Let $(A_n)$ a sequence of sets. How can I see geometrically $\limsup\limits_{n\to\infty} A_n$ ? I tried to make drawings, but I don't really see how it works.
I know that $x\in\limsup\limits_{n\to\infty }A_n$ if for all $n\in\mathbb N$ there is a $p\in\mathbb N$ such that $x\in A_k$ for all $k\geq p$, or in other words that $x\in A_k$ for an infinite number of $A_k$. But it doesn't say much. I would like a geometric vision if it's possible.
On
If $(A_n)$ is an increasing sequence of set (i.e. $A_{n}\subset A_{n+1}$), it's very easy to have an interpretation of $\bigcup_{i=1}^\infty A_n$. Same, if $(C_n)$ is decreasing (i.e. $C_{n+1}\subset C_n$), it's very easy to have an interpretation of $\bigcap_{i=1}^\infty C_n$. Both are very intuitif.
But if we are not in such ideal condition, we can try to get such condition in the way of the draw. The $B_i$ I have drawn are such that $(B_n)$ is decreasing, and thus $\bigcap_{n=1}^\infty B_n$ has now an intuitif sense, and we call it the $\limsup A_n$.
Now you have to imagine that our sequence of set is infinite and reproduce the same schema.
I hope it's more clear (sorry for the bad quality of my draws, put it's complicate to draw on the computer.)
I suppose the order on the sets is inclusion. Then $\displaystyle\sup_{k\ge n}A_k=\bigcup_{k\ge n}A_n$, consists in starting from the union of all $A_n$s, and removing one of these sets at a time: $$\bigcup_{k}A_k\supset\bigcup_{k\ge 1}A_k\supset\dotsm\supset \bigcup_{k\ge n}A_k\supset\dotsm.$$ Thus the limit of this non-increasing sequence of sets is their intersection: $$\limsup_n A_n=\bigcap_n\Bigl(\bigcup_{k\ge n}A_k\Bigr).$$