A question corcerning limsups and liminfs of sets

Question

Let $\{E_n \}$ be sets indexed by natural numbers. MY books define the following notions:

$$ \limsup E_n = \bigcap_{k=1}^{\infty} \bigcup_{n=k}^{\infty} E_n$$

$$ \liminf E_n = \bigcup_{k=1}^{\infty}\bigcap_{n=k}^{\infty} E_n$$

Why does it follow that

$$\limsup E_n = \{ x : x \in E_n \; \; \text{for infinitely many $n$ }\}$$

$$\liminf E_n = \{x : x \in E_n \; \; \text{for all but finitely many $n$}\} $$

???

I am having really hard time trying to understand this notions. I hope someone can explain to me why this is true. Thanks in advance.

Answer 1

If $x\in\limsup E_n$, then for each $k$, $x\in\bigcup_{n=k}^\infty E_n$. Therefore, for each $k$ there exists some $n>k$ such that $x\in E_n$. In particular this means that $x$ appears in infinitely many $E_n$'s.

On the other hand if $x$ appears in infinitely many $E_n$'s then for every $k$ there is some $n>k$ such that $x\in E_n$. So for each $k$, $x\in\bigcup_{n=k}^\infty E_n$, and therefore $x\in\bigcap_{k=1}^\infty\bigcup_{n=k}^\infty E_n$, as wanted.

The argument for $x\in\liminf E_n$ if and only if it is in all but finitely many is the same. If it is in $\liminf E_n$, then there is some $k$ such that $x\in\bigcap_{n=k}^\infty E_n$, and therefore $x\in E_n$ for all $n\geq k$, so it is in all but finitely many (finitely many could be $0$ too). I will leave you with the second inclusion as an exercise.

Answer 2

They are equivalent.

$$x\in\limsup E_n \iff \{ x : x \in E_n \; \; \text{for infinitely many $n$ } \} \\\iff \forall n\in \mathbb{N},\exists k>n, x\in E_k\iff x\in\bigcap_{k=1}^{\infty} \bigcup_{n=k}^{\infty} E_n$$

Similarly,

$$x\in \liminf E_n \iff \{x : x \in E_n \; \; \text{for all but finitely many $n$ } \} \\\iff \exists n\in \mathbb{N},\forall k>n, x\in E_k\iff x\in\bigcup_{k=1}^{\infty} \bigcap_{n=k}^{\infty} E_n$$

Answer 3

If x is in $\bigcap_{k=1}^{\infty }\bigcup_{n=k}^{\infty}E_n$ then it must be in each of the sets $\bigcup_{n=k}^{\infty}E_n$. If x is only in a finite number p of the $E_n$ then re-order the sets $E_1, ...E_p, E_{p+1},..$. Now x is not in the set $\bigcup_{n=p+1}^{\infty}E_n$, which is a contradiction. Try and come up with something similar for the other identity.