Why can't we define $\liminf$ as $\liminf_nA_n=\bigcap_{n\in\Bbb N}\bigcap_{k\ge n}A_k.$ I know that $\liminf$ is defined as 
$\liminf_nA_n=\bigcup_{n\in\Bbb N}\bigcap_{k\ge n}A_k.$
I understand that it means that it is a member of all but finitely many of $A_n$. 
My question is why don't we then define it as  
$\liminf_nA_n=\bigcap_{n\in\Bbb N}\bigcap_{k\ge n}A_k.$
Are the two definitions the same? I am unable to spot the difference. Please help.
 A: Suppose $A_0=\varnothing$. Then $\bigcap_{n\in\Bbb N}\bigcap_{k\ge n}A_k=\varnothing$, regardless of the values of the other $A_n$! That's bad. This extreme sensitivity to the first set in the sequence is the opposite of what we want from a limit operator.
A good limit definition should not depend at all on $A_0$. (For that matter, it should not depend on $A_1$, and it should not depend on $A_2$...)
A: Let $$ B_n = \bigcap_{k \ge n} A_k. $$
Then $B_1$ consists of every element that belongs to all of $A_1,A_2,\dots$. $B_2$ consists of every element that belongs to all of $A_1,A_2,\dots$ except possibly not to $A_1$. $B_3$ consists of everything that belongs to $A_1, A_2, \dots$ except possibly not to $A_1$ or $A_2$. Therefore
$$\bigcup_n B_n = \liminf A_n $$
consists of everything that belongs to all of $A_1,A_2,\dots$ except possibly not to finitely many of them. The key phrase is "all but finitely many."
On the other hand,
$$ \bigcap_n B_n = \bigcap_n A_n $$
is the set of everything that belongs to all of $A_1,A_2,\dots$.
For example, let $A_1 = \{1\}$ and $A_2 = A_3 = A_4 = \dots = \{1, 2\}$. Then $B_1 = \{1\}$ and $B_2 = B_3 = B_4 = \dots = \{1,2\}$. So
$$ \bigcap_n B_n = \bigcap_n A_n = \{1\} $$
because only $1$ belongs to all of $A_1,A_2,\dots$. But
$$ \bigcup_n B_n = \{1,2\} $$
because $1$ belongs to all of $A_1,A_2,\dots$ and $2$ belongs to all but finitely many of $A_1,A_2,\dots$ (all but $A_1$).
A: Take $A_n = \{-n,-(n-1),-(n-2),...\}$. Then the first definition gives the integers, but the second definition gives the natural numbers. So they are different.
