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 $j &gt;m$. So $ x\in \bigcap_{j=m}^{\infty} A_j$. It follows that $x \in \bigcup_{n=1}^{\infty}\bigcap_{j=n}^{\infty}A_j$
Your proof so far is perfectly fine.  All you need to do is prove the reverse inclusion and you'll be set.
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