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.
Best Answer
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.