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?

  • $\begingroup$ 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. $\endgroup$ 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. $\endgroup$ Jun 29 '17 at 22:12
  • 1
    $\begingroup$ 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 $x\not \in \lim \inf A_n.$ $\endgroup$ Jun 30 '17 at 6:04

(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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.