Let $\{A_n\}_{n\in\mathbb{N}}$ a family of subsets of a metric space
$X$. Define $$\lim\sup_n A_n=\cap_{n=1}^\infty
(\cup_{i=n}^\infty A_i)\quad;\quad \lim\inf_n A_n=
\cup_{n=1}^\infty(\cap_{i=n}^\infty A_i)$$ Show thata) $\lim\inf A_n\subset \lim\sup_n A_n$
b) If $A_n\subset A_{n+1}$ $\forall n\in\mathbb{N}$ then $$\lim\inf_n
A_n=\lim\sup_n A_n=\cup_{n=1}^\infty A_n$$
I followed some answers from here lim sup and lim inf of sequence of sets. and did
a) If $x$ is a member of $\cup_{n=1}^\infty(\cap_{i=n}^\infty A_i)$ then $x$ is a member of at least one of $\cap_{i=n}^\infty A_i$ what means that $x$ is member of all except a finite number of $A_i$, so $x$ is member of $\cup_{i=n}^\infty A_i$ and consequently $x$ is a member of $\cap_{n=1}^\infty
(\cup_{i=n}^\infty A_i)$
I'm not sure how to justify this, but it seems to me that something is missing.
b) Since $A_n\subset A_{n+1}$ then $\forall x\in A_n\Rightarrow x\in A_{n+1}$. Then $\cap_{i=n}^\infty A_n=A_n$. Thus
$$\lim\inf_n A_n=\cup_{n=1}^\infty(\cap_{i=n}^\infty A_i)=\cup_{n=1}^\infty A_n$$
I guess that I should proof that $$(1) \lim\inf_n A_n\subset \lim\sup A_n$$ and $$(2) \lim\sup_n A_n\subset \lim\inf_n A_n$$
But I'm stuck
In (1) if $x$ is member of $\cup_{n=1}^\infty(\cap_{i=n}^\infty A_i)$ then $x\in \cup_{n=1}^\infty A_n$
But how I can justify that $x$ is member of $\lim\sup_n A_n$ too?
Best Answer
(a) Let $x \in \bigcup_{n=1}^{\infty} \bigcap_{k=n}^{\infty} A_k$. Then for some $n$, we have that $x \in \bigcap_{k = n}^{\infty}A_k$. Thus we also have $x \in A_k$ for all $k \geq n$. This means that $x \in \bigcup_{m=i}^{\infty} A_m$ for all $i \in \mathbb{N}$. Therefore we have \begin{align*} x \in \bigcap_{i \in \mathbb{N}} \bigcup_{m=i}^{\infty} A_m \end{align*} This shows $\liminf A_n \subset \limsup A_n$.
(b) We already showed that $\liminf A_n \subset \limsup A_n$, and it is obvious that $\bigcap_{n=1}^{\infty} \bigcup_{k=n}^{\infty} A_k \subset \bigcup_{n=1}^{\infty}A_n$, since this set (on the right) is one of the sets we are taking an intersection over.
It remains to show that $\bigcup_{n=1}^{\infty} A_n \subset \liminf A_n$. To this end, let $x \in \bigcup_{n=1}^{\infty} A_n$, so $x \in A_n$ for some $n$. Since $A_n \subset A_{n+1}$, we must have $x \in A_k$ for all $k \geq n$. Thus, $x \in \bigcap_{k=n}^{\infty}A_k$ for this specific $n$. This means that $x \in \bigcup_{n=1}^{\infty} \bigcap_{k=n}^{\infty} A_k = \liminf A_n$.
We conclude that if $A_n \subset A_{n+1}$ for all $n \in \mathbb{N}$, we have \begin{align*} \liminf A_n \subset \limsup A_n \subset \bigcup_{n=1}^{\infty} A_n \subset \liminf A_n \end{align*} so all the sets are equal.