Liminf and Limsup of a sequence of sets

As Didier pointed out, the sequence is not increasing. For example, $\frac{1}{2}$ is in $A_2$ while it is not in $A_3$. There are much more examples, as we shall see. But let's first have a closer look at the definition of limit inferior and limit superior in the case of sets.

Suppose that $\{A_n\}_{n\in\mathbb{N}}$ is a sequence of subsets of some set $X$. Then define $$ \liminf_{n\to\infty}A_n:=\bigcup_{n\in\mathbb{N}}\bigcap_{m\geq n} A_n $$ and $$ \limsup_{n\to\infty}A_n:=\bigcap_{n\in\mathbb{N}}\bigcup_{m\geq n} A_n $$ Suppose that $x\in\liminf_n A_n$. This means that there exists an $M\in\mathbb{N}$ such that $x\in A_m$ for all $m\geq M$. Or, stated otherwise, $x$ is an element of all but finitely many of the sets $A_n$, as you have correctly noted.

Now suppose that $x\in\limsup_nA_n$. This means that for all $n\in\mathbb{N}$, there exists an $m\in\mathbb{N}$ with the property that $x\in A_m$. Or, stated otherwise, $x$ is an element of infinitely many of the sets $A_n$. Now back to the problem.

To see what that the limit inferior of $\{A_n\}$ is $\mathbb{Z}_{\geq 0}$, suppose that $\frac{p}{q}$ is a positive non-integer rational number (i.e. $q> 1$ and $p>0$), and that $n$ is relatively prime to $q$. Then it follows easily that $\frac{p}{q}$ is not in $A_n$. Since there are infinitely many numbers relatively prime to $q$, it follows that $\frac{p}{q}$ is not in $\liminf_n A_n$. The non-negative integers, however, in $\liminf_n A_n$. To see this, note that if $n$ is an integer and $m\geq n$, then $n=\frac{mn}{m}\in A_m$. Thus, we get $$ \liminf_{n\to\infty}A_n=\mathbb{Z}_{\geq 0}. $$

The limit superior of $A_n$ can be found in a similar way. Note that $\frac{p}{q}\in A_{nq}$ for any $n\in\mathbb{N}$ when $\frac{p}{q}$ is a non-negative rational. Thus, $$ \limsup_{n\to\infty}A_n=\mathbb{Q}_{\geq 0}. $$

It would be a good idea to write out a few of the sets $$A_n=\left\{\frac0n, \frac1n, \dots , \frac{n^2}n\right\}$$ explicitly. For starters, that would show that it’s not in general true that $A_n\subseteq A_{n+1}$.

$$\begin{align*} A_1&=\left\{\frac01,\frac11\right\}=\{0,1\}\\ A_2&=\left\{\frac02,\frac12,\frac22,\frac32,\frac42\right\}=\left\{0,\frac12,1,\frac32,2\right\}\\ A_3&=\left\{\frac03,\frac13,\frac23,\frac33,\frac43,\frac53,\frac63,\frac73,\frac83,\frac93\right\}=\left\{0,\frac13,\frac23,1,\frac43,\frac53,2,\frac73,\frac83,3\right\}\;, \end{align*}$$

and so on. It’s true that $A_1\subseteq A_2$, but clearly $A_2\nsubseteq A_3$.

It is easy to check that if $k\in\Bbb N$, then $k\in A_n$ for all $n\ge k$, so $\Bbb N\subseteq\liminf_n A_n$, and of course it’s very clear that $\limsup_n A_n\subseteq\Bbb Q\cap[0,\to)$, the set of non-negative rationals.

Suppose that $a,b\in\Bbb Z^+$ and $\gcd(a,b)=1$, so that $a/b$ is a positive rational in lowest terms. Suppose further that $b>1$, so that $a/b$ is not an integer. For what $n\in\Bbb Z^+$ is it true that $a/b\in A_n$? Clearly $n$ must be a multiple of $b$. Say $n=kb$; then $$\frac{a}b=\frac{ka}{kb}=\frac{ka}n\in A_n$$ if and only if $0\le ka\le n^2$. From here you should be able to describe fairly concisely exactly which $A_n$ contain $a/b$, and that will show you exactly what $\liminf_n A_n$ and $\limsup_n A_n$ must be.

Here are the actual answers, spoiler-protected; mouse-over to see them.

$\liminf_n A_n=\Bbb N$; $\limsup_n A_n=\Bbb Q\cap[0,\to)$.