Trying to prove $\liminf (A_n) \subseteq \limsup (A_n)$

Question

I am trying to show that for a sequence of sets $(A_n)$

$$ \liminf (A_n) \subseteq \limsup (A_n)$$

I can see why this is true intuitively as follows -

$x \in \liminf (A_n)$

$\implies x$ is an element of all but finitely many of the $A_n$'s

$\implies x$ is an element of infinitely many of the $A_n$'s

$\implies x \in \limsup (A_n)$

---

However I want to show it formally using set notation. Here is what I have so far.

$x \in \liminf(A_n)$

$\implies x \in \bigcup_n \bigcap_{k \ge n} A_k$

$\implies x \in \bigcap_{k \ge n_0} A_k$ for some $n_0$

$\implies x \in \big(\bigcup_{1 \le j < n_0} A_j\big) \bigcup \big(\bigcap_{k \ge n_0} A_k \big)$ for some $n_0$

$\implies x \in \big(\bigcup_{1 \le j < n_0} A_j\big) \bigcup \big(\bigcup_{k \ge n_0} A_k \big)$ for some $n_0$

But I am not sure how to proceed from here. It seems like the $n_0$ is now 'redundant' as we have a statement saying that $x$ is an element least one $A_k$ whether that $A_k$ is 'before, equal to or after' $A_{n_0}$ in the sequence $(A_n)$. But this leads to

$\implies x \in \bigcup_n A_n$

which just says $x$ is an element of at least one of the infinitely many $A_n$, whereas I need to show that $x$ occurs infinitely often.

Where am I going wrong?

Answer 1

Let us start from the line $x\in \bigcap_{k\geq n_0} A_k$. This implies that $x\in A_k$ for all $k\geq n_0$ and hence $x\in \bigcup_{k\geq n} A_k$ for all $n\geq 1$. Therefore also $x\in\limsup A_n$.

Answer 2

First, here are the traditional definitions for liminf and limsup for sets: $$\liminf_{n \to \infty}A_n = \bigcup_{n=1}^{\infty} \bigcap_{m=n}^{\infty}A_m \text{ and } \limsup_{n\to\infty}A_n=\bigcap_{n=1}^{\infty}\bigcup_{m=n}^{\infty}A_m.$$

I don't generally find these definitions too enlightening. Here's a better, equivalent way to read $\liminf_{n \to \infty}A_n$: $\omega \in \liminf_{n \to \infty}A_n$ if there exists an $n\in\mathbb{N}$ such that for all $m>n$, $\omega \in A_m$. Similarly, for $\limsup_{n \to \infty}A_n$, we have $\omega \in \limsup_{n \to \infty}A_n$ if for all $n\in\mathbb{N}$ there exists $m>n$ with $\omega \in A_m$. The difference is subtle, but important. If you read carefully, you'll realize this means we can define things like this: \begin{align*} \liminf_{n \to \infty}A_n &= \{\omega \in \Omega : \#\{n \in \mathbb{N} : \omega \not\in A_n\}<\infty\} \\ \limsup_{n \to \infty}A_n &= \{\omega \in \Omega : \#\{n \in \mathbb{N} : \omega \in A_n\}=\infty\}. \end{align*}

Put it another way: the liminf is the event where eventually all of the $A_n$ happen, whereas limsup is the event in which infinitely many of the $A_n$ happen. Written this way, the inclusion $$\liminf_{n \to \infty}A_n\subset \limsup_{n \to \infty}A_n$$ should be clear.

In case it isn't, here's an example of something in $\limsup_{n \to \infty}A_n$ but not in $\liminf_{n \to \infty}A_n$: let $\omega \in A_{2n}, \omega \not\in A_{2n+1}$ for $n\in \mathbb{N}$, so $\omega$ is in $A_2, A_4, \ldots$ but not in $A_1, A_3, A_5, \ldots$. Then check using any of the definitions above that $\omega \in \limsup_{n \to \infty}A_n$ but not in $\liminf_{n \to \infty}A_n$.