Definition of lim inf of a sequence of sets

Question

I have two questions concerning

1) Do I understand correctly that the union part does not change the subset that we get from the intersection part? I.e., the union says: give me those elements that belong to at least one set An. But since we have an intersection before, the union will just take all the elements obtained by the intersection since they'll all belong to all sets of An starting from a value m. Am I right?

2) Would it then be possible to define lim inf as follows: for m in N?

Thank you in advance!

Answer 1

As Henry pointed it out in the comments, the fact that $\cap_{n \geq m} A_n$ is dependent of $m$ and $\liminf\limits_{n \rightarrow \infty} A_n$ is a quite important problem.

An example : take $A_n=[0,n]$. $A_n$ is a striclty increasing sequence of sets which limit is $\mathbb{R}^+$, we have : $$\liminf\limits_{n \rightarrow \infty} A_n=\mathbb{R}^+.$$

But $\cap_{n \geq m} A_n= A_m=[0,m]$ -since the sequence is increasing strictly-, which is dependent of $m$, and will not be equal to $\mathbb{R}^+$ for any $m \in \mathbb{N}$. Then if we take the union of those sets, $\cup_{m\in \mathbb{N}}[0, m]=\mathbb{R}^+$ and everything is fine.

In conclusion, you will not be able in this case to find a $m$ such that $$\liminf\limits_{n \rightarrow \infty} A_n = \cap_{n \geq m} A_n.$$ In other words : $\forall m \in \mathbb{N},\liminf\limits_{n \rightarrow \infty} A_n \neq \cap_{n \geq m} A_n$

Answer 2

To help with the existing answer and comments, sometimes simply writing out what the notation says is the easiest way to understand things.

$$ \bigcup_{m=1}^{\infty} \left(\;\bigcap_{n=m}^{\infty} A_n \right) \;\; = \;\; \left(\bigcap_{n=1}^{\infty} A_n\right) \; \cup \; \left(\bigcap_{n=2}^{\infty} A_n\right) \; \cup \;\left(\bigcap_{n=3}^{\infty} A_n\right) \; \cup \;\left(\bigcap_{n=4}^{\infty} A_n\right) \; \cup \; \cdots$$

Thus, $\;x \in \liminf\limits_{n \rightarrow \infty} A_n \;$ if and only if

$(x$ belongs to each $A_n)$

or

$(x$ belongs to each $A_n$ except perhaps $A_{1})$

or

($x$ belongs to each $A_n$ except perhaps $A_{1}, A_{2})$

or

($x$ belongs to each $A_n$ except perhaps $A_{1}, A_{2}, A_{3})$

or $\; \ldots$

Therefore, $\;x \in \liminf\limits_{n \rightarrow \infty} A_n \;$ if and only if

$x$ belongs to at least one of these parenthesed descriptions, OR if and only if $x$ belongs to each $A_n$ except perhaps some finitely many of the $A_n$'s.