Show If $A_{1} \subset A_{2} \subset \cdots,$ then $\varliminf A_{n}=\varlimsup A_{n}=\cup_{n=1}^{\infty} A_{n}$ 
Show If $A_{1} \subset A_{2} \subset \cdots,$ then $\varliminf
A_{n}=\varlimsup A_{n}=\cup_{n=1}^{\infty} A_{n}$

My reasoning:
By definition $\varliminf A_{n}=\cup_{n=1}^{\infty} \cap_{k=n}^{\infty} A_{k}=\left\{\omega: \omega\right.$ is in all but finitely many $\left.A_{n} s\right\}$,$\overline{\lim } A_{n}=\bigcap_{n=1}^{\infty} \cup_{k=n}^{\infty} A_{k}=\left\{\omega: \omega\right.$ is in an infinite number of $\left.A_{n} s\right\} $.
Since $\varliminf A_n\subset \varlimsup A_n$, we only need to prove the other direction. Proof of the other direction: let $x\in\varlimsup A_n, x\in \cup^\infty_{n=1} A_n$, $x\in A_1 \cup A_2 \cup A_3 ,... $ where $A_1=\cap^\infty_{k=1} A_k$, $A_2=\cap^\infty_{k=2} A_k,...$ hence we have approved the opposite direction. We also know that $\cap^\infty_{n=k}A_n=A_k$,$\varliminf A_n= \cup^\infty_{n=1}A_n$ then follows.
My problem:
I feel I didn't really use the condition that $A_1\subset A_2 \subset$ properly as follow this logic I am unable to proof the other direction: If $A_{1} \supset A_{2} \supset \cdots,$ then $\varliminf A_{n}=\varlimsup A_{n}=\cap_{n=1}^{\infty} A_{n}$. Could someone please point out what I have been missing? I am also struggling to understand the difference.
 A: Your proof is rather unclear. I don't really understand what you try to prove. Nevertheless, notice that $$\liminf_{k\to \infty }A_n=\bigcup_{k=1}^\infty \bigcap_{n\geq k}A_n=\bigcup_{k=1}^\infty A_k\underset{(*)}{=}\bigcap_{k=1}^\infty \bigcup_{n\geq k}A_n=\limsup_{n\to \infty }A_n,$$
where $(*)$ comes from the fact that $\forall m\in \mathbb N$, $$\bigcup_{k=m}^\infty A_k=\bigcup_{k=1}^\infty A_k,$$
because $\{A_k\}$ is increasing.
A: As noted already, you only need to show that
$$
\bigcup_{n = 1}^\infty A_n \subseteq \liminf_{n \to \infty} A_n = \bigcup_{k = 1}^\infty \bigcap_{n = k}^\infty A_n  \, .
$$
Now since $A_j \subseteq A_{j+1}$ for all $j$, it follows that
$$
\bigcap_{n = k}^\infty A_n = A_k
$$
for all $k$.
To see this, note that clearly $\bigcap_{n = k}^\infty A_n \subseteq A_k$ for all $k$ and also $A_k \subseteq A_n$ for all $n \ge k$.
Therefore now
$$
\bigcup_{k = 1}^\infty \bigcap_{n = k}^\infty A_n  = \bigcup_{k = 1}^\infty  A_k 
$$
which proves everything.
A: As a general guideline, for two arbitrary sets $A$ and $B$, if you want to show $A \subset B$, then you can argue in this way: for any $\omega \in A$, show that $\omega \in B$. Usually this would make your proof much clearer.
Now take any $\omega \in \varlimsup A_n = \cap_{n = 1}^\infty\cup_{k = n}^\infty A_k$, this means for every positive integer $n$, $\omega \in \cup_{k = n}^\infty
A_k$. Just take $n = 1$, we have $\omega \in \cup_{k = 1}^\infty A_k$, then this implies there exists some $k_1 \geq 1$, such that
$\omega \in A_{k_1}$. By condition, $A_{k_1} \subset A_{k_1 + 1} \subset A_{k_1 + 2} \subset \cdots$, hence $\omega \in \cap_{k = k_1}^\infty A_k \subset \cup_{n = 1}^\infty \cap_{k = n}^\infty A_k = \varliminf A_n$. This completes the proof.
