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(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 $S$

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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 $k \geq K$.

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