Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 1 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.