Limit inf < Limit superior(proof)

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In summary, the author proves that there is something in the limit superior and that anything in the limit inferior is also in the limit superior.
  • #1
EinsteinKreuz
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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.
 

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  • #2
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##
 
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  • #3
Math_QED said:
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...
 
  • #4
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.
 
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  • #5
Math_QED said:
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.
 
  • #6
EinsteinKreuz said:
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.
 

1. What is the definition of "Limit inf < Limit superior(proof)"?

The limit inferior is the smallest limit that a sequence can approach, while the limit superior is the largest limit that a sequence can approach. The statement "limit inf < limit superior (proof)" means that the limit inferior is less than the limit superior for a particular sequence.

2. Why is "Limit inf < Limit superior(proof)" important in mathematics?

This statement is important because it helps us understand the behavior of a sequence. If the limit inferior and limit superior are equal, then the sequence converges to that limit. However, if they are not equal, the sequence may not converge at all, or it may converge to a different limit.

3. How is "Limit inf < Limit superior(proof)" used in real-world applications?

In real-world applications, this statement is often used in the analysis of data and trends. For example, in economics, the limit inferior and limit superior can help predict the behavior of stock prices or interest rates. In physics, they can be used to understand the behavior of a system over time.

4. What is the difference between "Limit inf < Limit superior(proof)" and "Limit inf = Limit superior(proof)"?

The difference between these two statements is that "limit inf < limit superior (proof)" indicates that the limit inferior is strictly less than the limit superior, while "limit inf = limit superior (proof)" means that they are equal. This difference is important because it tells us whether a sequence is converging or not.

5. How can one prove "Limit inf < Limit superior(proof)" for a given sequence?

There are several methods for proving this statement, depending on the specific sequence. One approach is to use the definition of limit inferior and limit superior and show that they are not equal. Another method is to use the properties of limits and show that the limit inferior is less than the limit superior. In some cases, using a specific example or counterexample can also help prove this statement.

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