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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1Type out the definition of lim sup and the rest is easy. – 2010-11-07
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1As pointed out in an answer, the "lim sup" operation requires some sort of limit - a sequence, for example, What do you mean by the lim sup of a set? – 2010-11-07
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0@Carl Mummert: It doesn't really make sense as a real-analysis problem with unions; it makes more sense as a problem of sequences of sets. So I would not exchange the set-theory tag for the real-analysis tag. – 2010-11-07
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0This question has absolutely *nothing* to do with set theory. People tend to misuse the set-theory tag for any question that involves sets, but that is just as silly as adding "abstract algebra" to any question that involves addition. – 2010-11-08
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0I have no idea whether the "sequences-and-series" tag would apply. @gaer: could you please clarify what the question is about? – 2010-11-08
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0I started a more general thread about the set-theory tag at http://meta.math.stackexchange.com/questions/1092/appropriate-uses-of-the-set-theory-tag – 2010-11-08
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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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0Assuming that A and B are sequences of sets, A union B will not be a sequence of sets unless A and B happen to be the same sequence. So $\limsup(A \cup B)$ wouldn't make sense. – 2010-11-08
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0@Carl Mummert: I think we can agree the statement is fairly messed up. I interpret it as $A=\{A_n\}$, $B=\{B_n\}$, and $A\cup B = \{A_n\cup B_n\}$. Obviously, I'm guessing. – 2010-11-08
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0I started the proof by writing down the definitions of limsup and I'm aware that I need to proof double inclusion, but what I'm missing is how to write the proof correctly. What I'm looking for is to see how to write a complete proof for this problem. – 2010-11-08
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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.