I Limit inf < Limit superior(proof)

[EinsteinKreuz]

Attached is a proof that $$ lim \inf \subset lim \sup $$ for an infinite sequence of non-empty sets. The basic idea is to use the axiom of choice/well ordering theorem to show that 1) there is something in the limit superior 2) there is something in the limit superior that is not in the limit inferior and 3) anything in the limit inferior is also in the limit superior. How does this one look? Is it correct? Feedback appreciated.

 

[member 587159]

I fail to see what's tricky:

$x \in \liminf A_n \iff \exists n_0: \forall n \geq n_0: x \in A_n$
$\implies \forall n \geq 0: \exists m \geq n: x \in A_n \iff x \in \limsup A_n$

 

[EinsteinKreuz]

I fail to see what's tricky:

$x \in \liminf A_n \iff \exists n_0: \forall n \geq n_0: x \in A_n$
$\implies \forall n \geq 0: \exists m \geq n: x \in A_n \iff x \in \limsup A_n$

That said, do you see any errors in the proof or no...

 

[member 587159]

Sorry, I did not want to read a whole-page proof of something that takes one line...

Moreover, if I understand correctly what you wrote you believe that $\liminf A_n \subsetneq \limsup A_n$. This is not true, even if all $A_n$ are distinct. For example
$\liminf [-n,n] = \limsup [-n,n] = \Bbb{R}$.

I completely fail to see where the axiom of choice comes in here.

 

[EinsteinKreuz]

I fail to see what's tricky:

$x \in \liminf A_n \iff \exists n_0: \forall n \geq n_0: x \in A_n$
$\implies \forall n \geq 0: \exists m \geq n: x \in A_n \iff x \in \limsup A_n$

That's not quite right. If x is in the limit inferior then x is in at least one infinite intersection for some n.
So what you have show is that for some n, if x lies in the infinite intersection of $$A_k| k\geq n$$ then $$x\in \bigcup_{k=n}^{\infty}(A_k)$$ Now the limit superior consists of elements that are common to all infinite unions. That is, $$x \in lim \sup \leftrightarrow x\in \bigcap_{n=1}^{\infty}(\bigcup_{k=n}^{\infty}A_k) $$ So you must show that every infinite union contains x for x to belong to the limit superior.

Here's how I did it:

$$ x\in lim\ sup(A_n)\leftrightarrow (\forall n)(\exists r\geq n )| x \in A_r $$

And so, $$ (\forall n)(!\exists r\geq n )| x \in A_r \leftrightarrow x \notin lim \ sup(A_n)$$

Also, $$
(\forall n)(!\exists r\geq n )| x \in A_r \leftrightarrow (\forall n)(\forall r\geq n )\ x\notin A_r
\Rightarrow (\forall n) x \notin \bigcap_{k=n}^{\infty}(A_k)\leftrightarrow x \notin lim\inf(A_n) $$

Thus, $$x \notin lim \sup(A_n) \Rightarrow x \notin lim \inf(A_n)$$ And it follows that
$$x \in lim\inf(A_n) \Rightarrow x \in lim\sup(A_n) \ (modus \ tollens)$$

Therefore,

$$ lim\inf(A_n) \subseteq lim\sup(A_n) $$

I initially attempted to prove what was written in the book(A course in Real Analysis ~ McDonald) which stated that the limit inferior is a proper subset of the limit superior, but I had some doubts about that doing the proof and your counterexample shows the falsity of that statement. But you're right that the axiom of choice is unnecessary to show that lim inf =< lim sup.

 

[member 587159]

That's not quite right. If x is in the limit inferior then x is in at least one infinite intersection for some n.

Yes, that's what I said? $x\in \liminf A_n$ implies there is $n_0$ such that $x \in \bigcap_{n\geq n_0} A_n$ so $x \in A_n$ for all $n\geq n_0$.

Unless I miss something, you are seriously overcomplicating the issue.