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I am in self-studying mode with the book of Resnick ("A probability Path 1"), and trying to solve a basic problem, namely, 1.9.6, so my apologizes for the basic question.

The problem states:

Assuming $a_n > 0$ and $b_n > 1$, $\lim_{n \rightarrow \infty} {a_n} = 0$, and $\lim_{n \rightarrow \infty} {b_n} = 1$.

Define, $A_n = \{x: a_n \leq x < b_n\}$.

Find, $\limsup_{n \rightarrow \infty} A_n$ and $\liminf_{n \rightarrow \infty} A_n$.

Using the definitions of both $\limsup_{n \rightarrow \infty} A_n$ and $\liminf_{n \rightarrow \infty} A_n$ I arrived to:

$\limsup_{n \rightarrow \infty} A_n = (0, 1)$

For $\liminf_{n \rightarrow \infty} A_n$, I understand that $\cap_{k=n}^{\infty} A_n = [a_n, b_n) \cap [a_{n+1}, b_{n+1}) \cap ... = [a_n, b_{n+1})$, since $\forall k \geq n, a_k > a_{k+1}$, and $b_k > b_{k+1}$. Thus, I would be arriving to $\liminf_{n \rightarrow \infty} A_n = (0, 1)$.

Am I doing it right? Thanks for possible tips/helps to correctly address the problem.

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    $\begingroup$ There is no hypothesis on the monotonicity of the sequence (it is not stated that $a_n$ is decreasing, nor that $b_n$ is increasing): hence you cannot deduce that $\cap_{k=n}^{\infty} [a_k,b_k) = [a_n,b_n)$. Moreover, we cannot deduce uniquely the lminf and limsup. There are two possibilities: $(0,1)$ or $(0,1]$ and both could be the answer. $\endgroup$
    – Crostul
    Jul 5, 2018 at 6:03
  • $\begingroup$ I am sorry, I made a typo, $b_n > 1$...maybe that clarifies that $b_n$ should decrease with $n \rightarrow \infty$ $\endgroup$
    – mgbacher
    Jul 5, 2018 at 6:26

2 Answers 2

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without $b_n > 1$ assumption

The best way of seeing $\limsup$ and $\liminf$ of a sequence of subsets $A_n$ is this : write down the definition of $\limsup$ as $\limsup A_n = \cap_n \cup_{k \geq n} A_k$.

More precisely, $x \in \limsup A_n \iff \forall n , x \in \cup_{k \geq n} A_k \iff \forall n , \exists k \geq n , x \in A_k \iff$ $x$ belongs in infinitely many $A_k$.

Similarly, $x \in \liminf A_n \iff x \in \cup_n \cap_{k \geq n} A_k \iff \exists n , \forall k \geq n, x \in A_k \iff$ $x$ is in $A_k$ for sufficiently large enough $k$.


Let $x \in (0,1)$. Let $\epsilon = \min \{x,1-x\}$. By definition, there exists $N \in \mathbb N$ such that $n > N \implies |a_n| < \epsilon$ and $|b_n - 1| < \epsilon$. Now, $x \geq \epsilon > |a_n|$ so $x > a_n$, while $x \geq 1 - \epsilon > 1 - |1-b_n|$, so $b_n > x$. Consequently, $x \in (a_n,b_n) \subset [a_n,b_n)$ for all $n \geq N$, or sufficiently large $N$.

Hence, $x \in \liminf A_n$, and therefore $x \in \limsup A_n$.


For $x \notin [0,1]$ (the closure of $(0,1)$), the same argument will show that $x \notin [a_n,b_n)$ for all sufficiently large $n$. So these points are not in $\limsup A_n$ since they do not belong to infinitely many $A_n$. Consequently, they do not belong to the limit inferior either.


For $x = 0$, note that $a_n > 0$ for all $n$, so $x$ does not belong to $A_n$ for any $n$. Consequently, $0$ is not in the limit superior or inferior.


For $x = 1$, we have options :

If $b_n = 1+\frac 1n$, for example, then $1 \in [a_n,b_n)$ for sufficiently large $n$ (those for which $a_n < \frac 12$, for example), so then $1$ lies in both the limit superior as well as inferior of $A_n$.

However, it is also possible that $b_n = 1 - \frac 1{2n}$, then $1 \notin [a_n,b_n)$ for sufficiently large $n$, then $1$ does not lie in $\liminf$ or $\limsup a_n$.

Finally, if $b_n = 1 - \frac 1{2n}$ for $n$ odd and $b_n = 1 + \frac 1n$ for $n$ even, then $1$ belongs in the limit superior (occurs in infinitely many $A_n$) but not in the limit inferior (does not occur always after sufficiently large A_n).

So, three possibilities are there : $\limsup A_n = \liminf A_n = (0,1)$ or $(0,1]$, or $\limsup A_n = (0,1]$ and $\liminf A_n = (0,1)$.


I leave you to figure out the "with assumption part" by yourself : one or more of the above cases will vanish.

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  • $\begingroup$ Thanks for your great answer!!! I made a typo before, $b_n > 1$, so I would be able to figure it out now...thanks!! $\endgroup$
    – mgbacher
    Jul 5, 2018 at 6:44
  • $\begingroup$ You are welcome! $\endgroup$ Jul 5, 2018 at 7:21
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Actually a formal proof should be tedious and we cannot assume $a_n$'s and $b_n$'s are monotone, no such assumptions mentioned before. But based on the theory of limits, there exists sub sequences of $a_n$'s and $b_n$'s that are monotonically convergent. By the virtues of the textbook, I propose the following proof scheme $$\underset{n\rightarrow\infty}{limsupA_n}=\cap_{n=1}^{\infty}\cup_{k=n}^{\infty}[a_k,b_k)=\cap_{n=1}^{\infty}(0,sup_{k\geq n}b_k)=(0,1]$$ since $limsup_{n\rightarrow\infty}b_n=inf_n (sup_{n\geq k} b_k)=1$ and an intersting property at Problem 1.11 is used.

Similarly $$\underset{n\rightarrow\infty}{liminfA_n}=\cup_{n=1}^{\infty}\cap_{k=n}^{\infty}[a_k,b_k)=\cup_{n=1}^{\infty}[sup_{k\geq n}a_k,1]=(0,1].$$

Remark: As previously stated, the conditions $a_n>0$ and $b_n>1$ matter in the process.

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