Just wondering if my answers of the following 2 part question are correct. If not i would much appreciate a layman explanation, where i stuffed up.
Cheers
Find lim sup $A_n$ and lim inf $A_n$ of the following sequence of sets:
$$A_n = \begin{cases}
x &0 \le x < 1 &\text{if n is odd} \\
x &1 \le x \le 2 &\text{if n is even}
\end{cases}$$
Answer Attempt lim sup $A_n = \{0, 2\}$ and lim inf $A_n = \emptyset$
$$E_n = \begin{cases}
x &-n \le x \le 0 &\text{if n is odd} \\
x &\frac{1}{n} \le x \le n &\text{if n is even}
\end{cases}$$
Answer Attempt lim sup $A_n = \{-\infty, 0\} \cup \{1/2, \infty\}$ and lim inf $A_n = \emptyset$
I am skeptical if i am doing this right as both of my lim inf were null sets. One would imagine out of only 2 exercises on lim sup and lim inf of sets at the end of the chapter one would have been different.
Any help would be much appreciated.
Definitions from the book.
For a sequence of subsets $A_n$ of a set $X$, the $\limsup A_n= \cap_{N=1}^\infty ( \cup_{n\ge N} A_n )$ and $\liminf A_n = \cup_{N=1}^\infty (\cap_{n \ge N} A_n)$.
Best Answer
I like to think about the set $\liminf_n A_n$ as the set $\{ x \in X \mid x \in A_n \mbox { for all but finitely many } n \}$, and $\limsup_n A_n$ as the set $\{ x \in X \mid x \in A_n \mbox {for infinitely many } n \}$.
This follows from your definitions quite clearly: $x \in \limsup_n A_n$ iff $\forall N \exists n \ge N: x \in A_n$, which says that for every index $N$ we can find a larger index $n$ such that $x \in A_n$, which holds precisely when there infinitely many of such indices. On the other hand, $x \in \liminf_n A_n$ iff $\exists_N \forall n \ge N: x \in A_n$, which says that there is some index $N$ from which we know that $x \in A_n$ for all larger ones, so $x$ only possibly misses the $A_n$ with $ n < N$, so $x$ is in all but finitely many of the $A_n$. Note also that this makes the inclusion $\liminf_n A_n \subseteq \limsup_n A_n$ obvious.
Then in (a), the set alternate between $[0,1)$ and $[1,2]$, so indeed no point is in all but finitely many of them, as the odd- and even-indexed sets are disjoint, and all points in $[0,1) \cup [1,2] = [0,2]$ are in infinitely many of them. So I get $\liminf_n A_n = \emptyset$ and $\limsup_n A_n = [0,2]$.
In (b), we again have sets such that $E_{2n} \cap E_{2n+1} = \emptyset$ for all $n$, so that no point can be in all but finitely many of them, so $\liminf_n E_n = \emptyset$, so agreed. Now, if $x \le 0$, then for $n \ge |x|$, $n$ even, we have that $x \in E_{n}$, so those $x$ are in infinitely many $E_n$, and if $x > 0$ then for $n > \min(\frac{1}{x}, x)$, $n$ odd, we know that $x \in E_{n}$ (as then $\frac{1}{n} < x < n$), again showing that $x$ is in infinitely many $E_n$. As $x \le 0$ or $x > 0$ holds for all $x \in \mathbb{R}$, $\limsup_n E_n = \mathbb{R}$.