Set sequences $\liminf$ and $\limsup$ – correct

elementary-set-theorylimsup-and-liminfsequences-and-series

I'm studying set theory for probability and statistics, and it's important, in order to work with a $\sigma$-algebra, to discuss the concept of $\liminf$ and $\limsup$ for sequences of sets.

But in doing so I'm not sure I got it well. For example:

When $A_n = \{(-1)^n\}$ we know: $$\liminf A_n = \bigcup_{n=1}^{\infty}\bigcap_{k=n}^{\infty}A_n = \emptyset $$ because no elements appear in all sets $A_n$ except for a finite number of them. However: $$\limsup A_n = \bigcap_{n=1}^{\infty} \bigcup_{k=n}^{\infty}A_n = \{-1, 1\}$$because elements $-1,1$ will end up appearing in all sets (and in the union of them) so the sequence does not converge.

Also, for a sequence such as:

$$ A_n = \left\{(x,y) \in \mathbb{R}^2 : \left(x-\frac{(-1)^n}{n}\right)^2 + y^2 \leq 1 \right\} $$

one can evaluate:

$$ \limsup A_n = \left\{(x,y) \in \mathbb{R}^2 : x^2 + y^2 \leq 1 \right\}= \liminf A_n$$ because, for the $\limsup$, the intersection of all unions starting with index $k=n$ will end up resulting in the circle centered in $(0,0)$. In a similar way, for the $\liminf$, the union of all intersections will end up coming as close as the circle centered in $(0,0)$, because $A_{n+1}\cap A_n \subset A_{n+2}\cap A_{n+1}$.

Is my track of thought logical? I am a bit afraid I didn't get the concept well…

Thank you!

Best Answer

To help your understanding, you can also consider the characterization of $\liminf A_n$ and $\limsup A_n$ using sequences in $A_n$

$\liminf A_n$ contains all the points $a$ such that exists a sequence $(a_n)$ converging towards $a$ with $a_n \in A_n.$

$\limsup A_n$ contains all the points $a$ such that exists a subsequence $(a_n')$ converging towards $a$ with $a_n' \in A_n'$

With your fist example, i.e., $A_n = \{(-1)^n\}$, clearly $\liminf A_n = \emptyset$ and $\limsup A_n = \{-1,1\}$

With your second example, $A_n = \left\{(x,y) \in \mathbb{R}^2 : \left(x-\frac{(-1)^n}{n}\right)^2 + y^2 \leq 1 \right\}$. Again, clearly $\liminf A_n = \left\{(x,y) \in \mathbb{R}^2 : x^2 + y^2 \leq 1 \right\} =\limsup A_n.$