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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 that

a) $\lim\inf A_n\subset \lim\sup_n A_n$

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?

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  • $\begingroup$ Why did you state that $\{A_n\}_{n\in\mathbb{N}}$ is a family of subsets of a metric space? Your question makes sense (and the answer is the same) for families of subsets of any set. $\endgroup$ – José Carlos Santos Jun 29 '17 at 22:07
  • $\begingroup$ @JoséCarlosSantos Well I'm the notes that I'm following $X$ is a metric space and in the question they say that $A_n$ are subsets of $X$, then I think that it's subsets of a metric space. $\endgroup$ – Roland Jun 29 '17 at 22:10
  • $\begingroup$ I suppose that you agree that neither your proof nor the one posted by Daniel Xiang make any mention to metric spaces. $\endgroup$ – José Carlos Santos Jun 29 '17 at 22:12
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    $\begingroup$ For any sequence $(A_n)_n$ of sets, we have $x\in \lim \inf A_n$ iff $\{n : x\not \in A_n\}$ is finite, and we have $x\in \lim \sup A_n$ iff $\{n:x\in A_n\}$ is infinite. For example if $x\in A_{2n}$ for all $n$ and $x\not \in A_{2n+1} $ for all $n$ then $x\in \lim \sup A_n$ and $x\not \in \lim \inf A_n.$ $\endgroup$ – DanielWainfleet Jun 30 '17 at 6:04
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(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.

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