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I was working on a proof about measure theory, where I was asked to show that for any sequence of subsets $\left(A_{n}:n\in \mathbb{N}\right)$ of some set $X$, if we created another sequence of subsets $\left(B_{n}:n\in \mathbb{N}\right)$ by dropping finitely many entries in $\left(A_{n}:n\in \mathbb{N}\right)$, then, despite of that, $\liminf_{n \to \infty} B_{n} = \liminf_{n \to \infty} A_{n}$ and $\limsup_{n\to \infty} B_{n} = \limsup A_{n}$. I tried to prove by using the fact that the limit inferior and limit superior use the limit of an increasing and decreasing sequence, respectively. I show my procedure verbatim:

Let $\left(A_{n}:n\in \mathbb{N}\right)$ be a sequence of subsets of some set $X$. Also, let $\left(B_{n}:n\in \mathbb{N}\right)$ be a sequence of subsets generated by dropping finitely many entires in $\left(A_{n}:n\in \mathbb{N}\right)$. Recall that both $\left(\bigcap_{k\geq n} B_{n} : n\in \mathbb{N}\right)$ and $\left(\bigcap_{k\geq n} A_{n} : n\in \mathbb{N}\right)$ are increasing sequences and so for any $n_{0}\in \mathbb{N}$ there is some integer $n_{1} > n_{0}$ such that $\bigcap_{k\geq n_{0}} B_{k} \subset \bigcap_{k\geq n_{1}} A_{k}$. Note that the same can be said for $\bigcap_{k\geq n_{0}} A_{k} \subset \bigcap_{k\geq n_{1}} B_{k}$. Therefore, $\liminf_{n\to \infty} A_{n} = \liminf_{n\to \infty} B_{n}$.

Also, note that both $\left(\bigcup_{k\geq n} B_{n} : n\in \mathbb{N}\right)$ and $\left(\bigcup_{k\geq n} A_{n} : n\in \mathbb{N}\right)$ are decreasing sequences and so for any $n_{0} \in \mathbb{N}$ there is some integer $n_{1} > n_{0}$ such that $\bigcup_{k\geq n_{1}} A_{k} \subset \bigcup_{k\geq n_{0}} B_{k}$. The same applies for $\bigcup_{k\geq n_{1}} B_{k} \subset \bigcup_{k\geq n_{0}} A_{k}$. Therefore, $x\in \bigcap_{n\in \mathbb{N}}\bigcup_{k\geq n} B_{n}$ is a necessary and sufficent condition for $x\in \bigcap_{n\in \mathbb{N}}\bigcup_{k\geq n} A_{n}$. Hence, $\limsup_{n\to \infty} A_{n} = \limsup_{n\to \infty} B_{n}$.

END OF PROOF

The author made a much simpler proof by using the fact that

$\liminf_{n\to \infty} A_{n} = \left\{x\in X: x\in A_{n} \text{ for all but finitely many } n \in \mathbb{N}\right\}$.

$\limsup_{n\to \infty} A_{n} = \left\{x\in X: x\in A_{n} \text{ for infinitely many } n \in \mathbb{N}\right\}$

as follows:

If $x \in \liminf_{n\to \infty} B_{n}$, then $x \in B_{n}$, for all but finitely many $n \in \mathbb{N}$ and hence $x \in A_{n}$ for all but finitely many $n\in \mathbb{N}$ and then $\liminf _{n\to \infty} A_{n}$. This shows that $\liminf_{n\to \infty} B_{n} \subset \liminf _{n\to \infty} A_{n}$. By the same argument we show that $\liminf_{n\to \infty} A_{n} \subset \liminf _{n\to \infty} B_{n}$ therefore we have $\liminf_{n\to \infty} A_{n} = \liminf_{n\to \infty} B_{n}$. We show by the same argument as above that $\limsup_{n\to \infty} A_{n} = \limsup_{n\to \infty} B_{n}$.

END OF PROOF

So I was wondering, what are the properties of these definitions such that they offer an easier alternative to work with the limit inferior and superior of sequences of subsets.

