Let $X$ be a set and let $(E_n)$ be a sequence of subsets of $X$. The $\liminf E_n$ and $\limsup E_n$ is defined as follows: $$\liminf E_n = \bigcup^{\infty}_{m=1} \bigcap^{\infty}_{n=m} E_n$$ and $$\limsup E_n = \bigcap^{\infty}_{m=1} \bigcup^{\infty}_{n=m} E_n $$ From here, it is an easy exercise to show that $\liminf E_n$ is the set of all $x$ such that $x$ belongs to all but finitely many $E_n$ and to show that $\limsup E_n$ is the set of all $x$ such that $x$ belongs to infintely many $E_n$. These two facts imply $\liminf E_n \subseteq \limsup E_n$. However, I wanted to show this fact using the "union-intersection" defintion. For a minute, I thought maybe I needed to use an induction argument (and maybe I do?). This is what I have in terms of a proof.
$\textbf{Proof:}$ Put $A_m= \bigcup^{\infty}_{n=m} E_n.$ Hence, $\limsup E_n = \bigcap^{\infty}_{m=1} A_m$. Notice that $(A_m)$ is a decreasing sequence. Now, let $x \in \liminf E_n$. Then, $x \in \bigcap^{\infty}_{n=m_0} E_n$ for some $m_0$. This implies $x \in A_m$ for all $m \geq m_0.$ Because $(A_m)$ is decreasing, we also have that $x \in A_m$ for all $m \leq m_0$. Indeed, $x \in A_m$ for all $m$. Therefore, we conclude $x \in \limsup E_n$. $\blacksquare$
Best Answer
Your proof is alright. Just a remark for a part of it in case you have made a mistake.
You say: "Then, $x\in\bigcap_{n=m_0}^{\infty} E_n$ for some $m_0$. This implies $x\in A_m$ for all $m \geq m_0$''
Well, this is true, provided that you are aware of the intermediate steps i.e.,
"Then, $x\in\bigcap_{n=m_0}^{\infty} E_n$ for some $m_0$. This implies $\exists m_0 \forall n\geq m_0,\,x\in\mathcal{E}_n$. Since $\mathcal{E}_m\subset A_m = \bigcup_{n=m}^{\infty} E_n$, we have $x\in A_m$ for all $m \geq m_0$."