# Defining limits for subsets of $\mathbb{R}$

Suppose that we have a sequence $\{a_n\}$ with $a_n\subseteq\mathbb{R}$ for all $n$. What I wonder is that is it possible to define a 'limit' for this kind of sequences?

Of course this 'limit' should make some sense. For example intuitively if $a_n=(\frac{-1}{n},\frac{1}{n})$, then $\lim a_n$ should be $\{0\}$, or if $a_n=(-n,n)$ then $\lim a_n$ should be $\mathbb{R}$, etc. I know that some kind of a limit process of curves is used to define functions that are continuous everywhere but differentiable nowhere or space filling curves. But I couldn't find any formal information about this limit.

My goal is to generalise this limit process to $\mathbb{R}^2$ to define limits for shapes. For example, a well-known proof of that $\pi$ is the ratio of the area of any circle to the square of its radius, includes drawing $n$-polygons inside circle and to limit the areas of these polygons. But maybe it is also possible to define a limit for these compact subsets such that the limit of $n$-polygons inside the circle, as $n$ is increasing, is the circle.

So on, I have no idea on how to define this limit. Any help is appreciated.

• like a filter? ${}{}$ – Jorge Fernández Hidalgo Apr 22 '16 at 19:18
• Are you familiar with the notions of liminf and limsup? These extend in a well-defined way to sequences of sets. – Bungo Apr 22 '16 at 19:18
• Here's a possible definition. – Git Gud Apr 22 '16 at 19:19
• Maybe Kuratowski convergence of sets or any of the many types of set convergence you can find in Set-Valued Analysis by Jean-Pierre Aubin and Hélène Frankowska. – Dave L. Renfro Apr 22 '16 at 19:30
• Say $a_n$ is a sequence of real numbers with $\lim a_n=a$. If we define a new sequence with $b_n=\{a_n\}$, with the definition you suggested it looks like $lim b_n=\varnothing$ (or maybe the limit does not exist?). But I think that this limit definition of subsets should also satisfy the limit properties of real numbers, so $\lim b_n$ should be $\{ a\}$. Do you think that it is possible? – Levent Apr 22 '16 at 19:30