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How to prove $$\limsup (A_k\cup B_k)=\limsup A_k \cup \limsup B_k \tag{1}$$ and does this remain true if I replace $\cup$ with $\cap$?

Here $A_k$ and $B_k$ are sets. The upper and lower limits are $$\liminf X_k=\bigcup_n\bigcap_{k>n}X_k,\qquad \limsup X_k = \bigcap_n\bigcup_{k>n}X_k$$


Property (1) is similar to $$\limsup \max(a_k,b_k) = \max(\limsup a_k, \limsup b_k)$$ for numerical sequences.

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closed as too localized by The Chaz 2.0, Lord_Farin, rschwieb, Davide Giraudo, Micah Apr 25 '13 at 18:03

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$\limsup$ is a number. What do you mean by $\limsup A_k \cup \limsup B_k$? –  Dennis Gulko Apr 25 '13 at 15:29
@DennisGulko: See this page for some information on the $\limsup$ of sets. –  Clayton Apr 25 '13 at 15:32
Hint: pick up an element of RHS, show it stay in LHS, and vise versa. –  Ma Ming Apr 25 '13 at 15:34
@noname1014 If it is the case, you should better explicitly state that the $A_k, B_k$ are sets of real numbers (and not real numbers). –  Hagen von Eitzen Apr 25 '13 at 15:38
I just learned of a second notion for $\liminf$ and $\limsup$ of sets. On the one hand, we have $\limsup X_k$ is the set of limit points of sequences $(x_k)$ with $x_k\in X_k$ and $\liminf$ is the set of limits of converging such sequences. It is however likely that the OP's definition is the case I already knew (corresponding to discrete topology): $\liminf X_k=\bigcup_n\bigcap_{k>n}X_k$, $\limsup X_k = \bigcap_n\bigcup_{k>n}X_k$. –  Hagen von Eitzen Apr 25 '13 at 15:46

2 Answers 2

up vote 1 down vote accepted

If $x \in \limsup A_k \cup \limsup B_k$, then $x$ is an at least in one of the two sets, say $\limsup A_k$. Then, for every $k$, there exists an $n \geqslant k$ such that $x \in A_n$, but $A_n \subset A_n \cup B_n$. Thus $\forall k$, $\exists n \geqslant k$, such that $x \in A_k \cup B_k$. We just proved that $x \in \bigcap_{k \geqslant 1} \bigcup_{n \geqslant k} \left( A_n \cup B_n \right) = \limsup \left( A_k \cup B_k \right)$.

Let's try the contrapositive proof of the other direction. Assume $x \not\in \limsup A_k \cup \limsup B_k$, then $x \not\in \limsup A_k \wedge x \not\in \limsup B_k$, then $\exists k_1, \forall n_1 \geqslant k_1, x \not\in A_{n_1}$, $\exists k_2$, $\forall n_2 \geqslant k_2$, $x \not\in B_{n_2}$. Take $k$ to be the max of $k_1$ and $k_2$, then $\forall n \geqslant k$, $x \not\in A_n \wedge x \not\in B_n$, which implies $x \not\in A_n \cup B_n$, thus $x \not\in \limsup \left( A_k \cup B_k \right)$

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why $\bigcap_{k \geqslant 1} \left[ \left( \bigcup_{n \geqslant k} A_n \right) \bigcup \left( \bigcup_{n \geqslant k} B_n \right) \right]\\ = \left( \bigcap_{k \geqslant 1} \bigcup_{n \geqslant k} A_n \right) \bigcup \left( \bigcap_{k \geqslant 1} \bigcup_{n \geqslant k} B_n \right)$ I cannot understand. –  noname1014 Apr 25 '13 at 16:20
First direction is right,but I cannot understand the other direction. –  noname1014 Apr 25 '13 at 16:25
you means $\forall k \exists n \geqslant k s.t. x \in A or x \in B$ we can conclude that $\forall k \exists n \geqslant k s.t. x \in A $ or $\forall k \exists n \geqslant k s.t. x \in B$.but I think it is not right ,because $\forall x(P(x) or Q(x))$we cannot conclude that$\forall x P(x)$ or $\forall x Q(x)$ . –  noname1014 Apr 25 '13 at 16:38
@noname1014 I tried a new proof by contrapositive (instead of proving $A$ implies $B$, I tried to prove not $B$, implies not $A$). Verify if my reasoning holds. –  Learner Apr 25 '13 at 17:01
I think it is right. –  noname1014 Apr 25 '13 at 17:31

Assume $x\in\limsup_k (A_k\cup B_k)$. Then there is a sequence $x_k\in A_k\cup B_k$ that has a subsequence $x_{n_k}$ converging to $x$. If $x_{n_k}\in A_{n_k}$ holds infinitely often, these infinite sub-subsequence also converges to $x$ and shows that $x\in\limsup_k A_k$. Otherwise, we must have $x_{n_k}\in B_{n_k}$ for all but finitely many $k$, hence $x_{n_k}\to x$ and $x\in\limsup_k B_k$. All in all, $x\in\limsup_k A_k\cup \limsup_k B_k$.

Conversely assume $x\in\limsup_k A_k\cup \limsup_k B_k$. Then $x$ is limit point of a sequence $(x_k)$ with $x_k\in A_k$ for all $k$, or of a sequence with $x_k\in B_k$ for all $k$. But then also $x_k\in A_k\cup B_k$ for all $k$, i.e. $x\in\limsup_k (A_k\cup B_k)$.

If you replace $\cup$ with $\cap$, the caim is no longer valid. For example, we can have $\limsup A_k=\limsup B_k\ne\emptyset$ although $A_k\cap B_k=\emptyset$ for all $k$.

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