Lim inf and lim sup of Infinite intersection of closed sets [0,1],[0,1/2],[0,1/3],....
I think an infinite intersection of all this set will be singleton set i.e. {0}
So this set does not have lim inf and lim sup.
But how to prove it??
If i am wrong then what will be lim inf and lim sup?
We have the sequence of sets $A_n = [0,\frac{1}{n}]$ that clearly are decreasing:
$$\forall n: A_{n+1} \subseteq A_n$$
Now the $\liminf_n A_n$ is always defined, namely as
$$\liminf_n A_n = \bigcup_m \bigcap_{n \ge m} A_n$$
And in this case all intersections $\bigcap_{n \ge m} A_n$ are just equal to
$\bigcap_n A_n = \{0\}$ and the union of these is still $\{0\}$ so that $\liminf_n A_n = \{0\}$.
The $\limsup_n A_n$ is also always defined, i.e. as
$$\limsup_n A_n = \bigcap_m \bigcup_{n \ge m} A_n$$
And for any fixed $m$, the union of all $A_m, A_{m+1},\ldots$ is just the first one, by the decreasingness of the sets. So $\bigcup_{n \ge m} A_n = A_m$ and so
$$\limsup_n A_n = \bigcap_m A_m = \{0\}$$
and hence $\limsup_n A_n = \liminf A_n = \{0\}$ and the sequence of sets is convergent to the set $\{0\}$.