to every subset of $\Omega$ its characteristic function.
- $\chi$ is bijective.
- $\chi$ is continuous, i.e. $\chi_{\lim\sup_{n\to\infty} A_n} = \lim\sup_{n\to\infty}\, \chi_{A_n}$ (pointwise limit)
- $\chi$ is a homomorphism, i.e. $\chi_{A \cup B} = \chi_A + \chi_B - \chi_A \chi_B$
Now your question reduces to the computation of an ordinary limit.