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My home work is: let $ A_{n},B_{n}$ be subsets of the sample space. Prove that

$$ \limsup_{n\to\infty} (A_{n}\cup B_{n}) = \limsup_{n\to\infty} A_{n}\cup\limsup_{n\to\infty} B_{n} $$

I managed to get to this:

$$ \bigcap_{1}^{n}\bigcup_{n\geq m}^{ } A_{m}\cup \bigcap_{1}^{n}\bigcup_{n\geq m}^{ } B_{m}\ = \bigcap_{1}^{n}\bigcup_{n\geq m}^{ } A_{m}\cup B_{m} $$

Really appreciate if anyone can help me with this

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  • $\begingroup$ Please use \limsup_{n\to\infty} or even, when inline, \limsup\lim_{n\to\infty}. $\endgroup$
    – Did
    Commented Sep 1, 2016 at 15:24

3 Answers 3

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Hint. Try to use the property that $x$ is in the limsup of a sequence of sets $A_n$ iff $x$ is an element of infinitely many of the sets $A_n$.

The inclusion $$\limsup_{n\to\infty} (A_{n}\cup B_{n}) \supseteq \limsup_{n\to\infty} A_{n}\cup\limsup_{n\to\infty} B_{n}$$ is trivial.

As regards the other one $$\limsup_{n\to\infty} (A_{n}\cup B_{n}) \subseteq \limsup_{n\to\infty} A_{n}\cup\limsup_{n\to\infty} B_{n}$$ you can show it by contradiction.

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We have $x \in \limsup_{n \to \infty}(A_{n} \cup B_{n})$ iff for every $n \geq 1$ there is some $N \geq n$ such that $x \in A_{N}$ or $\in B_{N}$; but, since $x$ is by definition independent of $n$ and $N$, this is equaivalent to the statement that for every $n \geq 1$ there is some $N \geq n$ such that $x \in A_{N}$ or for every $n \geq 1$ there is some $N \geq n$ such that $x \in B_{N}$, i.e. equivalent to $x \in \limsup_{n \to \infty}A_{n} \cup \limsup_{n \to \infty}B_{n}$.

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I think you can prove this using the distributivity laws of sets. $x \in \text{limsup} A_n \cup B_n \iff x \in \cap_n \cup_{ k \geq n} (A_k \cup B_k) \iff x \in \cap_n [ (\cup_{k \geq n} A_k) \cup (\cup_{k \geq n} B_k)] \iff x \in [\cap_n \cup_{k \geq n} A_k] \cup [\cap_n \cup_{k \geq n} B_k] \iff x \in \text{limsup} A_n \cup \text{limsup} B_n.$

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  • $\begingroup$ Unfortunately you last equivalence must be detailed because not necessarily the intersection of union of sets is the union of the individual intersections of such sets. $\endgroup$ Commented Nov 9, 2022 at 22:57
  • $\begingroup$ Just consider $X$ a nonempty set and set $A_k=\emptyset$ If $k$ Is odd and $A_k=X$ If $k$ Is even, then consider $B_k=X\setminus A_k$, we have that $\bigcap_k (A_k\cup B_k)=X$, but $\left(\bigcap_k A_k\right)\cup \left(\bigcap_k B_k\right)=\emptyset. $ $\endgroup$ Commented Nov 10, 2022 at 1:52

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