0

I am referring to the post: lim sup and lim inf of sequence of sets.

how would we prove that for $\limsup A_n $ $x$ is in infinitely many sets $A_i$?

If I define set $A$ as a set that consists of infinitely many sets $x \in $ $A_i$, what is its complement $A^C$ ?

Namely, I would like to prove that

$$ A^c \subseteq \bigcup_{N=1}^\infty \bigcap_{n\ge N} A_n^C $$

Community
  • 1
Zenga
  • 133

1 Answers1

0

In the answer to the post you mention it is explained very nicely that $$x\in\limsup A_n= \bigcap_{N=1}^\infty \bigcup_{n\ge N} A_n\iff x\text{ is contained in }A_n\text{ for infinitely many }n\in\mathbb N$$

So I don't really understand why you are asking this question here.

$x\in(\limsup A_n)^c$ then tells us that $x$ is contained in $A_n$ for a finite number of $n\in\mathbb N$ or equivalently that some $N$ exists with $n\ge N\Rightarrow x\in A_n^c$

That means $x\in\liminf A_n^c=\bigcup_{N=1}^\infty \bigcap_{n\ge N} A_n^c$ wich is also nicely explained in the answer.

Community
  • 1
drhab
  • 151,093
  • #drhab are you sure that $x\notin\limsup A_n$ tells us that $x$ is contained in a finite number of $A_i$ or equivalently that some $N$ exists with $n\ge N\Rightarrow x\in A_n^c$ ?

    Complement of infinitely many is not always finite, or am I wrong?

     – Zenga Oct 27 '14 at 12:27
  • ${n\in\mathbb N|x\in A_n}$ not infinite $\iff$ ${n\in\mathbb N|x\in A_n}$ finite $\iff\exists N\in\mathbb{N}\forall n\geq N\left[x\in A_n^{c}\right]$ – drhab Oct 27 '14 at 12:56
  • I have another stupid question. Can at the same time hold: $x \in A_n^c$ and $x \in A_n $ ? – Zenga Oct 27 '14 at 20:49
  • Are in the $A_n^C $ included all the other sets from the sequence? – Zenga Oct 27 '14 at 20:57
  • $x\in A_n^c$ can be interpreted as $x\notin A_n$ wich cannot hold at the same time as $x\in A_n$. I do not understand the question in your last comment. – drhab Oct 28 '14 at 07:16