2
$\begingroup$

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 $$\liminf_n (A_n\cup B_n)\supseteq \liminf_n A_n \cup \liminf_n B_n $$

$\endgroup$
12
  • $\begingroup$ @janmarqz the union of the two liminf''s $\endgroup$ Jul 30, 2020 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, 2020 at 0:42
  • $\begingroup$ @janmarqz: No, they are sets. $\endgroup$ Jul 30, 2020 at 0:42
  • 1
    $\begingroup$ 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. $\endgroup$ Jul 30, 2020 at 1:04
  • 1
    $\begingroup$ @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. $\endgroup$ Jul 30, 2020 at 3:03

1 Answer 1

3
$\begingroup$

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 $LHS\subseteq RHS$.

$\endgroup$
4
  • $\begingroup$ 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, $\liminf_n A_n = \liminf B_n =\emptyset$. $\endgroup$ Jul 30, 2020 at 1:13
  • $\begingroup$ You gave me an opposite conclusion of mine. I will check what is wrong. $\endgroup$ Jul 30, 2020 at 1:21
  • $\begingroup$ @CelineHarumi: No, he gave you the same inclusion as yours. $\endgroup$ Jul 30, 2020 at 3:03
  • $\begingroup$ @BrianM.Scott My initial sketch was wrong. Its conclusion was the opposite. $\endgroup$ Jul 30, 2020 at 14:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.