Equivalence of Definitions of lim inf of Sequence of Sets Prove : $\{w : w \in A_n \text{ for all $n$ except a finite number}\}= \bigcup_{n=1}^{\infty}\bigcap_{k=n}^{\infty}A_k$.
I am trying to prove these two definitions are equivalent but I am having trouble.

Proof so far: Let $x \in \{w : w \in A_n \text{ for all $n$ except a finite number}\}$. Then  $x$ is in all but a finite number of $A_n$. Now pick the largest $m$ such that $x$ is not in $A_m$. Then as $x \in A_j$ for all $j >m$. So $ x\in \bigcap_{j=m}^{\infty} A_j$. It follows that $x \in \bigcup_{n=1}^{\infty}\bigcap_{j=n}^{\infty}A_j$
 A: Your proof so far is perfectly fine.  All you need to do is prove the reverse inclusion and you'll be set.
A: The argument given in the question is correct and it shows that
$$
\{w : w \in A_n \text{ for all $n$ except a finite number}\} \subseteq \bigcup_{n=1}^\infty \bigcap_{k=n}^\infty A_k
$$
It remains to show that
$$
\{w : w \in A_n \text{ for all $n$ except a finite number}\} \supseteq \bigcup_{n=1}^\infty \bigcap_{k=n}^\infty A_k.
$$
Given that you've written what you've written so far, I suspect you can do that.
You wrote

Let $x \in \{w : w \in A_n \text{ for all $n$ except a finite number}\}$. Then  $x$ is in all but a finite number of $A_n$. Now pick the largest $m$ such that $x$ is not in $A_m$.

It seems to me that's a more complicated way of saying this than what is needed.  You could write

Suppose $w \in A_n \text{ for all $n$ except a finite number}$. Now pick the largest $m$ such that $w$ is not in $A_m$. [etc.]

A: Interesting. I always thought this was the definition of $\liminf$. The proof so far is right. I would prefer to add to the last sentence 'Since this holds for some $m \ge 1$' and then use m in the indices instead of n.

The reverse inclusion can be proven as follows:
Suppose $\omega \in \bigcup_{m\ge1}\bigcap_{n\ge m} A_n$.
Then $\omega \in \bigcap_{n\ge 1} A_n$ or $\omega \in \bigcap_{n\ge 2} A_n$ or $\omega \in \bigcap_{n\ge 3} A_n$...
Case 1: $\omega \in \bigcap_{n\ge 1} A_n$
Then $\omega \in A_n \forall n$ except for a finite number of $A_n$'s i.e. 0
Case 2: $\omega \in \bigcap_{n\ge 2} A_n$
Then $\omega \in A_n \forall n$ except for a finite number of $A_n$'s i.e. 1
$$\vdots$$
Case m: $\omega \in \bigcap_{n\ge m} A_n$
Then $\omega \in A_n \forall n$ except for a finite number of $A_n$'s i.e. $m-1$
$$\vdots$$
QED
