Please verify my solution for the following problem and let me know if there be anything wrong in my solution and also provide if there be any other solution in other way of the same problem.
Problem : Let $~A_n\subseteq\mathbb R~$ for $~n\ge1~$ and $~\mathcal X_n:\mathbb R\to\{0,1\}~$ be the function $$\mathcal X_n(x)=\begin{cases} 0 &\text{if}~~~~ x\not\in A_n \\ 1 &\text{if}~~~~ x\in A_n \end{cases}$$ Let $~g(x)=\lim_{n\to\infty}\sup\mathcal X_n(x)~$ and $~h(x)=\lim_{n\to\infty}\inf\mathcal X_n(x)~.$ Then which of the following option/s is/are true ?
$1)~~$If $~g(x)=h(x)=1~,$ then there exists $~m~$ such that for all $~n\ge m~$ we have $~x\in A_n~.$
$2)~~$If $~g(x)=1~$ and $~h(x)=0~,$ then there exists $~m~$ such that for all $~n\ge m~$ we have $~x\in A_n~.$
$3)~~$If $~g(x)=1~$ and $~h(x)=0~,$ then there exists a sequence $~n_1,~n_2,~\cdots~$ of distinct integers such that $~x\in A_{n_k}~$ for all $~k\ge1~.$
$4)~~$If $~g(x)=h(x)=0~,$ then there exists $~m~$ such that for all $~n\ge m~$ we have $~x\in A_n~.$
My Solution:
For option $\bf{(1)}$, If $~g(x)=h(x)=1~,$ then by the definition of the $~\lim\sup~$ and $~\lim\inf~,$ after some stage $~\mathcal X_n(x)~$ becomes $~1~$ i.e., there exists an $~m~$ such that for all $~n\ge m~$ we have $~x\in A_n~.$
Hence option $(1)$ is correct.
For option $\bf{(4)}$, If $~g(x)=h(x)=0~,$ then by the previous argument after some stage $~\mathcal X_n(x)~$ becomes $~0~$ i.e., there exists an $~m~$ such that for all $~n\ge m~$ we have $~x\not\in A_n~.$
Hence option $(4)$ is correct.
For option $\bf{(3)}$, If $~g(x)=1~$ and $~h(x)=0~,$ then there exists two sub-sequences $~\mathcal X_{n_k}(x)~$ and $~\mathcal X_{n_l}(x)~$ such that $~\mathcal X_{n_k}(x)~$ converges to $~1~$ and $~\mathcal X_{n_l}(x)~$ converges to $~0~.$
i.e., there exists a sequence $~n_1,~n_2,~\cdots~$ of distinct integers such that for $~k\ge m~,$ we have $~x\in A_{n}~.$
Hence option $(3)$ is correct.
And therefore only option $(2)$ is not true.
Your answers for 1), 3) and 4) are corect but you should also know why 2) is false. You need a counter-example to show that 2) is false. Take two disjoint sets, say $A=(0,1)$ and $B=(1,2)$. Let $A_n=A$ for $n$ even and $A_n=B$ for $n$ odd. Take $x \in A$. Then you can check that $g(x)=1$ and $h(x)=0$. But there is no $m$ such that $x \in A_n$ for all $n \geq m$.