Let $A_n$ and $B_n$ be two sequences of sets. How $(\liminf_n A_n \cup \liminf_n B_n)$ and $\liminf_n (A_n\cup B_n)$ are related?

Def. Given a sequence of sets $E_n$, the limit inferior of $E_n$ is defined as $$\liminf_{n\to\infty} E_n=\bigcup_{n=1}^\infty \bigcap_{k=n}^\infty E_k$$

Some thoughts

Write $\liminf_n A_n=\bigcup_{n}C_n$ and $\liminf_n B_n=\bigcup_{n}D_n$ where $C_n=\bigcap_{k=n}^\infty A_k$ and $D_n=\bigcap_{k=n}^\infty B_k$.

I will use a (intutive) result that requires a proof: $(\bigcup_{n\in\mathbb{N}}C_n) \cup (\bigcup_{l\in\mathbb{N}}D_l)=\bigcup_{n\in\mathbb{N}}C_n\cup D_n$.

On the other hand, for each $n$, $$C_n\cup D_n=\bigcap_{k=n}^\infty A_k \cup \bigcap_{l=n}^\infty B_l=\bigcap_{k=n}^\infty \left[ A_k \cup \left(\bigcap_{l=n}^\infty B_l \right)\right]\subseteq \bigcap_{k=n}^\infty A_k \cup B_k.$$ From these observations, we immediately have $$\liminf_n (A_n\cup B_n)\supseteq \liminf_n A_n \cup \liminf_n B_n $$

  • $\begingroup$ @janmarqz the union of the two liminf''s $\endgroup$ – Celine Harumi Jul 30 '20 at 0:41
  • $\begingroup$ I think $\lim\inf_n A_n \cup \lim\inf_n B_n \color{red} \subset \lim\inf_n A_n\cup B_n$ $\endgroup$ – Masoud Jul 30 '20 at 0:42
  • $\begingroup$ @janmarqz: No, they are sets. $\endgroup$ – Brian M. Scott Jul 30 '20 at 0:42
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    $\begingroup$ Suppose that $$A_n=\begin{cases}\varnothing,&\text{if }n\text{ is even}\\ [0,1],&\text{if }n\text{ is odd,}\end{cases}$$ and $$B_n=\begin{cases}\varnothing,&\text{if }n\text{ is odd}\\ [0,1],&\text{if }n\text{ is even.}\end{cases}$$ What are $\liminf_nA_n$, $\liminf_nB_n$, and $\liminf_n(A_n\cup B_n)$? This will give you a clearer idea of what the answer ought to be. $\endgroup$ – Brian M. Scott Jul 30 '20 at 1:04
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    $\begingroup$ @CelineHarumi: (Sorry to be so slow: my internet went out.) Yes, what you did is fine, and I really don’t think that the identity $$\bigcup_n(C_n\cup D_n)=\left(\bigcup_nC_n\right)\cup\bigcup_nD_n$$ needs proof. As an intuitive check, note that if $x$ is in either every $A_$ from some point on or every $B_n$ from some point on, then it’s certainly in every $A_n\cup B_n$ from some point on, so your final line is definitely true. $\endgroup$ – Brian M. Scott Jul 30 '20 at 3:03

We go to show that: $\liminf_{n}A_{n}\cup\liminf_{n}B_{n}\subseteq\liminf_{n}(A_{n}\cup B_{n})$.

Let $x\in LHS$, then $x\in\liminf_{n}A_{n}$ or $x\in\liminf_{n}B_{n}$. Suppose that $x\in\liminf_{n}A_{n}$, then there exists $n$ such that $x\in\cap_{k\geq n}A_{k}$. For each $k\geq n$, $x\in A_{k}\Rightarrow x\in A_{k}\cup B_{k}$. Therefore, $x\in\cap_{k\geq n}(A_{k}\cup B_{k})$. Hence $x\in\cup_{n}\cap_{k\geq n}(A_{k}\cup B_{k})=RHS$. Similarly, if $x\in\liminf_{n}B_{n}$, we can show that $x\in RHS$. This shows that $LHS\subseteq RHS$.

  • $\begingroup$ The reverse inclusion does not hold. For example. Let $A_n=\{0\}$ if $n$ is odd and $A_n=\{1\}$ if $n$ is even, $B_n=\{1\}$ if $n$ is odd and $B_n=\{0\}$ if $n$ is even. For each $n$, $ A_n \cup B_n =\{0,1\}$, so $\liminf_n (A_n \cup B_n)=\{0,1\}$. However, $\liminf_n A_n = \liminf B_n =\emptyset$. $\endgroup$ – Danny Pak-Keung Chan Jul 30 '20 at 1:13
  • $\begingroup$ You gave me an opposite conclusion of mine. I will check what is wrong. $\endgroup$ – Celine Harumi Jul 30 '20 at 1:21
  • $\begingroup$ @CelineHarumi: No, he gave you the same inclusion as yours. $\endgroup$ – Brian M. Scott Jul 30 '20 at 3:03
  • $\begingroup$ @BrianM.Scott My initial sketch was wrong. Its conclusion was the opposite. $\endgroup$ – Celine Harumi Jul 30 '20 at 14:51

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