# Determining the limit of a sequence of sets?

(a) Let $$\Omega = \mathfrak{R}$$. Define $$A_n = \big{[}0, \frac{n}{n+1}\big{)}$$. Determine if $$\lim_{n \to \infty} A_n$$ exists. If yes, what is it?

(b) Show that $$\lim_{n \to \infty} \big{[}0, 1 + \frac{1}{n} \big{)} = [0, 1]$$

Definitions:

$$\inf_{k \geq n}A_k = \bigcap_{k = n}^{\infty} A_k$$ $$\sup_{k \geq n}A_k = \bigcup_{k = n}^{\infty} A_k$$ $$\lim_{n \to \infty}\inf A_n = \bigcup_{n = 1}^{\infty}\bigcap_{k = n}^{\infty} A_k$$ $$\lim_{n \to \infty}\sup A_n = \bigcap_{n = 1}^{\infty}\bigcup_{k = n}^{\infty} A_k$$

The sequence of sets $$\{A_n\}$$ is said to converge to its limit A if: $$\lim_{n \to \infty}\inf A_n = \lim_{n \to \infty}\sup A_n = A$$

Question: Just looking at the sets in (a) and (b) and 'plugging' in $$\infty$$ I can obtain an answer, but I do not think that is the correct way to approach this problem. Is there a simple way to evaluate the infimum and supremum of a set?

I understand the infimum is the greatest element that is less than or equal to all elements of a set $$S$$ whereas the supremum is the least element that is greater than or equal to all elements of $$S$$

## 1 Answer

Sure, these all have relatively simple interpretations:

• $$\sup A_k$$: Union. It's the set of all points that are in $$A_k$$ for at least one $$k$$.
• $$\inf A_k$$: Intersection. It's the set of all points that are $$A_k$$ for all $$k$$.
• $$\limsup A_k$$: This is the set of points $$x$$ such that $$x$$ is in infinitely many $$A_k$$.
• $$\liminf A_k$$: This is the set of points $$x$$ such that $$x$$ is in all but finitely many $$A_k$$.

In general,

$$\liminf A_k \subseteq \limsup A_k.$$

This is because if $$x \in A_k$$ for all but finitely many $$k$$, then it must be in $$A_k$$ for infinitely many $$k$$. However, $$x$$ could be in $$A_k$$ for all even $$k$$ (infinitely many $$k$$) but not in $$A_k$$ for all odd $$k$$. Then $$x \in \limsup A_k$$ but $$x \notin \liminf A_k$$.

$$A := \lim A_k \text{ if } \liminf A_k = A = \limsup A_k.$$

This is a strong statement. Suppose $$\lim A_k$$ exists. Then if $$x \in A_k$$ for infinitely many $$k$$, there exist only finitely many $$k'$$ such that $$x \notin A_{k'}$$. In other words, there exists a $$K$$ such that $$x \in A_k$$ for all $$k \geq K$$.