Please confirm my lim sup and lim inf of sets answers

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 $\liminf A_n = \cup_{N=1}^\infty (\cap_{n \ge N} A_n)$.

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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 $x \in \mathbb{R}$, $\limsup_n E_n = \mathbb{R}$.

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I think there is a mistake in your part a lim inf as 1 is not in the set. Also there is no mention about n <= 0 or being real as n is n i just assumed the author must be referring to belong to Naturals. – Hardy Feb 11 '12 at 6:41
@Hardy the definitions of the sets changed while I was typing...! – Henno Brandsma Feb 11 '12 at 6:47
That's fine mate some one was kind enough to improve my type set so i do n't doubt u. Just as they stand now would my answers be correct then ? – Hardy Feb 11 '12 at 6:51
@Hardy no, you have type $\{0,2\}$ not $[0,2]$, e.g., and the last $\limsup$ is still all of $\mathbb{R}$ (I assume we are working with subsets of $\mathbb{R}$?) – Henno Brandsma Feb 11 '12 at 6:53
It is the formatting fellow i also assumed it was written correctly . but i get your confirmation that it should be an interval which is what i was expecting too. Thanks i also follow your other answer. – Hardy Feb 11 '12 at 6:58