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  • $\begingroup$ The relevant properties of these alternative (but equivalent) definitions are that they make it easier to reason about $\liminf$'s and $\limsup$'s. $\endgroup$ Feb 21 at 15:59
  • $\begingroup$ Oh I see. Could you give me some example to understand the way one could reason with them? $\endgroup$
    – Kr'aamkh
    Feb 21 at 16:02
  • $\begingroup$ What facts do you want to prove about them? It is a very general fact about mathematics that the more mental representations you have of a given mathematical object or operation the more likely you are to be able to prove what you want to prove. $\endgroup$ Feb 21 at 16:08
  • $\begingroup$ Indeed, I think that mahematics is just not about rigorousness (even some machines could make math proofs) but the human creativity and intuition is something invaluable in the process of understanding and creating mathematical objects. Therefore, I agree with you. I think I would like to prove that for some sequence of subsets if you generate another one by dropping finitely many entries, they both have the same inferior and superior limits. $\endgroup$
    – Kr'aamkh
    Feb 21 at 16:21

1 Answer 1

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I think I may understand a little better the reasoning behind the aformentioned definitions for limsup and liminf of a sequence of subsets. However, I encourage you, the reader,to correct me shall there be some mistake since I'm still a beginner.

Let $X$ be some nonemtpy set and $\left(A_{n}:n\in \mathbb{N}\right)$ be some sequence of subsets of $X$. First, let's consider the definition of liminf, namely,

$\liminf_{n\to \infty}A_{n} = \left\{x \in X:x \in A_{n} \text{ for all but finitely many } n \in \mathbb{N}\right\}$.

Note that the argument $x\in A_{n}$ for all but finitely many $n \in \mathbb{N}$ is the same as saying $x\in A_{n}$ for cofinitely many $n \in \mathbb{N}$. Basically, this means that there is some infinite set $B \subset \mathbb{N}$ such that its complement is finite and $x\in \bigcap_{n\in B} A_{n}$. For example, if $x\in \bigcap_{n \text{ is an odd positive integer}} A_{n}$, then this does not mean that $x\in \liminf_{n \to \infty} A_{n}$ (However it does not discard the posibility) since the set of the positive even integers is infinite.

On the other hand, considering the previous example, the fact that $x\in \bigcap_{n \text{ is an odd positive integer}} A_{n}$ implies that

$x\in \left\{x \in X:x \in A_{n} \text{ for infinitely many } n \in \mathbb{N}\right\}$.

since the set of the odd positive integers is inifinite. Therefore, $x\in \limsup_{n\to \infty} A_{n}$.

Now let's try to give a proof with this reasoning.

Show that for any sequence of subsets $\left(A_{n}:n\in \mathbb{N}\right)$ of some set $X$, if we created another sequence of subsets $\left(B_{n}:n\in \mathbb{N}\right)$ by dropping finitely many entries in $\left(A_{n}:n\in \mathbb{N}\right)$, then, despite of that, $\liminf_{n \to \infty} B_{n} = \liminf_{n\to \infty} A_{n}$ and $\limsup_{n\to \infty} B_{n} = \limsup_{n\to \infty} A_{n}$.

PROOF

Assume that there is some $x\in \liminf_{n \to \infty} B_{n}$. Then, $x\in B_{n}$ for all but finitely many $n\in \mathbb{N}$. Since $\left(B_{n}:n\in \mathbb{N}\right) \subset \left(A_{n}:n\in \mathbb{N}\right)$, it follows that $x\in A_{n}$ for all but finitely many $n \in \mathbb{N}$ and so $\liminf_{n \to \infty} B_{n} \subset \liminf_{n \to \infty} A_{n}$. On the other hand, if $x\in \liminf_{n\to \infty} A_{n}$, then $x\in A_{n}$ for all but finitely many $n\in \mathbb{N}$. Hence $x\in B_{n}$ for all but finitely many $n\in \mathbb{N}$ since we just dropped finitely many entries from $\left(A_{n}:n\in \mathbb{N}\right)$ to create $\left(B_{n}:n\in \mathbb{N}\right)$. Therefore, $\liminf_{n\to \infty} A_{n} \subset \liminf_{n\to \infty} B_{n}$ and so $\liminf_{n\to \infty} A_{n} = \liminf_{n\to \infty} B_{n}$.

Now, assume that some arbitrary $x\in \limsup_{n\to \infty} B_{n}$, then $x\in B_{n}$ for infinitely many $n\in \mathbb{N}$. Since $\left(B_{n}:n\in \mathbb{N}\right) \subset \left(A_{n}:n\in \mathbb{N}\right)$, it follows that $x\in A_{n}$ for infinitely many $n\in \mathbb{N}$ and so $\limsup_{n\to \infty} B_{n} \subset \limsup_{n\to \infty} A_{n}$. Now assume that $x \in \limsup_{n\to \infty} A_{n}$, then $x\in B_{n}$ for infinitely many $n\in \mathbb{N}$ since $\left(B_{n}:n\in \mathbb{N}\right)$ was created by dropping finitely many entries in $\left(A_{n}:n\in \mathbb{N}\right)$. Hence $\limsup_{n\to \infty} B_{n} = \limsup_{n\to \infty} A_{n}$. Note that we used the same arguments.

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