# limsup liminf of sequence of sets

Following up from the discussion here: Liminf and Limsup of a sequence of sets

I wanted to confirm my understanding of these concepts with another example. Suppose we have: $a_n>0$, $b_n >1$ and $$\lim_{n\rightarrow \infty}a_n =0,\ \lim_{n\rightarrow \infty}b_n =1$$ and $$A_n=\left \{ x:a_n \leq x <b_n \right \}$$ and we are trying to find liminf and limsup of $A_n$.

I view this as both $a_n$ and $b_n$ being decreasing sequences of real numbers. To help with enumeration, I have defined $a_n=\frac{1}{n}, b_n=1+\frac{1}{n}$.

Then $A_1= \left \{ x: 1 \leq x < 2 \right \} = [1,2),A_2= \left \{ x: 1/2 \leq x < 3/2 \right \} = [1/2,3/2)$ and so on.

To me, $A_n$ appears to be a sequence of half-open intervals of the form [a,b). I reason that this means the liminf and limsup (and hence the limit) are (0,1]. Is this line of thinking right?

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You cannot guarantee that $\langle a_n:n\in\Bbb N\rangle$ is a decreasing sequence: it might start out $$\left\langle 1,2,\frac1{10},\pi,\frac12,\frac13,\frac1{2^{100}},4,\frac1{100},\dots\right\rangle\;,$$ for instance. All you know is that all of its terms are positive, and for each $\epsilon>0$ there is some $n_a(\epsilon)\in\Bbb N$ such that $0<a_n<\epsilon$ whenever $n\ge n_a(\epsilon)$.
Similarly, all you can say for sure about $\langle b_n:n\in\Bbb N\rangle$ is that for every $\epsilon>0$ there is some $n_b(\epsilon)$ such that $1<b_n<1+\epsilon$ whenever $n\ge n_b(\epsilon)$. And above all you cannot assign specific values to the numbers $a_n$ and $b_n$: that’s changing the problem. (Of course, you can do so to look at an example or two in order to get a better idea of what’s going on, but that’s a different matter altogether.)
Now let’s take a look at $\liminf_n A_n$, where $A_n=[a_n,b_n)$: we want to determine which real numbers are eventually in the sets $A_n$, i.e., which are in all $A_n$’s from some point on. Here’s where those numbers $n_a(\epsilon)$ and $n_b(\epsilon)$ come in handy. For $\epsilon>0$ let $n(\epsilon)=\max\{n_a(\epsilon),n_b(\epsilon)\}$; then $0<a_n<\epsilon$ and $1<b_n<1+\epsilon$, and hence $$[\epsilon,1]\subseteq[a_n,b_n)\subseteq(0,1+\epsilon)\;.$$ for all $n\ge n(\epsilon)$. In other words, $[\epsilon,1]\subseteq A_n$ for all $n\ge n(\epsilon)$, and we conclude that for each $\epsilon>0$, $[\epsilon,1]\subseteq\liminf_n A_n$. It follows that $$\liminf_n A_n\supseteq\bigcup_{\epsilon>0}[\epsilon,1]=(0,1]\;.$$ On the other hand, if $x>1$, let $\epsilon=x-1$: for every $n\ge n(\epsilon)$ we have $1<b_n<1+\epsilon=x$, so $x\notin A_n$ whenever $n\ge n(\epsilon)$. This shows that $x$ isn’t even in infinitely many of the $A_n$’s, let alone in a tail of them, so $x\notin\limsup_n A_n$, and therefore certainly $x\notin\liminf_n A_n$. It’s also clear that no $x\le 0$ belongs to any of the $A_n$’s, so we’ve established that $\liminf_n A_n=(0,1]$, as you thought.
Along the way we’ve also seen that $\limsup_n A_n\subseteq(0,1]$, so $$\liminf_n A_n\subseteq\limsup_n A_n\subseteq(0,1]=\limsup_n A_n\;,$$ and it follows that $\limsup_n A_n=(0,1]$ as well, also as you thought.
It appears that you’re getting the concepts but might have a bit of difficulty actually writing down an argument to justify your reckoning of $\liminf$ or $\limsup$ of a sequence of sets.